Is x^x=0 solvable?
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- čas přidán 13. 09. 2024
- Let's discuss if the power tower equation x^x=0 is solvable or not? Of course, we will discuss the real cases and the complex cases. Then we talk about the limit.
The limit of x^x as x goes to 0+: • Limit of x^x as x goes...
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Finally 0^0 approaches 0:
czcams.com/video/X65LEl7GFOw/video.htmlsi=QRgxgdaL5If2FWnP
BPRP, how solve 1^x = 2?
@@costelnica3988 I think he has done it
0^0 = 1 so i^0 = 1
case 6 works with something akin to the squeeze theorem
tl;dr x^x as x approaches -infinity is in the form re^itheta where r approaches 0
if we rewrite x as re^itheta, as x approaches -infinity thats the same as x equals the limit as r approaches infinity of re^ipi
(re^ipi)^(re^ipi)=(re^ipi)^-r=(r^-r)(e^(i(-rpi)))
r^-r is a positive real number that approaches 0 as r approaches infinity, meanwhile e^(i(-rpi)) in the form e^itheta meaning its just polar form with a radius of r^-r
since its always on the circle on the complex plane with a radius of r^-r, and because the circle gets compressed to a point specifically down to 0, this means limit as x approaches -infinity of x^x should be considered 0
this is not the most rigorous proof but it works for me
@blackpenredpen Hey bprp. Can u do a formula for a tetrated by x plus b exponentiated by x plus c times x plus d equal to 0. Love ur videos. Bcos of u I learnt and mastered calculus at the age of 10.
If x is a complex number other than 0 we can write x=r*e^iθ and then we see that x^x=(re^iθ)^x=r^x e^ixθ
In order for this expression to be 0 we need either r^x=0 or e^ixθ=0. But since r>0 hence r^x != 0. Likewise e^ixθ != for any x. Thus x^x=0 has no solution in the complex numbers
wait what
@@TNTErick ?
probably true, but why can a^x not equal 0 for a!=0? With real numbers it's obvious, but with complex numbers I don't know
@@derda3209 If a != 0 then a^x=e^(x*log(a)) which is never 0 (exponential function is never zero over real numbers and complex numbers). Note that log(a) is defined for none zero complex numbers
I think, the easiest proof that the solution doesn't exist in the complex world is that there aren't any zero divisors in the complex world, so for x^a to be equal to 0 , x=0 , however contradiction.
He doesn't age man
Idk, I really miss his looooong beard
Maybe we age at the same rate as him 😂
He looked the same like years ago I watched him for the first time 😂
Well of course, he's asian
study math and get eternal life
If u want to know about the complex solution, u can find bprp's old video with title "the tetration of (1+i) and the formula (a+bi)^(c+di)"
You say "not my best video", but I really enjoyed it!
My high school teacher once declared 0^0 is undefined because powers can be written as fractions, so 0^0 equals 0/0. I never questioned it until now.
Your teacher's argument doesn't make sense. By the same logic, 0^1 is undefined, because
0^1 =
= 0^(2-1)
= (0^2)/(0^1)
= 0/0
= undefined
but of course we know that 0^1 = 0 is _not_ undefined.
What about 2^2 😂
@@yashrajtripathi4832 What about it?
@@yurenchu definitely not about your comment bro !
@@yashrajtripathi4832 Oops! My bad.
I think what the original commenter's high school teacher meant, is that in general,
a^b = (a^(b+1)) / a
which, in the case of a=2 and b=2 , would lead to
2^2 = (2^(2+1)) / 2 = (2^3)/2 = 8/2
which is correct (both sides equal 4); so nothing wrong there.
With a=0 and b=0 , this formula would lead to
0^0 = (0^(0+1))/0 = (0^1)/0 = 0/0
which on the righthandside is undefined (and therefore, according to his teacher, 0^0 should be undefined).
However, the rule a^b = (a^(b+1)) / a is not valid for a=0 , because for example with a=0 and b=1, it leads to
0^1 = (0^(1+1))/0 = (0^2)/0 = 0/0
which is clearly wrong, because the lefthandside is _not_ undefined; 0^1 equals 0.
Therefore, the high school teacher's argument as to why 0^0 is undefined, is wrong, as it doesn't hold water; he applied a rule that shouldn't be applied in this case.
For me 0^0 = 1/2, take the average value of 0 and 1, im sure everyone will be happy 😊
😂😂😊
Nope.
a^b = c means
multiplying any constant K by the factor a for a number of b times, is equal to multiplying the same constant K by the factor c.
Hence,
0^0 = c means
multiplying any constant K by the factor 0 for a number of 0 times, is equal to multiplying the same constant K by the factor c.
The former side gives K (because multiplying K by 0 for a number of 0 times, gives K), the latter side gives K*c (because multiplying K by the factor c gives K*c), so
K = K*c , valid for any value of K ==>
c = 1
Therefore,
0^0 = c = 1
Q.E.D.
@@yurenchuwhat the sigma
how can 0^0 be 1/2 ??? If we say that "it is defined", and these 2 zeros are identical, then the solution must be 1, because it assymptoticaly goes to 1. But more strictly it is not defined. So in none of case it could be 1/2.
the average value of 0 to 1 how??
Hey there BPRP, been watching for some 6 years now. Good stuff man
Thank you!!
Lots of love to you and to your iconic pens❤@@blackpenredpen
Yo también lo sigo desde hace 6 años, me puse a pensar en ello al leer tu comentario.
Must be the first vid without "of course, otherwise, how could I make this video?" Love the subversion of expectations
What do you mean not your best video? I enjoyed every single second. Good job
I believe as long as x is real, case 6 with the limit works. lim x->-inf (x^x) = lim x->inf ((-x)^(-x)).
(-x)^(-x)=(-1)^(-x) * x^(-x)
The latter factor always approaches zero as x approaches infinity. The former factor is cyclical and the absolute value is always equal to one, so (-x)^(-x) always approaches 0 as x approaches infinity and thus x^x approaches zero as x approaches negative infinity.
I think there’s an extension of this argument if you allow x to approach negative infinity from paths in the complex plane, but I haven’t fully formulated it yet.
This is deceptively simple but it makes sense to me.
There are problems trying to define a^x for non-integer values when a is negative.
E.g. (-1)^(1/3) should equal (-1)^(2/6) since 1/3 = 2/6. The first expression would give the cube root of -1 which is -1 (note that the cube root function *is* defined for negative real values). However, the second expression would give the 6th root of (-1)^2, which equals 1.
See en.wikipedia.org/wiki/Exponentiation#Real_exponents for more information.
@@DutchMathematician you have to simplify exponent fractions first. Values that you can get when it's not simplified don't matter. He showed this in one of his "1=2 proof" videos.
@@zachansen8293
There is no need to simplify fractions first. If f(x) = a^x is a well-defined function, then f(1/3) = f(2/6) *should* hold, no matter how we supply the same argument to the function f. If we give a function two equivalent expression then the outcome should be the same.
@@DutchMathematician Go watch the video, you absolutely do. That's how he proves that 1=2 or 1=-1 or whatever it was. You have to simplify fractions in an exponent (or you can do them in a specific order if it isn't). If you don't you are likely to get incorrect results.
edit: Go search for the video named "Hate to be that guy but I need the extra credit! " - it's the last chapter starting at 3:48. He shows that i=1 because i = i^(4/4) = 1
edit2: I believe if it's not simplified you can do the denominator first, but you cannot do the numerator first. But the easiest thing to remember is to just always simplify fractions that are exponents.
last edit: and of course you should run it through wolfram alpha to see that it doesn't get your "solution"
For the limit x^x you can do:
lim x^x = lim exp(ln(x)*x) = exp(lim ln(x)*x) because exp is a continuous function. Then lim ln(x) * x = lim ln(x) / (1/x) = lim (1/x) / (-1/x^2) = lim 1 / (-1 / x) = lim -x = 0 (using de l'Hopital) and therefore lim exp(ln(x)*x) = lim x^x = 1.
I think you need to assume x>0 ( If you use x Real) , in order to write x^x= e^{xlnx}, as lnx is defined only in (0, oo) for x Real.
I'd argue -inf works without djscreetization because the negative ultimately only changes the complex sign and the magnitude still goes to 0 and 0 with any complex sign is still 0
Why does function y=x^x does not have any values on the left side of the graph? I mean for x=-1, f(-1)=(-1)^(-1), which would be equal to -1, and so on for other arguments?
Counter-example: what do you do when x = -1/2? You're getting imaginary numbers for anything but x in Z* (I think)
It does have values,
For x that is in the form of even/odd, its |x|^x
For x in the form of odd/odd, its -|x|^x
For x in the form of odd/even or irrational, it’s imaginary, if you allow imaginary numbers, forget everything above and just do complex calculations using e^(xlnx). Anyway, overall, the function jumps between positive, negative and imaginary/undefined infinitely many times in any finite region, so it couldn’t be graphed using normal methods, I would graph it is x^x for x = 0 or x>0, ±|x|^x for x < 0
There are problems trying to define a^x for non-integer values when a is negative.
E.g. (-1)^(1/3) should equal (-1)^(2/6) since 1/3 = 2/6. The first expression would give the cube root of -1 which is -1 (note that the cube root function *is* defined for negative real values). However, the second expression would give the 6th root of (-1)^2, which equals 1.
See en.wikipedia.org/wiki/Exponentiation#Real_exponents for more information.
Actually, the negative part of the graph is just unconnected points for the integers. Because -1/2 for example.
Try graphing (|x|)^x or (|x|)^(|x|). The second one is a little weird.
Love how postmodern math is caught in a fantastic sceptical philosophy. He shows using 10 different reality based ways how x^x cannot be zero but proceeds with conclusion that « in a complex world » it « might ». Bake skepticism into the choice of words to really screw with all brains for 200 years and expect good results.
Well, it's easy to prove that I cannot find a solution. It's way harder to prove that there is definitely no solution... Or weirder: that even if a solution exists it is impossible for someone to find it.
@@Kleyguerth the one with the burden of proof is the one making a claim, an assertion. If you say it is possible, what proof do you present? Until you do, all of the refutations of this statement stand and you must shut up.
@@geekonomist Yes, I'm agreeing with you, I just repeated it in simple english lol
2 AM blackpenredpen? Dont mind if i do!
Woah my time when it got released was 9 am. I guess you're somewhere close to United States while I am in Europe
Thumbnail said this video isn't your best, but it's still one of your best to me :)
The reasoning works for when a is the Gaussian integers or Q[i] but when it is all C I have to deal with x^i =0 so yeah..
Thx for pointing out. 😊
@@AndyBaiduc-iloveu thanks 🤗
I understand that mathematician say : "It has no solution, because negative Infinity is not a real or complex number"... But, it is obvious that assymptoticly -inf (in integer) converges to solution. It could be really interesting to make some physical experiment, where the nature say if the -inf is the solution of this equation.... For me, from technicial - engineering aspect, it is solid solution (but not ideal); The only question for me is, if the nature supports the infinity .. :)
The magnitude of x^y for real numbers y, is |x|^y, and if we have the magnitude of a limit approaches 0, then we can safely say that the limit is zero. the limit as x-> -\infty of |x|^x, is also equal to the limit as x->\infty x^{-x} = 0, meaning that the magnitude is 0, therefore the limit as x-> -\infty x^x = 0
I have a question concerning 0⁰ :
We know that :
X¹ = X, and X⁻¹ = 1/X.
We also know that :
X¹ * X⁻¹ = X⁽¹⁻¹⁾ = X⁰
And X¹ * X⁻¹ = X¹ / X¹ = X / X
So, X⁰ = X / X.
And I've learned that there are convention saying that 0⁰ = 1.
So, does that mean that 0 / 0 can be solved, using that logic specifically ?
Or is something wrong here, like the proof that 1 = 2 ?
There is no finite complex solution
If z = r* e^it, r and t real
z^z = r^z * e^itz
So the equation becomes r^z=0 or e^itz=0
Case 1
r^z=0=>|r^z|=0
=>r^r =0, which has no solution as r is a positive real number
Case 2
e^itz=0
=> |e^itz|=0
=>e^-ity=0
Since t is between - pi and pi, y has to be - inf, thus no finite complex solution
This also requires
Instead of fields like Q, R or C, what about using a ring containing zero divisors? Does x^x=0 have solutions there?
Here comments is more difficult than videos. Please get this man a medal 🏅🏅. Salute professors.
Hello sir, I just wanted to thank you for making these videos because they've helped me out a lot.
To me 0^0 is undefined even if we take a seventh grade approach.
5^0 = 5^(m-m) = 5^m / 5^m.
If we take 0 instead of 5 we get:
0^m / 0^m
m can be any number yet we still have undefined fraction of 0/0 which could be be approached if we have some context, like in sinc function: sin(x)/x when x -> 0. But here we don't have that context, so it remains undefined.
P.S. I also was amazed that in English this limit doesn't have a special name. For example in Russian this limit is called "Первый замечательный предел" or "first remarkable (or wonderful) limit".
And the second such limit is (1+1/x)^x when x-> infinity.
With x = a + ib we have z = x^x = (a + ib)^(a+ib) = exp([½ a ln(a²+b²) - b theta] + [½b ln(a²+b²) + a theta] i), where theta = arg(a,b). We want │z│= exp(½ a ln(a²+b²) - b theta) = 0. To me it seems this is the case if a goes to minus infinity independent of b. (Or possibly if b goes to plus infinity ???). So my tentative answer would be x = - infinity + i b with b arbitrary. I checked (-50)^(-50) is very small, i.e. close to zero. (-50)^(-50) = 1.126 · 10^(-85). So it seems plausible.
I have a solution for this question, rewrite 0 as ln(1) and rewrite 1 as e^2πin, and the equation become x^x=2πin, and now you can clearly solve it whit W lambert function and take natral log on both side and become xln(x)=ln(2πn)+ln(i), and using lambert W function it become ln(x)=W(ln(2πn)+e^i(π/2+2πm)) and teh solution of the equazion x^x=0 is x=e^W(ln(2πn)+e^i(π/2+2πm)) with n,m integer
What happens when n and m are 0?
n cannot be equal to 0 because ln(0) is undefined
But that doesn't work because you've changed branch. You need the branch corresponding to n=0 to solve the initial equation.
@@xinpingdonohoe3978 i don't know bro but wolfram alpha said this is allowed
Take two complex numbers z and w. If z,w≠0, then z^x=w can be solved for x. If z=0 and w=0, then x>0. If z=0 and w≠0, x has no solutions. If z≠0 and w=0, x has no solutions.
x^x=0 should have no solutions. Sure, if we take the limit as x→-∞, where -∞=inf R travelled along the negative real axis and not the general complex ∞, then we'd get a complex number of magnitude decreasing to 0, but we want numbers, not extended numbers.
I know as x->0, x^x goes to 1, it takes a dip to x=1 then shoots up past that. I know it only exists at negative integers for real solutions. The only way it could possibly get to 0 is at x=-inf. Right? How does it behave in the complex plane?
x^x = e^(x*(ln(-x)+iπ)) = e^(x*ln(-x) + xiπ)
So as you go into increasingly negative numbers along the real line, you get an ever-shrinking spiral around the x-axis, which... yes, I'm pretty sure converges to 0 as you approach x=-∞.
Well there's no finite solution but if you let x->-inf, you will approach 0
nice exploration of techniques!
I think 💭 the first video I watched from was pi^e vs e^pi.
Back then I think was entering masters in math/stats and today I’m a year away from PhD in statistics.
Always inspiring!
Btw did you ever did a video on Riemann-Stieltjes integration????
I recall it but can’t find it
Intuitively I knew that before your attempts to find a solution. My Calc teacher may have gone over this with me in college and I am just recalling the lesson.
A negative real number to the power of itself gives a number with a definite magnitude, if not a definite angle in the polar form. I think if the magnitude goes to 0 as x goes to negative infinity, that is sufficient to say that the limit is 0.
Exploring Solutions in Modular Arithmetic: If we think about our number system as being modular (clock) arithmetic, it’s easy to show that there is a solution to `x^x=0`. Simply use the `modulus x`, because when `x` is raised to the power of `x`, it will always be congruent to `0 modulo x`. In other words, if we work in `modulo x`, then `x^x≡0`.
No one specified what kind of numbers we were talking about, right? 😄
Please make a video on how to manually calculate complex answers of Lambert w function
As for more (fruitless) ways to solve it, there were a couple I came up with (assuming you're using the extended complex plane)
The first is pretty simple: if z^z = 0, then e^(z ln z) = 0, and thus z ln z = -infinity (we're in the extended complex plane so this is fine)
The only value this works for is z = -infinity, so that's our solution
The second is just substitution; set z = x+1, so (x+1)^(x+1) = 0
Then (x+1)^x * (x+1) = 0, then x * (x+1)^x + 1 * (x+1)^x = 0
So x * (x+1)^x = -(x+1)^x, meaning ln(x) + x ln(x+1) = ln(-1) + x ln(x+1)
Assuming x =/= -1, we can remove x ln(x+1) from both sides, giving us ln(x) = ln(-1), which lets us conclude x = -1, which is a contradiction of our assumptions and thus the equation has no solutions
Nice video, a bit in the style of your videos of a few years ago. Thank you!
What about x^(x-a)=0 when a is a positive integer?
0=x^x=(e^ln(x))^x=e^(xln(x)). No complex power of e can make the result zero. There's no solution.
i can bet everyone that watched this know that
This is correct, and a simple proof that e^z is non-zero for every complex z is the fact that e^-z = 1/e^z, and e^z is defined and finite on the entire complex plane, so if some complex number c were a zero of e^z, then e^-c would be infinite.
I see why e^-c can’t work, but why not e^c?
@@gamemakingkirb667 What I'm saying is that if z^c=0 for any complex number c then z^-c would be 1/0 or infinity, contradicting the fact that e^z is an entire function (no poles.)
I mean , just say there aren't any zero divisors in the complex world, so there aren't any solutions in the complex world.
Also if you define that x^x =0 and make a new ring , would technically be a solution..
actually for no. 6, the 2n limit, it even works for x = n for n integer, but it approacges 0 from up then down, up the down, and so on (much like sin(x)/x as x approaches infinity
would i be correct in thinking that x to the power or x is not a continuous function?
It is continuous on the real interval (0,infinity).
But if you take two rational approximations of -sqrt(2) arbitrarily close to each other where both have odd denominators but one has an even numerator and the other has an odd numerator then you will get a positive and negative answer more than some specific distance apart, and therefore we know that x^x is not continuous on its maximal domain that keeps the domain and range as subsets of the real numbers.
It is continuous along the positive real numbers. It is true that it is not continuous over the whole domain of real numbers.
@@Dreamprism sometimes i think that functions such as this require some kind of analytical continuation, i mean the undefined is not necessarily undefined , it's just that we define it that way :)
@@Dreamprism wo.... i never thought of a discontinuance that way, but yes, i guess it is that way :)
@@Dreamprism wo.... i never thought of a discontinuance that way, but yes, i guess it is that way :)
I want you to discuss the hailstone sequence and Collatz conjecture and your opinion about it
May be the answer of 0⁰ is the friends we made along
It's interesting that if you input a very very small number raised to that same number in your calculator, the result approaches 1.
The 0^x -> 0 limit is only onesided. If you look at both sides the limit doesn't exist
Well, if you consider x>0, then we can use x^x= e^{x*lnx}, and e^y is never 0. Same goes for the Complex Exponential; 1 is the only value e^z doesn't take. I think this does it for X^x=0.
no solutions in the complex, because if z is not 0 then by definition z^z = e^(z*ln z), and e^w is nonzero for any w complex
I wonder if there are any hypercomplex solutions? Or matrix solutions?
the x value needed to receive the lowest possible number with x^x, is e^(-1), and ( e^(-1) ^ (e^(-1)) ) is equal to 0.69...
so x^x is impossible for 0.4, 0.5, 0.6 as well
What I thought of is multiplying both sides by x so then we have x^x•x=0•x =>. x^(x+1)=0 but x^x is also 0 so x^(x+1)=x^x take ln both sides then cancel the lnx because x can't be one and then we have x+1=x witch definitely has no solutions
you cannot multiply both sides by x when one side is 0, if you could, nothing would make sense anymore, take x^2-1=0 and multiply both sides by x and you got x^3-x=0 and that doesnt work
@@nakelekantoo if you don't include zero in the solutions Im pretty sure you can do that
You cannot take the ln at both sides when both sides equal 0 .
For example, solve (x-1)^x = 0 . Clearly x=1 is a solution. But if we raise both sides to the power of 3:
(x-1)^(3x) = 0 = (x-1)^x
(x-1)^(3x) = (x-1)^x
Take ln at both sides:
ln( (x-1)^(3x) ) = ln( (x-1)^x )
3x * ln(x-1) = x * ln(x-1)
2x * ln(x-1) = 0
2x = 0 OR ln(x-1) = 0
x = 0 OR x-1 = 1
x = 0 OR x = 2
which are clearly not valid solutions to the original equation. Instead, these are extraneous solutions that were introduced by raising both sides to the power of 3 (before taking the ln ); but that's not the problem. The real problem is that the solution x=1 that we wanted to find, is now suddenly _gone_ . And this disappearance is caused by the step of taking the ln (when both sides equal 0).
Pow(x,x ) is undefined for all Re(x)0 , will be defined ob Re(x)
I have seen people on Quora get really angry when told that 0^0 is not one.
They should as the definition of the power is 0 multiplications from the original definition of 1. This is a formulation for the inductive proof of every single number. The fact that every other positive power of 0 is 0 is irrelevant to what the zeroth power of zero is.
@@ronaldking1054
0^0 = 1 so i^0 = 1
@@ronaldking1054 0^0 is not a calculateable thing, you just define it, there is no real answer unless everyone agrees on it.
@@mhm6421 When the definition of power by simple integer is set up with 1 as the basis, it makes little sense to claim that 0^0 is not 1. You'd be putting in a branch for no reason. In other words, it is simpler.
@@ronaldking1054That definition was not just randomly defined, it comes from a pattern seen with exponents. For example, 2^4 = 2*2*2*2. 2^3 = 2*2*2. Based on this, we could assume that x^n = (x^(n+1))/x. By this logic, 0^0 = 0^1 / 0. Division by zero is undefined, and so this has no answer. We could define one, but not everyone will agree on the definition.
There are 2 complex numbers that satisfy the equation. Before I state what they are, is it worth publishing?
when you talk about log(0), shouldn't it be negative infinity and not undefined?
I wish all the "0^0 undefined" enthusiasts stopped using summation notation for polynomials and power series since evaluating them at 0 requires 0^0=1 to be a thing.
This makes as much sense as
Lim x->inf of 1/(x*0)
Cant even prove whether or not the limit equals the value at the parameter.
This is probably the best video you have ever done. It’s every way to approach the solution to an equation disguised as a search for the answer to x^x=0. Bravo. Beautiful.
Yeah, we arrived at the same conclusion. Lim -> - infinity but I didnt think of rational numbers, saying x belongs to Z solves it the best we can do. Now I have a question, cam we do lim in Complex numbers?
I enjoyed this video a lot!!
if you put it into desmos, it will show a weird looking graph on the negative infinity side, and if you move it in any way it disappears!!! idk why
I was thinking about higher dimensional number systems that have zero divisors and how you can prove that for example the complex numbers can't have them. Could you prove that no "zero exponents" can exist in these higher dimensional systems?
man do you even age? I see the same enthusiasm but a little turned down! Also please show cat on main channel
Who cares if (-infinity)^infinity has the potential to be a [non-real] complex number?
1 divided by any sequence of complex numbers that surpass arbitrarily large magnitudes will approach 0 regardless of which directional infinity or mix of them is involved.
So, I think we can take each n to be a rational number when we say n approaches negative infinity and we consider what n^n approaches.
Irrational numbers have larger problems, as (-sqrt(2))^sqrt(2), for example, does not have a proper definition within the complex numbers, but we could lazily think of n^n as 1 over some sort of "indeterminate infinity" as our irrational n approaches negative infinity. Maybe this is being a little sloppy, but I'm sure we could have a mathematical system where this works, kind of like we could have a math system where 0/0 gives some object analogous to all potential solutions to 0x=0 and do various math with it.
Perhaps this indeterminate infinity is even applicable to the rational powers if we think of (-1201/12)^(-1201/12), for example, as giving all 12 complex answers instead of focusing on a principal root.
Anyway... my point is that 1 divided by an indeterminate infinity will still be 0 regrdless.
So, I think there is a reasonable formal system in which the actual real-number limit (maybe even complex number limit?) of n^n as n approaches negative infinity is 0.
It also kind of just makes sense that since e^(-infinity) = 0 we could very lazily say this meabs we want x•ln(x) = -infinity and check that x=-infinity itself satisfies this equation.
Why would it satisfy that? Well, ln(-infinity) would lazily be ln(-1) + ln(infinity), which is i•pi + infinity, and the finite part would essentially disappear, leaving just infinity.
Thus, -infinity • ln(-infinity) = -infinity • infinity = -infinity.
0/0 and 0^0 is simply a form of (0,0) the origin or simply the zero vector. It is a point of rotation, point of symmetry, a point of reflection and in some cases a point of inflection that is a change of the sign or change in direction in either the Horizontal-Real (cos(t)) component or within the Vertical-Real & Complex(Imaginary) Component.
Every Integer Fraction Y/X including 0 for either Y or X Exclusive OR but not for both, is the vector from the origin (0,0) to the point (X,Y) that lies on the circumference of a circle with (0,0) being the center point of that circle AND Y/X is also the slope and orientation of that vector which is the tangent of that angle. Y/X = Sin(t)/Cos(t) when the Cosine approaches 0. the Sine Approaches 1 within their range.
The Sine extending into the first quadrant is an Increasing Odd Function. When Sine is either 1 or -1 it is at its maximum or minimum. The Cosine extending into the first quadrant is a Decreasing Even Function. When the Cosine is 0 it is at the Midpoint of its Range It. Their ranges are from [-1, 1] with [0] being the mid-point. If we look at their ranges as being a vector as well, then [-1,1] is the vector extending from the origin (0,0) of the equation y = -x. The ranges of the sine and cosine functions in the form of a vector has a slope of -1 and an angle of rotation by 135 degrees from the +X. The line y = x has a slope of 1 and an angle of 45 degrees. These lines are perpendicular. 135 - 45 = 90. 90 Degrees is perpendicularity, it is orthogonality. They are normal to each other. The reflection point of [-1,1] about the Vertical axes or the Y component is the vector [1,1] and [1,1] lies on the line y = x.
There is no coincidence in the sine and cosine having a range of [-1,1]. Their Domains are the Set of All Reals.
We can treat 0/1 simply as being 0 slope based on sin(0)/cos(0) = 0/1 = tan(0) = 0.
We can also treat 0/1 as the vector (1,0) extending from (0,0).
When we rotate this vector (1,0) by 90 degrees we end up at (0,1)
This gives us the slope 1/0 from sin(90)/cos(90) = 1/0. This has an infinite slope which is tan(90).
We cannot think of 1/0 as being "undefined".
We need to look at this as being the ratio of the rate of change in Verticality with respect to the Horizontal. 0/1 is commonly accepted as 0. We have 0 translation in the Y from the sine being 0 at +/-90 and +/-270 degrees.
Yet, people tend to throw their hands up in a fit of anger with 1/0 "IT"S UNDEFINED". I am telling all of them, the entire world if needs be, they are wrong!
1/0 does exist. It's not undefined.
1/0 is the unit vector (1,0) being rotated to the vector (0,1). However, it's more than just (0,1). Simultaneously it is also (0,i).
Here, we have translation in the vertical, and the horizontal is fixed.
Within the context of Integer Division, Modulus based on the additive identities. 1+0 = 1, and 1-0 = 1. Here with 1 being the left-hand-side operand, 0 does not change 1. As for Integer Division, Modulus Arithmetic division by 0 is infinite slope. And again, they argue that vertical slope is undefined, or that tan(90) or 180*N where N is an integer increments of are also undefined because of their misconception Y/0. They fail to realize that Y/0 are the vectors (0,Y), (0, i*Y), or (0, sin(t)), (0, i*sin(t)).
The cosine exists in the Real. The Sine exists in both the Real and the Complex. The two functions are perpendicular functions. They are 90-degree rotations, and they are 90 degree horizontal translations of each other, They create a right angle from each other. They are perpendicular functions. The tangent is the ratio proportion between them. Both the sine and cosine are continuous transcendental periodic functions. Even when they are in a denominator and their range becomes 0, they still persist. So, when we have a 90 degree angle or rotation, division by 0 we have the form sin(90)/cos(90) = 1/0 Yet since the sine is also within the complex domain, we can also have i/0 which is the imaginary unit vector (0,i).
Multiplying a value by i or the sqrt(-1) is the same as rotating by 90 degrees. Here's a simple list of i^n translations.
A * i^0 = A
A * i^1 = i*A
A * i^2 = -A
A * i^3 = -i*A
A * i^4 = -A
And every for other integer multiple exponents of i repeats this pattern.
Here's the degree or radian association of them:
i^0 = 0 degrees, 0 radians
i^1 = 90 degrees, -270 degrees, PI/2 radians or -3*PI/2 radians
i^2 = +/- 180 degrees or +/-PI radians
i^3 = 270 degrees, -90 degrees, 3*PI/2 radians or -PI/2 radians
i^4 = +/-360 Degrees, +/- 2PI Radians {full circle}
And i^0 and i^4 are Congruent to Each other.
So, when the cosine of the angle becomes 0 even when it's in the denominator doesn't mean that the cosine or the sine stops functioning. It's a phase shift between the sine and cosine. And the nature of the Sine appearing in both the Real and Imaginary components simultaneously where the Cosine only appears in the Real component is akin to the nature or duality of the wave-particle duality we see in nature based on probability densities within quantum mechanics. Think of this as super position.
We cannot think in terms of limits here because limits impose discrete finite limitations. We have to think in terms that there are no limits. We have to think in terms of this type of arithmetic and the association of their properties and relative relationships are Analog, not Discrete. Numbers themselves that are not 0 or the zero vector are circular.
1 * i^0 = 1
1 * i^1 = i
1 * i^2 = -1
1 * i^3 = -i
1 * i^4 = 1
Without people realizing it, division by 0 when the Numerator is Not 0 so for all A/0 where A does not equal 0. Will have a Quotient of +/- Infinity based on the Sign of A and will have a constant remainder that is the absolute value of A. And that it is also equivalent to a 90 degree rotation (translation) which is the same as multiplying by i or by sqrt(-1).
This infinity in the quotient is the tangent function extending out into infinity at infinity which for us within "discrete" is unreachable. Yet within Analog both the Sine and Cosine Have Always Been There. They are Infinite Functions. This is why we have a hard time understanding how to get a perfect square that results in -1.
1/0 is the same as rotating the vector (1,0) by 90 to the point(s) (0,1) and (0,i). 1*i = i = sqrt(-1).
The Sqrt(-1) is also division by 0. It has Infinite Complexity seen within tan(90). Verticality that is perpendicular, orthogonal, normal to the Horizontal.
As for 0/0 and 0^0 they are the Indeterminate forms. Because they are the origin (0,0). A point of rotation, reflection, symmetry and even inflection. For any unit vector extending from (0,0) for any and all rotations about that origin, there will never exist a point on the circumference of the circle where the slope of that vector Y/X has its point on that circle at (X,Y) where Y/X = sin(t)/cos(t) = tan(t) where sin(t) and cos(t) both evaluate to 0. This will never happen when rotating the vector about the origin. This is because the zero vector and any arbitrary unit vector are perpendicular, orthogonal, and normal to each other yet by degree of separation of translation becoming two endpoints they generate a line segment that has an angle of 180 degree or PI radians. So yes (1,0) with respect to (0,0) has perpendicularity. Yet the vector V = P0(0,0) + P1(1,0) is also 180 degrees. 90 * 2 = 180. There are 2 points. This duality of perpendicularity and 180 degrees defines Linearity. This Linearity is also equivalent to the summation of the three interior angles of any 2D Euclidean planar triangle. This is why the Pythagorean Theorem, and the Equation of the Circle are of the Same Form. And it is from these that we can also generate the Distance and Midpoint Formulas. Every value can be expresses by a combination of any number of zero and unit vectors. 3/5? ((1,0) + (1,0) + (1,0)) / ((1,0) + (1,0) + (1,0) + (1,0) + (1,0)) == (5,3) And with this we can see the following:
0/1 = (0,0) / (1,0) = (1,0)
1/0 = (1,0) / (0,0) = (0,1), (0,i)
I'll have to try and find a workaround.
x^x should have a unique analytic continuation to C, no? what is it?
what if, we instead use (2n+1)? lim (2n+1) {[2n+1]^(2n+1} also approaches 0 (from the negative side), just as 2n approaches from the positive side
Can it be proven whether or not y=x^x 8s continuous except at 0?
Been yelling "MINUS INFINITY" two minutes in. Glad I got close enough, too bad i hate complex numbers
Poser 0^0 = 1 par convention. Il n'y a pas de raison de ne pas prolonger par "continuité" en 0 la fonction f : x ---> 0^x
Ainsi, 0^0 = 1^0 = 2^0 = .... = 1
On a donc posé :
f(0) = lim f(x)
(x ---> 0)
Hey what if we make a new complex number especially for 0⁰? Same for the square root of negative 1 it also inteecept so why not make a new imaginary number for it?
This equaction has no solution bc I x would have to be similar to ln(0) which itself has no solution
Why can't you just do this?
1) X^X = 0
2) (X^X)^(1/x) = 0^(1/X) ... that is, take the x-th root of each side of the equation. That leads to...
3) X = 0^(1/X) , or X = 0.
Let u = xln(x). Then you have e^u = 0, and this has no solution, even in complex world. However any branch of the log makes xln(x) onto (I think) so you wouldn't be able to find x in any case. Thus there is no solution
For the limit case : you don't have to put x =2n as this also works in the complex world, the modulus of 1/(-inf)^inf would be 0 no matter the branch
It's 5am why am I watching this lol.
4am here. Other than I like math, I also don't know why I'm watching this.
I don't see why we can't just have x = -infinity .
x^x =
= (-inf)^(-inf)
= 1/[(-inf)^inf]
... let exponent inf = K + m , where K = "integer infinity" and real m satisfies 0
Wait why is 2^0=1? Shouldn't it be 2? Because that's basically saying two times itself zero times is one, when it is logical to say that two times itself zero times is still two, since you haven't done anything to the base.
Clever!
I can’t stop saying this regardless that I love math…0 is not a number. That is why we have so many conflicts and undefined answers…
It is though, you have the mentality of someone from the year 4
@@the_linguist_ll It is accepted by most...not all. And I of course use it as a number. But zero means nothing.... no 0.000000000000000000000000000...1. It means nothing. And again, the fact that zero causes so many conflicts in Math, Physics is enough to alarm you that we probably have to apply some different law about zero that would satisfy even the 1/0 and not stick forever to the undefined. In order to argue about something, we have to have a definition. So, the mentality of the 4 year old is mostly to those who do not like to hear truths that abolish their beliefs... Do you get it? Someone came up with i=√-1 and even though it doesn't exist, it works. I just wonder if anyone will ever come up and define 1/0 and flip upside down math!!!
@@websparrow I said from the year four, not four years old, but anyways
Imaginary numbers had a clear and natural approach to their discovery because they had a gap to fill that was perfectly explained by extending the number line a dimension (and more)
There’s no such gap for division by zero. And I get the whole “you won’t know until it happens” argument you’ll have, but come on, that can be applied to anything. There needs to be an argument more than “but maybe”
Check out the video series “Imaginary numbers are real”, it gives a good sense of the natural discovery of imaginary numbers, and the gap that existed before them. I’d like you to show me modern mathematicians with the perspective that zero is not a number.
And finally, it’s alright for a number to have different properties, and that isn’t unique to zero. Having to set exclusions for certain numbers happens all the time, it’s just how numbers are. We had to change the definition of prime to exclude 1, just to eliminate the need for a clause in equations to account for it (which I personally disagree with, in my mind just dealing with extra clauses is more natural than excluding 1 from the set of primes, but I recognize I’m in the minority for that)
And anything that relies on the oddness of primes needs to exclude 2 from consideration. Every number is part of at least one set of numbers which will have properties that need to be excluded from something for it to work, that isn’t unique to zero.
10s ago is CRAZYYYY~~~
Maybe one or more solutions exist, but we cannot prove them because of Gödels theorem.
It is not. The only way to have 0 via an exponent is having 0 to the power of x where x is different than 0. Any other number, as little as they are, will not be equal to 0 (and 0 to the power of 0 is 1, yes.)
?
by definition, no.
any number to the power of 0 is 1.
Any higher than that, it'll increase. Any lower than that it'll also 1/(increase).
So, this shouldn't be possible.
6 hours ago is majestic
(a+bi)^(a+bi) can be rewritten as (a+bi)^a*(a+bi)^bi now we know that either (a+bi)^a=0 or (a+bi)^bi=0 then (a+bi)^bi can be expressed as e^(bi*Log(a+bi)) where Log is the complex log (if f(z)=e^z then f^-1(z)=Log(z)) let's bring the formula for the complex Log: Log(z)=ln(|z|)+arg(z)i now now let's simplify e^(bi*ln(|a+bi|)+arg(a+bi)i) simplify a little bit more (e^bi)^(ln(|a+bi|)+arg(a+bi)i) now simplify
e^(bi*ln(|a+bi|))*e^(bi*arg(a+bi)i) then just solve the equation e^(bi*ln(|a+bi|))*e^(bi*arg(a+bi)i) = 0 we get (cos(b)+sin(b)i)^ln(|a+bi|)*e(-b*arg(a+bi)) now (cos(b)+sin(b)i)^ln(|a+bi|) can't equal 0 because |cos(b)+sin(b)i|=1 for any b (= R and and for this to equal 0 |cos(b)+sin(b)i| must equal 0 now the second part e^(-b*arg(a+bi)) |e| =e so this is not equal to 0 for the same resons now let's come back to our original problem: either (a+bi)^a=0 or (a+bi)^bi=0 but we proven that (a+bi)^bi/=0 so (a+bi)^a=0 then |a+bi| = 0 then a+bi=0 then a=0 and b=0 then we get 0^0 which is either 1, undefined or has infinitly many values but for all cases 0^0/=0 so (a+bi)^(a+bi) has no solution
Extra: let's see if equation z^z has any solutions in set denoted as C(epsilon, phi) (I will type j for epsilon and k for phi) where j^2=0 where j /=0 j/j=1 , k=1/j and k^2 is undefined firstly let's define exponents: e^z=sum from: n=0 to: infinity z^n/n! then e^jz=sum from: n=0 to: infinity (jx)^n/n! = 1+jx+0+0+0+... =1+jx now the ln: ln(a+bj)=c+dj then e^(c+dj)=a+bj then simplify (e^c+(e^c)dj) = a+bj then form two equations e^c=a and e^c(dj) = bj solve the first one c = ln(a) substitute a(dj) = bj now ad=b now d = b/a so ln(a+bj)=ln(a)+(b/a)j
Now I don't want to to waste another 30 minutes of my life solving this equation so if someone sees that please solve this problem
Definitely no solution.
If x≠0, then using the definition of complex exponentiation (valid for x≠0), xˣ=exp(x ln x), and exp(z) is never zero, as exp(z)exp(-z)=exp(0)=1.
If x≠0, then for real exponentiation (not always the same as complex exponentiation when the base is negative) we have
|xˣ|=|x|ˣ=exp(xln|x|), always >0, so never zero.
Finally, 0⁰=1, not zero.
So xˣ=0 has no solution for real or complex x.
Can we start solving it using laplace transform?? I believe that I have solved something similar in my memory.😮😮may be I am wrong…..🤔🤔
One day I will find solutions to all of these undefined questions
Also if you have x=z=a+bi and take a limit of b->±∞, where a≠0, then z^z will also aproach 0... but otherwise you don't have any more solutions for the equation i think.
Board k samne trasnlarion aa rha h, baord saaf nhi dikh rha
wouldnt any complex number z with r
No, because
lim_{r--> 0} z^z = 1
try the quarternions?
Hi there! 😁
Como é possível que seus vídeos antigos sejam idênticos aos recentes?
if 1*log(1)=0
but we know a*log (b) = b^a
then 1*log(1) = 1^1= 1
how this is possible please explain
You wrote " a*log(b) = b^a ", which is not correct. The correct identity is
a*log(b) = log(b^a)
which is valid for positive values of b.
What about using dual numbers?
Use hypercomplex (dual) numbers?
Joder, bro me doy cuenta que te sigo desde hace 6 años 😮
Is this a possible solution?
x^x = 0
x√x^x = x√0
x = 0
x^x = 0 at x = nevative infinity