@@JB-ym4up 4n/x for any integer x != 0 arguably produces solutions, and for x = 1 unarguably produces solutions. Any operations on an integer m that leave it as an integer won't effect that, so 4n/(4m-1) is just as valid as 4n/(4m+1) or even 4n/(3m^2-26)
Thanks. I actually have been using my phone to record in 1080P, 60FPS. My old videos (when I was recording in classrooms) were recorded on my MacBook Air, 720p!
The main problem with polar coordinate (which pops up often in your videos) is that you can assign multple values to get the same results. The form x = (4n) / (4m+1) is equivalent to taking the (4m+1)th root of i^(4n), for which 1 is a possible solution (but you know there are 4m+1 answers in total, so another 4m possible answers, and those will not be 1, rather complex numbers). A more strict definition of polar coordinates restricts the argument to be in the [-PI, PI) range (or some other 2-PI length interval, e.g [0, 2PI) which is perhaps more practical) and avoids somewhat whacky results like this.
but also in a 2pi range multiple solutions exist e.g. for x^5=1. so doesn't the ambiguity remain? i.e. 5th root of 1 can be mapped to different values in polar coordinates even in a 2*pi range..
even if you define such a restriction so only one answer pops up. in my eyes, the info that other answers exist, should be stated. Not for compitionisms sake, but because, such an interval is not the same for every problem, or even have a formula for every different problem. So the only reason to do it is for visualization in a class or smthng. Such debates should exist if this is not your Calc 1,2 etc. For me its just that the question is poorly stated. There should be an interval in the question or further explanation. Its not something new. Poorly stated questions are asked all the time in math. And create fun or not so, problems.
I think i^x just raises more questions when you forget that exponents have different branches in their function. Also, by looking at branches I think you can get infinitely many answers from i^(4/pi)...
Maybe the ambiguity comes from the fact that taking a root of one can yield multiple values. Take the square root of one: technically, there are 2 real answers to this: 1 and -1. Negative one times negative one also equals one. By just saying that 1^(1/pi) = 1, you may be missing other complex solutions to the pi'th root of 1. If you put 1^(1/3) into wolfram alpha, you get 3 complex solutions (and maybe more if you add 2pi, idk).
tbh you don't need that much to enjoy most math. You can learn up to calc 3 in a few months if you have good resources and aren't bad at learning new concepts in general. What's hard is being a good problem solver because that needs a lot of practice to be able to recognize a lot of patterns in all sorts of different topics.
an equation is well defined if it is required that each of its solutions make the equality true: an equation can have multiple solutions but not solutions that solve the equality or not in an arbitrary way. So in the second equation the only acceptable solution is x = 4n. The polar representation of complex numbers, in equations, must be used when it does not give rise to ambiguity, that is, when it effectively allows to identify all the solutions, not when it produces expressions with multiple results that give rise to "partial" equalities.
The main thought-error happend at 2:06. exp(2·PI·n)=x with x==1 works only for n=0, but not for all integers. E.g. the expression provides for n=1 an x=23.1407, etc. The resolution is quite simple: Eulers polar form of complex numbers only work in the C-world. In the moment, i is removed from the exp(i)-term, everything is shifted back to R. In R even sqrt(-1) is undefined. Things will become clearer, by using complex math strictly. While it is technically legal to eliminate any number by ist inverse (e.g. k·(1/k)=1), in the C-world context matters.
only if m tends to infinity the error or difference from zero and 8m + 2 (reduced to the range 0 to 2 * Pi) in each turn could eventually give values close to zero and only if m tends to infinity will the error be zero. in a continuous circle of dots
The property doesn't apply to multivalue power function consider let p q φ real numbers exp(iφ)^p = exp(iφp + 2πinp) (exp(iφ)^p)^q = exp(i(φ+2πn)pq + 2πimq) It depends on order of applying power. For φ=π/2 we have 1 when n=0 m=-1 q=1/π and can't have 1 for q=4.
I agree with WFA. The best way to avoid any confusion is to implement principal values and principal functions for multivalued situations. It's the best way to communicate your intentions in math. This means we have to sacrifice the relation (z^a)^b = z^(a*b). It is no longer guaranteed in the complex world, unless you take proper precautions! The proper way to multiply exponents is, unless z is purely a positive real number, first rewrite z = |z| e^(Arg(z) i) where the principal argument Arg(z) lies in the interval (-pi, pi] THEN multiply the exponents. WFA, wikipedia, and my math textbook agrees. It is also what my advanced calculator does. If I type sqrt[ e^(-pi i) ], it doesn't return (-i) = e^(-pi/2 i). It returns i = e^(pi/2 i). WFA does the same. This is because they both rewrite e^(-pi i) = e^(pi i) using the principal argument -pi < Arg(z) ≤ pi before multiplying the exponents. This is of course just a convention, but it's a good convention, and everyone should know about it. Then, the principal value of i^(4/5) is [e^(pi/2 i)]^(4/5) = e^(pi/2 i * 4/5) = e^(2/5 i pi) = 0.31 + 0.95i. i^(4/5) =/= (i^4)^5, and is not a valid solution to i^x = 1. The mistake that leads to the extraneous 4n/(4m+1) solutions is that you didn't express i in its principal polar form before multiplying the exponents. The proper way to do it would've been: i^x = 1 e^(pi/2 i x) = e^(2pi i m), m is an integer x/2 = 2m x = 4m
Great video! I don't really know how we should define complex numbers raised to a non-integer exponent, as if we just treat them like usual then we could say that i^ANYTHING=1 as we could write the ANYTHING as (ANYTHING/4)*4 so we'd have (i^4)^(ANYTHING/4) which supposedly is 1.
The confusing part is the fact that if you compute it as (i^4)^(1/anything), you will always get 1 because i^4 is 1 and 1 to the power of anything is 1. So isn't the answer (4n/b) where n is an integer and b =/= 0?
I figured something out. If you take 4n/q where n and q are integers, it still works. Because qth root of 1 will give q solutions out of which 1 will always be one of them.
Counterexample: let n=1, q=8, then 4n/q = ½, but what is i^(½)? In fact {4n/q | n,q are integers} is just Q, so you are saying that _i_ raised to any rational power is 1.
@@VedantArora, yes, I believe that would then work! I reasoned this out to myself as follows, thinking geometrically about the n'th roots of unity and how you can decompose any of the n'th roots of any complex number into the product of the number's principal n'th root and any of the n'th roots of unity: Let n be any positive integer. We'll consider the n'th roots of unity, as well as the principal value of i^(1/n); that is, the n'th root of _i_ that has the smallest positive argument, namely the number z = e^(iπ/(2n)). Then raising z to any multiple of 4 gives a complex number which is equal to one of the n'th roots of unity; let integer m be given, then this number is y=z^(4m). Since y is equal to an n'th root of unity, there must be another n'th root of unity whose angle/argument "fills the gap" between y and 1. Judiciously choosing _w_ to be this other root of unity (that is, _w_ is the n'th root of unity such that w^(4m) has argument equal to 2π-arg(y)), we get y·w^(4m) = 1. In particular, observe: y·w^(4m) = z^(4m) w^(4m) = (zw)^4m, and _zw_ is any of the n'th roots of _i,_ so the above equals i^(4m/n). I am now trying to think about what it is in this chain of logic that n being a multiple of 4 doesn't work with.
🤔 Wait a minute! When both m and n are negative, the fraction 4m/(4n+1) turns into a 4p/(4q-1), where p=-m and q=-n. That means fraction can be generalized to 4m/(2n+1).
For x=4n/(4m+1) where n & m belong to Z, i^x=i^[4n/(4m+1)]=i^(4*n/(4m+1))=(i^4)^(n/4m+1). As i^4=1, (i^4)^(n/4m+1)=1^(n/4m+1). Therefore, whatever is the value of n and m, i^x will always be equal to one and there is no other output.
The issue with the second question is that it is ambiguously phrased. *For* *fractional* *x* , i^x is multi-valued, and since in the complex world there is no good way to define principal root [e.g., which is the principal root among the first question's answer e^(2nπ)?], the second question "i^x = 1 " is as dubious/incorrect as "1^(1/2) = -1" (the latter because ±1 ≠ 1). So, the second question ought to be changed to “For what values of x is 1 *_a_* value of i^x ?". On the other hand, because the first question’s LHS turns out to collapse into a single value 1, its LHS indeed properly equals its RHS. So, the first question is coherent and valid as it is.
been subscribed for two years now, great video as always. i'm graduating next week with a BS in electrical engineering. as for the answer to your question: I'm an engineer, i will just round it to 1 anyways lmao. actually though, i don't think it is wrong to say both solutions are correct, as long as it is understood why the 2nd method can produce multiple values. cheers
It depends of the question. If the question is "find X s.t. i^x=1", of course is correct 4n/(4m+1). If the question is "find X s.t. i^x =1 and only 1", the answer is 4n. I don't give a s... if i^(4/5) have multiple answers as long as 1 is one of the answers. This is the difference between "or" or "exclusive or".
on the other hand, if I perform the operation first for the exponent 4. the result converges directly to 1. but operations with irrational factors will only be achieved at infinity.
Not quite, since 1 has 5 5th roots, the number 1, and 4 more complex numbers, all of which can be represented in polar form with many solutions, hence 4m+1 solutions in either case, if you don't limit your domain :)
I mean, I saw the derivation fine, but why isnt the denominator any odd number? Any odd root (2m+1) of 1 should give 1, not just 4m+1. And if you want to accept the multiple answer paradigm, why not all integers for denominator? Sqrt(1) will give 1 and others. Why x=4n/4m+1 and not x=4n/2m+1 or even x=4n/m?
Hello, if I get your point , I think m =0 in the beginning, but it’s change to n= 1 when you let n=m , so in the last equation( m i #/2 x )= 2i# n / yo start from #=0, not in rad not 2# .. I hope this will help, if doesn’t, please let us know the answer. Not( my English language is so bad) . Maybe I don’t understand what you say .
choose any x, choose any y such that |y|=1 i^x = i^(4*(x/4)) = (i^4)^(x/4) = 1^(x/4) = 1 now sice y*1 = y, then y * i^x = y so let z be such that i^z=y, we get i^(x+z) = y, now we define w=x-z repeating the same trick from before, i^x = i^(w+z) = i^z * i^w = y * 1 = y
Let z€C. i^z = i^(4* (z/4)) = (i^4)^(z/4) = 1^(z/4) = 1. Since our choice of z was arbitrary, i^z = 1 for all z€C. I think the Wolfram Alpha answer is correct.
Since the angle 8m + 2 is multivalued within 2 * Pi, then by increasing m it will cover points at the angles from 0 to 2 * Pi. then when m tends to infinity we will have a reduced angle equal to 0 or 2 * Pi. only then Pi is irrational.
i didnt have clue so i just looked at the second one. i = sqrt-1, so i^2 = -1 . -1 * -1 = 1 . Since i has to be times by itself twice (i mean ^2) to get -1, and then -1 is squared to get 1, then surely i^4 = 1
I would accept (4n/4m+1) because even computing the denomitor first, 1 is always a possible answer. We know i^4n is always 1, but to add to that, i^(4m+1) is always i. That being said, if we write that out as i^(4m+1)=i and take the (4m+1) root of both sides we also find that i=i^(1/4m+1) as one of the solutions. It's just important to know its not the ONLY solution and there are 4 other ones no matter what.
I would say that 4n/(4m+1) is technically correct because it accounts for all possible solutions. However, using the form of (i^4n)^(1/(4m+1)), like you did with (i^4)^0.2, justifies the WFA answer of just x=4n because the 1/(4m+1) is redundant.
I think when you cancel π and 1/π in step 2, you cut the possibility of other roots. In (i^4)^1/4, if we just cancel 4 and 1/4 we get just one value of the expression, i. And then you do it like (i^4)^1/4= 1^1/4=(1^4)^1/4=1. What i =1? The problem lies in just taking one root at one time and then another root at another time. Multiplication of exponents (or cancellation of factors) here can give two results, so caution needs to be taken.so (i^4)^1/4 and i^(4/4) can mean two different things i^(4/4)=i, but (i^4)^1/4 has four different values i, -i, 1,-1.so we can wrongly say i=i^1 = i^(4/4)= (i^4)^1/4= 1^1/4=(1^4)^1/4=1. In the last step of i=i^1 = i^(4/4)= (i^4)^1/4= 1^1/4=1, I'm taking just one value of 1^1/4 which creates the problem. You are taking i^(4/π)= (i^4)^1/ π = 1^ 1/π Does 1^ 1/π have just one value Then the same expression i^ (4/π)= [i^1/π]^4=[(sth^π)^1/π]^4 Here by cancelling π and 1/ π in (sth^π)^1/π you are taking just one value of the expression which just isn't equal to another value of the same expression.
I think I have realised the mistake (without checking any comment in case anyone else has already answered this). The correct solution to i^x = 1 must be a rational number. So only rational numbers ( of the form [ 4n/(4m + 1) ] with integer values of m and n ) will result in 1. Hence in bonus case of 4/pi, the solution ideally must not result in 1.
I have just cleared my 10th standard but I am watching your videos for about 1 year and I can now solve most of the calculus and complex no. Problems that students upto class 12th can't do. Tan(x) a lot
Wouldn't any exponent that has a multiple of four in the numerator satisfy this as long as the denominator was non-zero? i^(4n/k) = (i^4)^(n/k) = 1^(n/k) = 1
4/5 = 0.8 < 1 if you convert the fraction into decimal , i^(4/5) would surely not equal to 1, but rather it would equal (cos(72 degree) + i*sin(72 degree)) (sorry for using degree, because it's much easier to type). But if you do systematically by raising i^4 first then root it down then i^(4/5) equal 1, that's why Wolfram alpha only give 4n as the answer so there would be no confusion. Though if you want the 4n/(4m+1) as the answer, you need to set the condition, those condition which beside m = 1 you would do m = (n-1)/4
I think we should only consider the principal root, and if not, it should be noticed anywher, in this situation you can have √1=1 and if you want to consider the "other solution" then it's just ±√1, and similarly, if we have x²=1 we CANT just take the sqrt on both sides, well we can but √1=1 and √x²≠x, it's just ±x. All that stuff is because 1¹ is obviously 1, and that is also obviously 1^(2/2), so if we take multiple values we'll get 1=±1. Also we could say 1^(2/2)=(√2)² wich in case it's multivaluated, it gives 1 also, but then i^(4/5) wouldn't be 1
Question: if we were willing to accept x=4n/(4m+1), why stop there? isn't x= 4n divided by any real number going to give you 1 as a solution for i^x? for example x=4/7 doesn't fit in the original x=4n/(4m+1), but i^(4/7)=⁷√(i^4)=⁷√1=1?
@@vp_arth apologies, you're right, i missed that. but what about an irrational number in the denominator, such as x=4/e? 1^(1/e) is still 1 so surely this solution should be accounted for as well?
How different a debate is this from the debate that 4^(1/2) should equal 2 because it also equals -2? f(x) = x^n for most integer n values map two different x onto the same y, so taking the inverse function you will get one x mapping to two different y. There’s nothing really special about i here, most numbers to the 4/5 will have 5 possible answers because the “to the 1/5” is the inverse of “to the 5” and the logic I said above about having multiple answers applies. I personally think if you accept you need to plus the plus/minus when you square root both sides, you should except all values of a fractional exponentiation, but that’s more of an opinion that consistency should be maintained than a good mathematical argument. I’d be curious to hear what the real truth on how fractional exponents are evaluated in formal mathematics; is there a process for taking “one correct value” or do you need to accept them all.
I guess X=4n where n is an integer is a more appropriate form as it will always result in 1. Even in your expression X=4n/4m+1 if we keep m=1 it will result in X=4n and will return 1. But if we take some arbitrary values like n=m=2 then we end up with X=8/9. Here if we evaluate using (i^8)^1/9 then we end up with 1 but if we take (i^1/9)^8 then we don't get 1. If n=0 then i^X is always 1. If we take n=1 and m=2 then we end up with X=4/9 and again if we evaluate using (i^4)^1/9 we get 1 but not in the other case. Therefore irrespective of what values we keep in n and m there is always a possibility that we don't get 1. So in my opinion Wolfram Alphas answe is more appropriate and generalized cause it completely eliminates the denominator by keeping m=1 in your expression. By the way big fan of yours and love watching your videos.
I don't have any kind of advanced math knowledge but I love the solution is 4n. In the extra content, 2nd case, why didn't you did [cos(8m+2)+isin(8m+2)]^4 ? multiplying real numers are not the same than imaginary numbers.
I'm not sure I understand your question, but he is multiplying two expressions containing only real quantities there: 4 * [(4m+1)/2] = 8m + 2 (for m in Z - although to be honest, the same would be true for m in C) the multiplication by 4 _is_ the raising to the 4th power in e^{[iπ(4m +1)/2]^4/π} - the complication is that if m is NOT an integer then you can't simply multiply the exponents.
There should be some notation to distinguish whether i^x= 1 means that any value of i^x should equal 1, or that the principal value of i^x should equal 1.
So take i^(4/n) where n is any real number (other than 0). We take the power of the numerator, and then the root of the denominator. Regardless of n, we can only get i^(4/n) = 1. As n is any non-zero real value, 4/n also represents every non-zero real value, so let's call it k. And as we already know, x^0 = 1 for all non-zero x, thus i^0 = 1 anyway. So, it is shown that i^k = 1 for all real values of k. ;)
if you have i^(4n/(1+4m) you can rewrite it as (-1)^((4n/(1+4m))/2) then the fraction cn be converted to 2n/(1+4n) which always has the solution 1 and like some other comments pointet out that any fraction with an multiple of 4 on top and any odd nuber at the bottom is an solution Ps: sorry if my language is a bit confusing im a 12th year math stundent in germany and english isnt one of my mayors i hope that my idea is understandabile
i'd argue that wfa's answer is incomplete: let's say nth root is the inverse operation of nth power. then 5th root of 1 has multiple solutions as well as x^5=1 can be solved e.g. by e^(2*pi/5). so i'd conclude, that nth root can't be interpreted as a function in this context (meaning mapping to just one solution for each x is wrong here) and multiple solutions can't be avoided without ambiguity
The problem in your demonstration is actually when you write (1)^(1/5)=1, which is clearly true in R, but when we are in C, x^5=1 has five solutions (which are exp(2.i.k.pi/5), where k is an integer. The question you raise is about the definition of a function x---->x^q (where q is a rational) in C. The problem is particularly tough when x is not a real, so we know that there isn't any real solution. So how to choose between these five values? I haven't studied the question deeply but I don't think we can defiine a continuous function on C.
How about using the Taylor series e^z = 1 + z + (z^2)/2 + (z^3)/6 + ... Let a = m/n, and insert z = i*a*pi*(4k+1)/2 into the formula. So that we can get only one anwser for a k value?
For the last part. Can you even conclude that? From the start of that problem, i to the 4/pi, you are already assuming that m cannot be an integer because you plugged in pi. Therefore when concluding your answer that cos(8m+2)+isin(8m+2) cannot equal 1, it’s incorrect to state that m is an integer there.
For the bonus part, I think that the second method is better. If we consider the first method, we could extend the solutions of i^x = 1 to all the complex numbers. Whichever x you choose (except 0), you'll always be able to write it with the form 4/(4/x). Therefore, when you apply the first method, you get i^x = (4/x)-th root of 1, which is 1. Maybe I'm wrong but it should work for every complex number other than 0. So, this method has something wrong in it, I'd rather use the second one.
I think the second method in the bonus part is the proper way to calculate irrational complex roots. More explicitly, converting the complex number to exponential form then simplifying the exponents, not calculating each exponent separately.
5:44 "Is this legitamite? What do you think?" I think the answer x=4n/(4m+1) is not legitamite for two reasons (well: mainly the second). 1) I think analoguous to the sqrt-function (sqrt(x) typically the principle branch k=0 for x^0.5), you shouldn't be allowed to change the branch (nor the branch cut) of the complex logarithm within an equation, so you can't end up using n and m other than n = m, else you could confuse different numbers, which share the same representation (think of sqrt(2)=-sqrt(2), with the second sqrt(x) using the branch k=1 for x^0.5). But don't get me wrong here; that wouldn't change the resulting formular here; you would need an additional step: i^x = 1 (Def. complex exp) e^(i (4k+1)/2 π x) = e^(i 2 π k) (periodic over angle with period 2 π) e^(i (4k+1)/2 π x) = e^(i 2 π (k+l)) (n := k+l; m := k) e^(i (4m+1)/2 π x) = e^(i 2 π n) 2) Expanding numbers 'wildly' includes the risk of resizing the branch cut or changing the used branch unintentionally. If you allow that, then you would be able to assign any complex number representation to a solution x for i^x = 1. Some examples: e^(i 0.5(4k+1) π x) = i^x = 1 = e^(LN(1)) = e^(i 2 π k) = e^(i 2 π k LN(e)) = e^(i 2 π k (i 2 π k)) = e^(-1 2^2 π^2 k^2) ==> i 0.5(4k+1) π x = -1 2^2 π^2 k^2 x = i 8 π k^2 / (4k+1) e^(i 0.5(4k+1) π x) = ... = e^(LN(1) (LN(e))^2) = e^(-i 2^3 π^3 k^3) ==> i 0.5(4k+1) π x = -i 2^3 π^3 k^3 x = -1 2^4 π^2 k^3 / (4k+1) e^(i 0.5(4k+1) π x) = i^x = 1 = 1*1 = e^(-1 2^2 π^2 k^2) * e^(-i 2^3 π^3 k^3) = e^(-1 2^2 π^2 k^2 (1 + i 2 π k)) ==> i 0.5(4k+1) π x = -1 2^2 π^2 k^2 (1 + i 2 π k) x = 2^3 π k^2 (2 π k - i)/(4k+1) Definitions i used: branch k ∈ ℤ; branch cut B_k:=[2 π k, 2 π (k+1)); α_k ∈ B_k; r ∈ ℝ; y, z:= r e^(i α) ∈ ℂ; ln: logarithm over real numbers; arg_k(z) := α_k := α_0 + 2 π k; LN(z) := ln(|z|) + i arg_k(z); y^z := e^(z LN(y))
I also wanted to point out, that i don't see any need for the definitions at timestamp 9:38 - just use the definiiton of complex exponentiation, to see those terms are all equal (which also could be used to calculate all solutions): (i^(1/n))^m := e^(m LN(i^(1/n))) := e^(m LN(e^((1/n) LN(i)))) := e^(m (1/n) LN(i)) = e^((m/n) LN(i)) =: i^(m/n) (i^m)^(1/n) := e^((1/n) LN(i^m)) := e^((1/n) LN(e^(m LN(i)))) := e^((1/n) m LN(i)) = e^((m/n) LN(i)) =: i^(m/n)
Well to be fair, if you take out 4 you end up with 4(1/4m+1) and then you substitute that bracket part with anything lets say A, you again end up with a multiple of 4 because the final part is in form of 4A.
Ok, I’ve had a think about this and I think that the assumption that you can chain the exponents is flawed. Here’s my reasoning: 1. Exponentiation at its heart is multiplying a number by itself a certain number of times. Yes, we’ve extended that definition to allow for rational and irrational exponents, but essentially that’s the core of it. 2. Multiplying any number (defining a point) in the complex plane by i rotates that point by pi/2 radians counterclockwise. 3. Therefore, raising i to any power should result in i being rotated, resulting in a point on the unit circle. By taking the pi root first, you end up with a number with both real and imaginary components. When you take that to the 4th power, you are no longer doing a rotation about the origin - there’s rotation, scaling and translation going on. Doing it the other way doesn’t work either. I think the answer is -cos(2 + 2n pi) + i sin (2 + 2n pi), as you’re rotating i by 4/pi * pi/2 radians.
Personally, I would say that it depends on how the problem is presented/phrased. For instance, if you simply had "Solve for x" where i^x = 1, then I would argue either answer is a valid solution to the conditions presented. Even a simplistic answer of x=0 or x=4 would be a valid solution for x. However, if the problem was presented as "Solve for all possible values of x" --- or a similar phrasing --- it would REQUIRE the more general solution of x = (4n)/(4m+1) as this returns the set of all possible solutions for x.
The reason this is confusing is because we have not been rigorous in defining what (s1)^(s2) means beforehand for complex s1 and s2. If we go with exp[s2·log(s1)] for nonzero s1, then x^i := = exp[i·log(x)] = 1, and i^x := exp[x·log(i)] = 1. For the former, i·log(x) = 2·m·π·i, so log(x) = 2·m·π, implying x = exp(2·m·π), which is the result in the video. However, for the latter, exp[x·log(i)] = 1, this results in exp(x·π·i/2) = 1, implying x·π·i/2 = 2·m·π·i, so x = 4·m. It would be a mistake to say that i^(4/5) = (i^4)^(1/5), because in reality, fractional exponentiation is almost always defined as [i^(1/5)]^4, making it single valued. And this is not interchangeable with (i^4)^(1/5). Remember: mathematics is all about the definitions. You COULD define exponentiation in such a way as to make it set-valued, rather than complex-number-valued, but no mathematician would actually do this. x^y should ALWAYS, unless explicitly stated otherwise, understood to be an abbreviation for exp[y·log(x)] for nonzero x, where log(x) = ln(|x|) + i·atan2[Im(x), Re(x)].
Maybe I'm wrong but there is something I don't understand. In De Moivre's formula, when we raise e^(i pheta) (with pheta the angle) to the power n, such that we have (e^(i pheta))^n, n must be an integer. But at 1:40 you raise both sides to the power of 1/i, which is not an integer. Therefore your demonstration must be wrong. If someone could explain me if I am wrong, I would be grateful.
I think the values of x that give multiple solutions should be allowed. I tend to think of branch cuts as arbitrarily destroying the beautiful Riemann surface structure of complex numbers (and complex exponentials in particular).
In this sense, isn't every real number a solution? For example: i^3 = i^(12/4) = (i^12)^(1/4) = 1^(1/4) = 1 (and 3 other roots). Although obviously i^3 = -i
There's a problem I've been banging my head against a while that might interest you... what do all the solutions of "X to the power Y equals Y to the power X" look like if you allow X and Y to be complex?
Wouldn't you necessarily need to take the values of x that satisfy the equation otherwise the multi valued nature of the function interferes with your solution? I think it's maybe a situation where you have to be careful to choose a solution that avoids the branch cuts of the complex function or else you get contradictory answers like this, but I'm not sure either. This is interesting!
Hm maybe you can take some limit of maxima(cos(8x+2), x is integer) or something Also, maybe you can't really cancel out the pi with the 1/pi in that expression?
my first thought about i^x = 1 was 4Z (the same as x = 4n as n is in Z) and in fact i^x = (e^i(pi/2)) ^x = e^(i x pi/2) this is equal to 1 if (x pi/2) = 2kpi as k is in Z, so x = 4k. I think the problem with the second solution is that for example 1^(1/5) equals 1 only in the Real numbers, while in Complex numbers 1^(1/5) gives 5 different solutions, namely e^i(2/5)pi, e^i(4/5)pi, e^i(6/5)pi, e^i(8/5)pi, e^i(10/5)pi = e^i2pi = 1. p.s.: writing this stuff on my phone while in bed is hell
x^n for noninteger n is neither surjective nor injective in the complex numbers. As such, we would need to assign a principal value to an expression like i^(4/pi) in order for it to have a well-defined value; what we assign the principal value to can be 1 or cos(2)+isin(2), or something else. There's probably already a convention for this, which is why WolframAlpha does not output, say x=4/5, as a solution to i^x=1 (see edit below). Knowing this, we may wonder why we never get x=4/pi as a solution to i^x=1. I believe the error lies in (e^((4m+1)/2*pi*i))^x e^((4m+1)/2*pi*i*x). This does not hold for noninteger x when using principal values (e.g. (e^(2*pi*i))^(1/2)=1^(1/2)=1 but e^(2*pi*i*1/2)=e^(pi*i)=-1). -a very, very inexperienced mathematician edit: In some sense, I think it's like asking if x=1 is a solution to sqrt(x)=-1. We define the principal value of sqrt(1) to be 1, so in this case, x=1 is an extraneous solution. However, if we defined the principal value of sqrt(1) to be -1, there would be a solution. Similarly, we can define the principal value of a number to be such that i^(4/5)=1 in some cases and in other cases not. WolframAlpha says that i^(4/5)=0.309+0.951i, so I'm guessing some convention has been defined such that i^(4/5) != 1 and 4/5 is not a solution to i^x=1.
Amazing! I didn't know that the identity (e^x)^y= e^xy does not hold for complex x and y. The exponentiation page of wikipedia explains it well. Since you are using this identity in both calculations, both are wrong. I think you are eliminating answers in both calculations
Are n and m in x = 4n / (4m + 1) really independent variables? It'd seem to me, intuitively at least, that they aren't, since I presume the angles on both sides of an "=" sign to be linked. In that case, m = n + 2πk + π/2, which would result in x = n / ( n + 2πk + π/2 + 1/4 ). Ouch...
Okay, we got solution with 4m+1 in the denominator and we checked the solution 4/5 , but what if we just check for example 4/7? i^4/7= (i^4)^1/7 =1, and 4m+1=7 has no solution in integers. Where is a mistake in my reasoning?
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
The disappointment on his face after realizing that no one paused the video to solve the equation
😆
@@blackpenredpen
The disappointment on Oreo's face after realizing that no one paused the video to solve the equation🐰
I did! I at least attempted to do the second one
i did and it was fun(?
No, I paused to solve it, but still incorrect lol
This is the greatest video of a math debate, not like the 6÷2(3) stuff.
😆
By the way, the answer is 9 right? :P
@@pbj4184 the answer is 0
@@pbj4184 you multiply first because of the bracket
You mean Mind you Decision?
Got to love those edge cases where the commutativity of operations that works perfectly fine for real numbers becomes a trap in the complex world.
4n is included in 4n/(4m+1) because 4n is just when m is equal to 0
True
@@JB-ym4up 4n/x for any integer x != 0 arguably produces solutions, and for x = 1 unarguably produces solutions. Any operations on an integer m that leave it as an integer won't effect that, so 4n/(4m-1) is just as valid as 4n/(4m+1) or even 4n/(3m^2-26)
You pronounced "Euler's formula" quite well! Greetings from Germany :)
Thanks.
I'd like to know the average wage in Germany. Could you tell me that?
@@Mnemonic-X €22 per hour.
Is pronounced wheeeeler
Hello fellow LP fan
I don't know if you upped the res or I changed the settings on my phone, but that's one high quality vid. Great stuff as always!
Thanks. I actually have been using my phone to record in 1080P, 60FPS. My old videos (when I was recording in classrooms) were recorded on my MacBook Air, 720p!
The main problem with polar coordinate (which pops up often in your videos) is that you can assign multple values to get the same results. The form x = (4n) / (4m+1) is equivalent to taking the (4m+1)th root of i^(4n), for which 1 is a possible solution (but you know there are 4m+1 answers in total, so another 4m possible answers, and those will not be 1, rather complex numbers).
A more strict definition of polar coordinates restricts the argument to be in the [-PI, PI) range (or some other 2-PI length interval, e.g [0, 2PI) which is perhaps more practical) and avoids somewhat whacky results like this.
but also in a 2pi range multiple solutions exist e.g. for x^5=1. so doesn't the ambiguity remain? i.e. 5th root of 1 can be mapped to different values in polar coordinates even in a 2*pi range..
even if you define such a restriction so only one answer pops up. in my eyes, the info that other answers exist, should be stated. Not for compitionisms sake, but because, such an interval is not the same for every problem, or even have a formula for every different problem. So the only reason to do it is for visualization in a class or smthng. Such debates should exist if this is not your Calc 1,2 etc. For me its just that the question is poorly stated. There should be an interval in the question or further explanation. Its not something new. Poorly stated questions are asked all the time in math. And create fun or not so, problems.
I think i^x just raises more questions when you forget that exponents have different branches in their function. Also, by looking at branches I think you can get infinitely many answers from i^(4/pi)...
Pi-rt(I) 💀💀💀💀💀
I return to this channel nearly after a year and damn you got a nice beard now!
Maybe the ambiguity comes from the fact that taking a root of one can yield multiple values. Take the square root of one: technically, there are 2 real answers to this: 1 and -1. Negative one times negative one also equals one. By just saying that 1^(1/pi) = 1, you may be missing other complex solutions to the pi'th root of 1. If you put 1^(1/3) into wolfram alpha, you get 3 complex solutions (and maybe more if you add 2pi, idk).
Despite being a medical student, I really enjoy your videos! ❤️
Thanks, future dr.!
I'm in the same situation Tazrian T!!!! His videos are awesome!!!
tbh you don't need that much to enjoy most math. You can learn up to calc 3 in a few months if you have good resources and aren't bad at learning new concepts in general. What's hard is being a good problem solver because that needs a lot of practice to be able to recognize a lot of patterns in all sorts of different topics.
an equation is well defined if it is required that each of its solutions make the equality true: an equation can have multiple solutions but not solutions that solve the equality or not in an arbitrary way. So in the second equation the only acceptable solution is x = 4n. The polar representation of complex numbers, in equations, must be used when it does not give rise to ambiguity, that is, when it effectively allows to identify all the solutions, not when it produces expressions with multiple results that give rise to "partial" equalities.
The main thought-error happend at 2:06. exp(2·PI·n)=x with x==1 works only for n=0, but not for all integers. E.g. the expression provides for n=1 an x=23.1407, etc. The resolution is quite simple: Eulers polar form of complex numbers only work in the C-world. In the moment, i is removed from the exp(i)-term, everything is shifted back to R. In R even sqrt(-1) is undefined. Things will become clearer, by using complex math strictly. While it is technically legal to eliminate any number by ist inverse (e.g. k·(1/k)=1), in the C-world context matters.
only if m tends to infinity the error or difference from zero and 8m + 2 (reduced to the range 0 to 2 * Pi) in each turn could eventually give values close to zero and only if m tends to infinity will the error be zero. in a continuous circle of dots
The property doesn't apply to multivalue power function
consider
let p q φ real numbers
exp(iφ)^p = exp(iφp + 2πinp)
(exp(iφ)^p)^q = exp(i(φ+2πn)pq + 2πimq)
It depends on order of applying power. For φ=π/2 we have 1 when n=0 m=-1 q=1/π and can't have 1 for q=4.
I agree with WFA. The best way to avoid any confusion is to implement principal values and principal functions for multivalued situations. It's the best way to communicate your intentions in math.
This means we have to sacrifice the relation (z^a)^b = z^(a*b). It is no longer guaranteed in the complex world, unless you take proper precautions! The proper way to multiply exponents is, unless z is purely a positive real number, first rewrite z = |z| e^(Arg(z) i) where the principal argument Arg(z) lies in the interval (-pi, pi] THEN multiply the exponents. WFA, wikipedia, and my math textbook agrees.
It is also what my advanced calculator does. If I type sqrt[ e^(-pi i) ], it doesn't return (-i) = e^(-pi/2 i).
It returns i = e^(pi/2 i). WFA does the same.
This is because they both rewrite e^(-pi i) = e^(pi i) using the principal argument -pi < Arg(z) ≤ pi before multiplying the exponents. This is of course just a convention, but it's a good convention, and everyone should know about it.
Then, the principal value of i^(4/5) is [e^(pi/2 i)]^(4/5) = e^(pi/2 i * 4/5) = e^(2/5 i pi) = 0.31 + 0.95i.
i^(4/5) =/= (i^4)^5, and is not a valid solution to i^x = 1.
The mistake that leads to the extraneous 4n/(4m+1) solutions is that you didn't express i in its principal polar form before multiplying the exponents. The proper way to do it would've been:
i^x = 1
e^(pi/2 i x) = e^(2pi i m), m is an integer
x/2 = 2m
x = 4m
Great freaking video again! Soo close to 700k
Thanks.
I see you in Michael Penn as well
thank you for all the content you put out, man! You keep me interested in math
Great video! I don't really know how we should define complex numbers raised to a non-integer exponent, as if we just treat them like usual then we could say that i^ANYTHING=1 as we could write the ANYTHING as (ANYTHING/4)*4 so we'd have (i^4)^(ANYTHING/4) which supposedly is 1.
if we consider (4n)/(4m+1) to be the set of solutions, shouldn’t we also consider (4n)/(4m+3)?
It’s included in that expression. Since n and m are integers. For example, to get 4/3, you just pick n=m=-1 in my expression.
Then take (4n)/(4m+(2k+1)) as a set of solutions where k=integers😂😂
I've got a question, in the end m could be any real number as 4n=1 if n is integer.
@@blackpenredpen Thank you !
The confusing part is the fact that if you compute it as (i^4)^(1/anything), you will always get 1 because i^4 is 1 and 1 to the power of anything is 1. So isn't the answer (4n/b) where n is an integer and b =/= 0?
I figured something out. If you take 4n/q where n and q are integers, it still works. Because qth root of 1 will give q solutions out of which 1 will always be one of them.
Thanks, even I didn't thought such
Counterexample: let n=1, q=8, then 4n/q = ½, but what is i^(½)?
In fact {4n/q | n,q are integers} is just Q, so you are saying that _i_ raised to any rational power is 1.
@@JivanPal Considering your point. Can we add the statement “where q is not divisible by 4”??
@@VedantArora, yes, I believe that would then work! I reasoned this out to myself as follows, thinking geometrically about the n'th roots of unity and how you can decompose any of the n'th roots of any complex number into the product of the number's principal n'th root and any of the n'th roots of unity:
Let n be any positive integer. We'll consider the n'th roots of unity, as well as the principal value of i^(1/n); that is, the n'th root of _i_ that has the smallest positive argument, namely the number z = e^(iπ/(2n)).
Then raising z to any multiple of 4 gives a complex number which is equal to one of the n'th roots of unity; let integer m be given, then this number is y=z^(4m). Since y is equal to an n'th root of unity, there must be another n'th root of unity whose angle/argument "fills the gap" between y and 1.
Judiciously choosing _w_ to be this other root of unity (that is, _w_ is the n'th root of unity such that w^(4m) has argument equal to 2π-arg(y)), we get y·w^(4m) = 1.
In particular, observe:
y·w^(4m)
= z^(4m) w^(4m)
= (zw)^4m,
and _zw_ is any of the n'th roots of _i,_ so the above equals i^(4m/n).
I am now trying to think about what it is in this chain of logic that n being a multiple of 4 doesn't work with.
@@JivanPal damn, love your calculations dude!
🤔 Wait a minute! When both m and n are negative, the fraction 4m/(4n+1) turns into a 4p/(4q-1), where p=-m and q=-n. That means fraction can be generalized to 4m/(2n+1).
For x=4n/(4m+1) where n & m belong to Z,
i^x=i^[4n/(4m+1)]=i^(4*n/(4m+1))=(i^4)^(n/4m+1).
As i^4=1, (i^4)^(n/4m+1)=1^(n/4m+1).
Therefore, whatever is the value of n and m, i^x will always be equal to one and there is no other output.
Because raising e to a complex exponent is not 1 to 1, you cannot really say that if e^(ix) = e^(iy) then x=y
This is the step at 4:30
The issue with the second question is that it is ambiguously phrased.
*For* *fractional* *x* , i^x is multi-valued, and since in the complex world there is no good way to define principal root [e.g., which is the principal root among the first question's answer e^(2nπ)?], the second question "i^x = 1 " is as dubious/incorrect as "1^(1/2) = -1" (the latter because ±1 ≠ 1).
So, the second question ought to be changed to “For what values of x is 1 *_a_* value of i^x ?".
On the other hand, because the first question’s LHS turns out to collapse into a single value 1, its LHS indeed properly equals its RHS. So, the first question is coherent and valid as it is.
been subscribed for two years now, great video as always. i'm graduating next week with a BS in electrical engineering. as for the answer to your question:
I'm an engineer, i will just round it to 1 anyways lmao.
actually though, i don't think it is wrong to say both solutions are correct, as long as it is understood why the 2nd method can produce multiple values.
cheers
That's exciting Brian! Congrats!!!
It depends of the question. If the question is "find X s.t. i^x=1", of course is correct 4n/(4m+1).
If the question is "find X s.t. i^x =1 and only 1", the answer is 4n.
I don't give a s... if i^(4/5) have multiple answers as long as 1 is one of the answers. This is the difference between "or" or "exclusive or".
on the other hand, if I perform the operation first for the exponent 4. the result converges directly to 1. but operations with irrational factors will only be achieved at infinity.
Not quite, since 1 has 5 5th roots, the number 1, and 4 more complex numbers, all of which can be represented in polar form with many solutions, hence 4m+1 solutions in either case, if you don't limit your domain :)
I mean, I saw the derivation fine, but why isnt the denominator any odd number? Any odd root (2m+1) of 1 should give 1, not just 4m+1.
And if you want to accept the multiple answer paradigm, why not all integers for denominator? Sqrt(1) will give 1 and others.
Why x=4n/4m+1 and not x=4n/2m+1 or even x=4n/m?
Hello, if I get your point , I think m =0 in the beginning, but it’s change to n= 1 when you let n=m , so in the last equation( m i #/2 x )= 2i# n / yo start from #=0, not in rad not 2# .. I hope this will help, if doesn’t, please let us know the answer. Not( my English language is so bad) . Maybe I don’t understand what you say .
"namely: multiples of n" -> "namely: multiples of 4" ~ @ 2:30
choose any x, choose any y such that |y|=1
i^x = i^(4*(x/4)) = (i^4)^(x/4) = 1^(x/4) = 1
now sice y*1 = y, then y * i^x = y so let z be such that i^z=y, we get i^(x+z) = y, now we define w=x-z
repeating the same trick from before, i^x = i^(w+z) = i^z * i^w = y * 1 = y
blackpenredpenbluepen
Nice work! You destroyed Wolfram Alpha! 😄
😆
Let z€C. i^z = i^(4* (z/4)) = (i^4)^(z/4) = 1^(z/4) = 1. Since our choice of z was arbitrary, i^z = 1 for all z€C.
I think the Wolfram Alpha answer is correct.
Since the angle 8m + 2 is multivalued within 2 * Pi, then by increasing m it will cover points at the angles from 0 to 2 * Pi. then when m tends to infinity we will have a reduced angle equal to 0 or 2 * Pi. only then Pi is irrational.
If 4n/4m+1 is a solution isn't then 4n/m m0 also a solution? Because 1 is then always one solution
wow I remember when you did your 100k video, glad to see you're getting close to a million
Seeing you stumped on this mystery was unnerving but kind of exciting.
i didnt have clue so i just looked at the second one. i = sqrt-1, so i^2 = -1 . -1 * -1 = 1 . Since i has to be times by itself twice (i mean ^2) to get -1, and then -1 is squared to get 1, then surely i^4 = 1
I would accept (4n/4m+1) because even computing the denomitor first, 1 is always a possible answer. We know i^4n is always 1, but to add to that, i^(4m+1) is always i. That being said, if we write that out as i^(4m+1)=i and take the (4m+1) root of both sides we also find that i=i^(1/4m+1) as one of the solutions. It's just important to know its not the ONLY solution and there are 4 other ones no matter what.
I would say that 4n/(4m+1) is technically correct because it accounts for all possible solutions. However, using the form of (i^4n)^(1/(4m+1)), like you did with (i^4)^0.2, justifies the WFA answer of just x=4n because the 1/(4m+1) is redundant.
I think when you cancel π and 1/π in step 2, you cut the possibility of other roots.
In (i^4)^1/4, if we just cancel 4 and 1/4 we get just one value of the expression, i. And then you do it like (i^4)^1/4= 1^1/4=(1^4)^1/4=1. What i =1?
The problem lies in just taking one root at one time and then another root at another time.
Multiplication of exponents (or cancellation of factors) here can give two results, so caution needs to be taken.so (i^4)^1/4 and i^(4/4) can mean two different things
i^(4/4)=i, but (i^4)^1/4 has four different values i, -i, 1,-1.so we can wrongly say i=i^1 = i^(4/4)= (i^4)^1/4= 1^1/4=(1^4)^1/4=1.
In the last step of i=i^1 = i^(4/4)= (i^4)^1/4= 1^1/4=1, I'm taking just one value of 1^1/4 which creates the problem.
You are taking i^(4/π)= (i^4)^1/ π = 1^ 1/π
Does 1^ 1/π have just one value
Then the same expression i^ (4/π)= [i^1/π]^4=[(sth^π)^1/π]^4
Here by cancelling π and 1/ π in (sth^π)^1/π
you are taking just one value of the expression which just isn't equal to another value of the same expression.
I think I have realised the mistake (without checking any comment in case anyone else has already answered this).
The correct solution to
i^x = 1 must be a rational number. So only rational numbers ( of the form
[ 4n/(4m + 1) ] with integer values of m and n ) will result in 1. Hence in bonus case of 4/pi, the solution ideally must not result in 1.
Amazing Video. Tomorrow Is My Physics Test. But My Mind Is Focusing More On Your Video. 😰😅😂
I have just cleared my 10th standard but I am watching your videos for about 1 year and I can now solve most of the calculus and complex no. Problems that students upto class 12th can't do.
Tan(x) a lot
i just started my math studies this fall and you are a source of constant inspiration for me
Wouldn't any exponent that has a multiple of four in the numerator satisfy this as long as the denominator was non-zero?
i^(4n/k) = (i^4)^(n/k) = 1^(n/k) = 1
Yeah I really didn't understand why 4m+1 specifically while ^4/6 for example will satisfy the equation
4/5 = 0.8 < 1
if you convert the fraction into decimal , i^(4/5) would surely not equal to 1, but rather it would equal (cos(72 degree) + i*sin(72 degree)) (sorry for using degree, because it's much easier to type). But if you do systematically by raising i^4 first then root it down then i^(4/5) equal 1, that's why Wolfram alpha only give 4n as the answer so there would be no confusion.
Though if you want the 4n/(4m+1) as the answer, you need to set the condition, those condition which beside m = 1 you would do m = (n-1)/4
i^0.8 has 5 values. And one of them is 1
@@vp_arth how do u get the other values?
Not sure for now. May be I got them as solutions of x^5=i^4
Black pen red pen, yay!!!
7:08 By the same logic the solutions are 4n/(2m+1)
Because i^(4/7)=1...
I think we should only consider the principal root, and if not, it should be noticed anywher, in this situation you can have √1=1 and if you want to consider the "other solution" then it's just ±√1, and similarly, if we have x²=1 we CANT just take the sqrt on both sides, well we can but √1=1 and √x²≠x, it's just ±x.
All that stuff is because 1¹ is obviously 1, and that is also obviously 1^(2/2), so if we take multiple values we'll get 1=±1.
Also we could say 1^(2/2)=(√2)² wich in case it's multivaluated, it gives 1 also, but then i^(4/5) wouldn't be 1
Question: if we were willing to accept x=4n/(4m+1), why stop there? isn't x= 4n divided by any real number going to give you 1 as a solution for i^x? for example x=4/7 doesn't fit in the original x=4n/(4m+1), but i^(4/7)=⁷√(i^4)=⁷√1=1?
it fits, n=-1, m=-2
@@vp_arth apologies, you're right, i missed that. but what about an irrational number in the denominator, such as x=4/e? 1^(1/e) is still 1 so surely this solution should be accounted for as well?
@@omkarnagarhalli5217 you cannot simplify x^(4/e) as (x^4)^(1/e), you can do that only for q/n where both q and n are coprime integers.
How different a debate is this from the debate that 4^(1/2) should equal 2 because it also equals -2? f(x) = x^n for most integer n values map two different x onto the same y, so taking the inverse function you will get one x mapping to two different y.
There’s nothing really special about i here, most numbers to the 4/5 will have 5 possible answers because the “to the 1/5” is the inverse of “to the 5” and the logic I said above about having multiple answers applies.
I personally think if you accept you need to plus the plus/minus when you square root both sides, you should except all values of a fractional exponentiation, but that’s more of an opinion that consistency should be maintained than a good mathematical argument. I’d be curious to hear what the real truth on how fractional exponents are evaluated in formal mathematics; is there a process for taking “one correct value” or do you need to accept them all.
it really depends on what angle u take it as, how many rounds rotate
so wolfram alpha took the limit of 2pi(n) to 1 because that's the only way you don't end up with a denominator.
I guess X=4n where n is an integer is a more appropriate form as it will always result in 1.
Even in your expression X=4n/4m+1 if we keep m=1 it will result in X=4n and will return 1.
But if we take some arbitrary values like n=m=2 then we end up with X=8/9. Here if we evaluate using (i^8)^1/9 then we end up with 1 but if we take (i^1/9)^8 then we don't get 1.
If n=0 then i^X is always 1.
If we take n=1 and m=2 then we end up with X=4/9 and again if we evaluate using (i^4)^1/9 we get 1 but not in the other case.
Therefore irrespective of what values we keep in n and m there is always a possibility that we don't get 1.
So in my opinion Wolfram Alphas answe is more appropriate and generalized cause it completely eliminates the denominator by keeping m=1 in your expression.
By the way big fan of yours and love watching your videos.
I don't have any kind of advanced math knowledge but I love the solution is 4n. In the extra content, 2nd case, why didn't you did [cos(8m+2)+isin(8m+2)]^4 ? multiplying real numers are not the same than imaginary numbers.
I'm not sure I understand your question, but he is multiplying two expressions containing only real quantities there:
4 * [(4m+1)/2] = 8m + 2 (for m in Z - although to be honest, the same would be true for m in C)
the multiplication by 4 _is_ the raising to the 4th power in e^{[iπ(4m +1)/2]^4/π} - the complication is that if m is NOT an integer then you can't simply multiply the exponents.
There should be some notation to distinguish whether i^x= 1 means that any value of i^x should equal 1, or that the principal value of i^x should equal 1.
So take i^(4/n) where n is any real number (other than 0). We take the power of the numerator, and then the root of the denominator. Regardless of n, we can only get i^(4/n) = 1. As n is any non-zero real value, 4/n also represents every non-zero real value, so let's call it k. And as we already know, x^0 = 1 for all non-zero x, thus i^0 = 1 anyway. So, it is shown that i^k = 1 for all real values of k. ;)
if you have i^(4n/(1+4m) you can rewrite it as (-1)^((4n/(1+4m))/2) then the fraction cn be converted to 2n/(1+4n) which always has the solution 1 and like some other comments pointet out that any fraction with an multiple of 4 on top and any odd nuber at the bottom is an solution
Ps: sorry if my language is a bit confusing im a 12th year math stundent in germany and english isnt one of my mayors i hope that my idea is understandabile
i'd argue that wfa's answer is incomplete: let's say nth root is the inverse operation of nth power. then 5th root of 1 has multiple solutions as well as x^5=1 can be solved e.g. by e^(2*pi/5). so i'd conclude, that nth root can't be interpreted as a function in this context (meaning mapping to just one solution for each x is wrong here) and multiple solutions can't be avoided without ambiguity
The problem in your demonstration is actually when you write (1)^(1/5)=1, which is clearly true in R, but when we are in C, x^5=1 has five solutions (which are exp(2.i.k.pi/5), where k is an integer. The question you raise is about the definition of a function x---->x^q (where q is a rational) in C. The problem is particularly tough when x is not a real, so we know that there isn't any real solution. So how to choose between these five values? I haven't studied the question deeply but I don't think we can defiine a continuous function on C.
Would the answer be different if 1 was expressed in the polar form at 11:13, like in 01:32?
WHAT? DID YOU WATCH THIS IN THE PAST???????????????????
@@aashsyed1277 yes many can
@@krishbishwanath HOW?
@@aashsyed1277 If i had to guess, i'm sure members have access to videos early
Interensting video
How about using the Taylor series e^z = 1 + z + (z^2)/2 + (z^3)/6 + ...
Let a = m/n, and insert z = i*a*pi*(4k+1)/2 into the formula.
So that we can get only one anwser for a k value?
For the last part. Can you even conclude that? From the start of that problem, i to the 4/pi, you are already assuming that m cannot be an integer because you plugged in pi. Therefore when concluding your answer that cos(8m+2)+isin(8m+2) cannot equal 1, it’s incorrect to state that m is an integer there.
Why is this video invisible? I only found it with the link from “Introduction to Power Series for Calc2”, but it doesn’t appear on your channel
LVE STREAM STARTING!
But this comment is a day old
Wait a little time treveler
@@asamenechbayissa553 NO HE IS NOT A TIME TRAVELER
@@asamenechbayissa553 Indeed I am. In the future, Steve Jobs dies of Ligma
For the bonus part, I think that the second method is better. If we consider the first method, we could extend the solutions of i^x = 1 to all the complex numbers. Whichever x you choose (except 0), you'll always be able to write it with the form 4/(4/x). Therefore, when you apply the first method, you get i^x = (4/x)-th root of 1, which is 1. Maybe I'm wrong but it should work for every complex number other than 0. So, this method has something wrong in it, I'd rather use the second one.
I think the second method in the bonus part is the proper way to calculate irrational complex roots. More explicitly, converting the complex number to exponential form then simplifying the exponents, not calculating each exponent separately.
5:44 "Is this legitamite? What do you think?"
I think the answer x=4n/(4m+1) is not legitamite for two reasons (well: mainly the second).
1) I think analoguous to the sqrt-function (sqrt(x) typically the principle branch k=0 for x^0.5), you shouldn't be allowed to change the branch (nor the branch cut) of the complex logarithm within an equation, so you can't end up using n and m other than n = m, else you could confuse different numbers, which share the same representation (think of sqrt(2)=-sqrt(2), with the second sqrt(x) using the branch k=1 for x^0.5).
But don't get me wrong here; that wouldn't change the resulting formular here; you would need an additional step:
i^x = 1
(Def. complex exp) e^(i (4k+1)/2 π x) = e^(i 2 π k)
(periodic over angle with period 2 π) e^(i (4k+1)/2 π x) = e^(i 2 π (k+l))
(n := k+l; m := k) e^(i (4m+1)/2 π x) = e^(i 2 π n)
2) Expanding numbers 'wildly' includes the risk of resizing the branch cut or changing the used branch unintentionally.
If you allow that, then you would be able to assign any complex number representation to a solution x for i^x = 1.
Some examples:
e^(i 0.5(4k+1) π x) = i^x = 1 = e^(LN(1)) = e^(i 2 π k) = e^(i 2 π k LN(e)) = e^(i 2 π k (i 2 π k)) = e^(-1 2^2 π^2 k^2)
==> i 0.5(4k+1) π x = -1 2^2 π^2 k^2
x = i 8 π k^2 / (4k+1)
e^(i 0.5(4k+1) π x) = ... = e^(LN(1) (LN(e))^2) = e^(-i 2^3 π^3 k^3)
==> i 0.5(4k+1) π x = -i 2^3 π^3 k^3
x = -1 2^4 π^2 k^3 / (4k+1)
e^(i 0.5(4k+1) π x) = i^x = 1 = 1*1 = e^(-1 2^2 π^2 k^2) * e^(-i 2^3 π^3 k^3) = e^(-1 2^2 π^2 k^2 (1 + i 2 π k))
==> i 0.5(4k+1) π x = -1 2^2 π^2 k^2 (1 + i 2 π k)
x = 2^3 π k^2 (2 π k - i)/(4k+1)
Definitions i used:
branch k ∈ ℤ;
branch cut B_k:=[2 π k, 2 π (k+1));
α_k ∈ B_k;
r ∈ ℝ;
y, z:= r e^(i α) ∈ ℂ;
ln: logarithm over real numbers;
arg_k(z) := α_k := α_0 + 2 π k;
LN(z) := ln(|z|) + i arg_k(z);
y^z := e^(z LN(y))
I also wanted to point out, that i don't see any need for the definitions at timestamp 9:38 - just use the definiiton of complex exponentiation, to see those terms are all equal (which also could be used to calculate all solutions):
(i^(1/n))^m := e^(m LN(i^(1/n)))
:= e^(m LN(e^((1/n) LN(i))))
:= e^(m (1/n) LN(i))
= e^((m/n) LN(i))
=: i^(m/n)
(i^m)^(1/n) := e^((1/n) LN(i^m))
:= e^((1/n) LN(e^(m LN(i))))
:= e^((1/n) m LN(i))
= e^((m/n) LN(i))
=: i^(m/n)
Well to be fair, if you take out 4 you end up with 4(1/4m+1) and then you substitute that bracket part with anything lets say A, you again end up with a multiple of 4 because the final part is in form of 4A.
Ok, I’ve had a think about this and I think that the assumption that you can chain the exponents is flawed. Here’s my reasoning:
1. Exponentiation at its heart is multiplying a number by itself a certain number of times. Yes, we’ve extended that definition to allow for rational and irrational exponents, but essentially that’s the core of it.
2. Multiplying any number (defining a point) in the complex plane by i rotates that point by pi/2 radians counterclockwise.
3. Therefore, raising i to any power should result in i being rotated, resulting in a point on the unit circle.
By taking the pi root first, you end up with a number with both real and imaginary components. When you take that to the 4th power, you are no longer doing a rotation about the origin - there’s rotation, scaling and translation going on.
Doing it the other way doesn’t work either.
I think the answer is -cos(2 + 2n pi) + i sin (2 + 2n pi), as you’re rotating i by 4/pi * pi/2 radians.
Personally, I would say that it depends on how the problem is presented/phrased. For instance, if you simply had "Solve for x" where i^x = 1, then I would argue either answer is a valid solution to the conditions presented. Even a simplistic answer of x=0 or x=4 would be a valid solution for x. However, if the problem was presented as "Solve for all possible values of x" --- or a similar phrasing --- it would REQUIRE the more general solution of x = (4n)/(4m+1) as this returns the set of all possible solutions for x.
The reason this is confusing is because we have not been rigorous in defining what (s1)^(s2) means beforehand for complex s1 and s2. If we go with exp[s2·log(s1)] for nonzero s1, then x^i := = exp[i·log(x)] = 1, and i^x := exp[x·log(i)] = 1.
For the former, i·log(x) = 2·m·π·i, so log(x) = 2·m·π, implying x = exp(2·m·π), which is the result in the video.
However, for the latter, exp[x·log(i)] = 1, this results in exp(x·π·i/2) = 1, implying x·π·i/2 = 2·m·π·i, so x = 4·m.
It would be a mistake to say that i^(4/5) = (i^4)^(1/5), because in reality, fractional exponentiation is almost always defined as [i^(1/5)]^4, making it single valued. And this is not interchangeable with (i^4)^(1/5).
Remember: mathematics is all about the definitions. You COULD define exponentiation in such a way as to make it set-valued, rather than complex-number-valued, but no mathematician would actually do this. x^y should ALWAYS, unless explicitly stated otherwise, understood to be an abbreviation for exp[y·log(x)] for nonzero x, where log(x) = ln(|x|) + i·atan2[Im(x), Re(x)].
In last part both are correct because we know x⁴ must have 4 roots so, i^(4/π) =1, cos (8m+2) + i sin(8m+2) and 2 more roots.
Am I right?
Maybe I'm wrong but there is something I don't understand. In De Moivre's formula, when we raise e^(i pheta) (with pheta the angle) to the power n, such that we have (e^(i pheta))^n, n must be an integer. But at 1:40 you raise both sides to the power of 1/i, which is not an integer. Therefore your demonstration must be wrong. If someone could explain me if I am wrong, I would be grateful.
I think the values of x that give multiple solutions should be allowed. I tend to think of branch cuts as arbitrarily destroying the beautiful Riemann surface structure of complex numbers (and complex exponentials in particular).
In this sense, isn't every real number a solution? For example: i^3 = i^(12/4) = (i^12)^(1/4) = 1^(1/4) = 1 (and 3 other roots). Although obviously i^3 = -i
THANK YOU SO MUCH THIS MADE SO MUCH CLICK IN MY BRAIN
There's a problem I've been banging my head against a while that might interest you... what do all the solutions of "X to the power Y equals Y to the power X" look like if you allow X and Y to be complex?
okay heres a question: why is x not just equal to (4n)/m ? (where n and m are integers)
because
(i)^((4n)/m)
(i^4)^(n/m)
1^(n/m)
1
I take your answer! It's absolutely obvious if you consider roots.
The best maths teacher who catches pokemon also 😆
Wouldn't you necessarily need to take the values of x that satisfy the equation otherwise the multi valued nature of the function interferes with your solution? I think it's maybe a situation where you have to be careful to choose a solution that avoids the branch cuts of the complex function or else you get contradictory answers like this, but I'm not sure either. This is interesting!
Hm maybe you can take some limit of maxima(cos(8x+2), x is integer) or something
Also, maybe you can't really cancel out the pi with the 1/pi in that expression?
8:18 From Where U Have Got i^8=8
Much as quaternions do not, in general, commute under multiplication, it seems complex numbers do not, in general, commute under exponentiation.
You know it can be really simple
i^x=1
i^x=i^4n n€N
so
x=4n
and then you can write it in roster form as (4,8,12,………)
my first thought about i^x = 1 was 4Z (the same as x = 4n as n is in Z) and in fact i^x = (e^i(pi/2)) ^x = e^(i x pi/2) this is equal to 1 if (x pi/2) = 2kpi as k is in Z, so x = 4k.
I think the problem with the second solution is that for example 1^(1/5) equals 1 only in the Real numbers, while in Complex numbers 1^(1/5) gives 5 different solutions, namely e^i(2/5)pi, e^i(4/5)pi, e^i(6/5)pi, e^i(8/5)pi, e^i(10/5)pi = e^i2pi = 1.
p.s.: writing this stuff on my phone while in bed is hell
The problem I have is that i^((4n)/(4m-1)) = 1 as well. Any odd number will be coprime 4, after all.
If n and m are negative then 4n/(4m+1) = 4 * |n| / (4 * |m| - 1) so all these values are covered as well
I'm watching this while in middle school, idk why but I want to understand what you're doing lol
Haha nice I did that too
Same here
thanku for introducing us from Brilliant
x^n for noninteger n is neither surjective nor injective in the complex numbers. As such, we would need to assign a principal value to an expression like i^(4/pi) in order for it to have a well-defined value; what we assign the principal value to can be 1 or cos(2)+isin(2), or something else. There's probably already a convention for this, which is why WolframAlpha does not output, say x=4/5, as a solution to i^x=1 (see edit below).
Knowing this, we may wonder why we never get x=4/pi as a solution to i^x=1. I believe the error lies in (e^((4m+1)/2*pi*i))^x e^((4m+1)/2*pi*i*x). This does not hold for noninteger x when using principal values (e.g. (e^(2*pi*i))^(1/2)=1^(1/2)=1 but e^(2*pi*i*1/2)=e^(pi*i)=-1).
-a very, very inexperienced mathematician
edit: In some sense, I think it's like asking if x=1 is a solution to sqrt(x)=-1. We define the principal value of sqrt(1) to be 1, so in this case, x=1 is an extraneous solution. However, if we defined the principal value of sqrt(1) to be -1, there would be a solution. Similarly, we can define the principal value of a number to be such that i^(4/5)=1 in some cases and in other cases not. WolframAlpha says that i^(4/5)=0.309+0.951i, so I'm guessing some convention has been defined such that i^(4/5) != 1 and 4/5 is not a solution to i^x=1.
But by this logic we can make i^ any rational number = 1
Eg: i^7 = (i^4)^7/4 = 1^7/4 = 1
it doesn't work. We can state z^(q/n)=(z^q)^n only for coprime integer q and n
Amazing! I didn't know that the identity (e^x)^y= e^xy does not hold for complex x and y. The exponentiation page of wikipedia explains it well. Since you are using this identity in both calculations, both are wrong. I think you are eliminating answers in both calculations
If you include all the multiple branch answers for both calculations, some of them become equal to each other
Are n and m in x = 4n / (4m + 1) really independent variables? It'd seem to me, intuitively at least, that they aren't, since I presume the angles on both sides of an "=" sign to be linked. In that case, m = n + 2πk + π/2, which would result in x = n / ( n + 2πk + π/2 + 1/4 ). Ouch...
What about x = 4/3 or 4/7 or really 4 over anything? It's not in the solution but still works!
Sir please solve integration of cos(x)^2
Another examples for the nice and astonishing effects of complex numbers.
From the start m and n are both integers, when u considered m as pi, you've only violated the domain value. So that's why it wasn't true
Okay, we got solution with 4m+1 in the denominator and we checked the solution 4/5 , but what if we just check for example 4/7? i^4/7= (i^4)^1/7 =1, and 4m+1=7 has no solution in integers. Where is a mistake in my reasoning?