Integral of x^i
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- čas přidán 21. 08. 2024
- Learn how the complex integral of x^i can help us find the integral of sin(ln(x)) and cos(ln(x)) without using integration by parts. Complexifying the integral is an integration technique that is usually not taught in calculus 2.
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"Killing two birds with one stone" is a strangely appropriate metaphor, seeing as "calculus" is the Latin word for "stone". :-)
@VeryEvilPettingZoo 😂
you are wrong, it is killing two eggs with one nail
@Bob Trenwith calculus means simply small stone. Then by extension it was used for those pebbles used for counting.
@Bob Trenwith From Charlton T. Lewis, Charles Short, A Latin Dictionary:
calcŭlus , i, m. dim. 2. calx; cf. Paul. ex Fest. p. 46.
I. In gen., a small stone, a pebble.
In modern italian “calcoli” means both calculations and kidney stones
6:55 "I don't like to be on the bottom, I shall be on the top "... your choice man!
Olivier L. Applin isn't it?
Smash Boy
He meant something else
It's his choice (most indians will get this)
@@morjithmattapalli9531 czcams.com/video/ftPn8nZIt9A/video.html
@@morjithmattapalli9531 lol
I feel that teaching with colors as you do aids a lot in the grasping of these problems. When reading textbooks, it isn't always clear how they get from one step to the next. Your demonstrations make it much clearer.
Luke van Eyk thank you!!! And Oreo made it cuter too!!!
Y E S
Yankee
Echo
Sierra
Y E S
The colors help *so much* and I love BPRP for it.
I started doing this in class after I'd been watching his videos. It really adds to the clarity! My students seem to appreciate it at least :-)
Joren Heit you just have to master the single-handed marker swap technique!
I just found out why it is called black pen red pen
This channel is awesome *isn't it?*
Suave Atore thanks!
Yes it is
i(sn't + t)
Ofcourse.
@my pp is hot it(derivative of ns + 1)
This is pure joy.
kujmous
I love your channel. I'm currently in high school, working on my second year of calculus, and I love challenging myself with your higher level content, and your explanations are easy to follow and in depth enough, but not too in depth to the point of being boring. Keep up the great videos.
I recommend you to see the graph of the integral of sin(ln(x)). It is really cool, the graph repeats itself in ever larger scales as x grows to infinite.
WHoZ Wow, it's looks like a kind of fractal
Tecnically it's not a fractal. The right name to this property is "self-similarity".
In that the specific function the property works as following: f(xe^(2pi)) = e^(2pi)*f(x)
Kiritsu The first time I saw this function I also thought it was "kind of fractal" hahahaha
math.stackexchange.com/questions/2407743/is-the-function-fracx2-cdot-sin-lnx-cos-lnx-a-fractal
10:35 "if you are obsessed with +c" i laughed so hard
Yeah lmao nobody likes c
One last thing though - technically, you would need to prove the power rule in fact is valid for complex numbers to make this airtight, though I suppose you're going on that that's already been done: nonetheless, it wouldn't necessarily be justifiable at, say, the usual Calc II-like level where this seems to target, to do that.
But that is easily remedied: you can first assume _as a heuristic or hypothesis_ that the power rule will work for n = i (seems reasonable, no?) and then after carrying through with it, go back and _differentiate your final answer_ to see that you do indeed get cos(ln(x)) + i sin(ln(x)), thereby proving that not only does the rule apply for that imaginary power but also that you have indeed integrated the two integrals that you wanted to "with one stone". This is not circular because you did not reference the conclusion as justification; rather simply only took it as a hypothesis to then later be verified.
I think you could prove it using Cauchy's residue theorem, but that's college complex analysis stuff... very beautiful though, because it makes a lot of tough integrals melt away magically.
Just do it with implicit differentiation. Let y=x^n, where n is element of all complex numbers.
y=x^n |take the ln
ln(y)=ln(x^n) = n ln(x) |differentiate with respect to x
(1÷y)×y'= n÷x |multiply both sides by y
y' = ny÷x | y= x^n
y'=nx^n÷x =n×x^n-1
That's how you derive the power rule, for all integers, and if you integrate you just do the reverse
@@CatchyCauchy I think it's bit heavier than that. Problem is that all the standard calculus results (real) relate to gradients and area .. and that is how the fundamental theorem of calculus (which is where the n+1 rule emerges) comes from. Now as soon as you make the function non-real all those concepts disappear so I guess everything needs to be re-derived? Fortunately most of the standard results still hold in complex analysis ... and you get a whole lot of (beautiful) extra embellishments :)
and some people say imaginary numbers aren't useful
AndDiracisHisProphet Actually wireless transmition can't be explained without system of complex numbers
Yes, of course it could. It is just more convenient to do it this way.
It could (with an asterisk), but not as concisely and as accurately as you can do with complex numbers.
When dealing with any electronics that uses an AC signal (mains power, audio electronics, RF electronics, etc.) you're no longer dealing with simple real quantities like resistance, capacitance, and inductance. You're now also dealing with complex quantities like capacitive reactance and inductive reactance, which are the imaginary parts that combine with the real part resistance to form the complex value of inductance. In the face of an AC signal, capacitors and inductors exhibit a form of resistance to the time-variant signal. But it's not real resistance. It's imaginary resistance--i.e. reactance. They also induce phase shifts to the AC signal as well.
Electrical engineering very, very quickly dives deeep down the calculus rabbit hole when you venture into the AC world. In the DC world, 90% of the math you do boils down to basic arithmetic. And there's still plenty of that in the AC world, too! But there's also calculus in the AC world. Lots of it.
And it's all math that is presently above my head, but I'm still doing what I can to suss it out in my own free time in lieu of being able to afford to go to university. :) (Heck, I'll probably have a better, more intuitive understanding of it all than what any university course could probably teach me. Leg up for when I go back to school!)
How would you describe the vectors purely numerically? You could probably use the linear algebra approach of 2x1 matrices, but then you have to remember all the additional mathematics that comes with linear algebra as well. The alternative approach is to describe those vectors numerically using complex numbers. Or if you need a notation that describes the vector's length and angle, you'll need z e^(i theta), and you'll need to know how to convert between the two. Remembering that z e^(i theta) = z(cos(theta) + i sin(theta)) is a lot easier than having to remember a whole bunch of extra stuff regarding linear algebra and matrix algebra. :P
Bob Trenwith It seems possible, but very inconvenient
The title of this video should not be "integral of x^i" but "how to solve integral of sin(ln x) and of cos (ln x) at the same time and much easier than using substitution and integral by parts"
I love how happy you got after seeing this patern. I've recently stopped focusing on mathematics, because of my software engineer carrier, but that video reminded me the way I felt after solving things. Thank you for the video.
at the end, you missed the "i" at the final result
this is the most wholesome channel on youtube
i love how this dude sits and works out crazy-ass maths problems on his own at home, and when he comes up w something great, he shares it w everyone on CZcams. his excitement is infectious, and i love watching his channel
I've become addicted to this guy's videos!
Just now seeing this. Very, very clever! Well done, sir.
Thank you! : )
love how he switches so seamlessly between marker colors
One of my favorite things about this channel is that you respond to EVERY SINGLE COMMENT. You really do make CZcams a greater place not only because of your mathematical genius, but because you are just a wonderful person
Hi Strafe, thanks for your awesome comment! I do try to respond to as many comments as possible. I also feel sorry that when I can't respond to everyone since I have been getting LOTS of comments per day. I love them all, but I just can't get to every single one of them. Thank you again. Reading comments always make my days brighter!
Can we take a second to appreciate the pen lid removal at 6:16
ToastyBread great catch!
I'm more excited about your excitement than the solution itself. :D But I understand the genuise way (I could never figure out by myself). Really good job! Booth thumbs up.
Daniel Gschösser thanks!!!! I am very glad to hear!!
I've to say thanks. Your work for society in science is unpayable. Cheers man!
Another mind blowing video. I teach calculus myself right now and I did not anticipate that equating of coefficients using real and imaginary parts in the end.
When my students are ready (They are still working on limits) I will certainly introduce them to your channel.
Andrew Stallard thank you!! I should be doing a better job in terms of organizing the topics that I am doing. Right now, I am just doing videos on whatever it's in my mind. Lol
Andrew Stallard oh, I do have some videos on limit already, plus my site www.blackpenredpen.com
+blackpenredpen Perhaps in time, you can make playlists by topic or theme? :)
BlackFiresong I have the playlists (for HW sol for my students) on my calc resource site. But for my YT channel now... it's very unorganized haha. I guess it's just a math person's habit...
math always seems to amaze me
(never ceases to )*
"Tonight I figured this out" omg dude, you nights must be pretty intense.
Ay he's rocking a supreme jacket 🔥💯👌
Isn't it
My dawg fresh af
Your enthusiasm is infectious. I can feel the joy you felt when you worked this out.
(x^alpha)*(cos(beta*lnx)), (x^alpha)*(sin(beta*lnx)) are solutions of cauchy-euler second order differential equations when the "characteristic equation"(per se, its not called that) has complex conjugate roots(alpha+/-i*beta). There you encounter the need to solve x^complex number.-------- To anyone wondering where u have the need for solving x^ complex number. (1 instance).
This is without doubt one of the coolest methods I have ever seen.
My man really out here teaching calculus in supreme, respect
1. Assume that i is a constants
2. calculate the integral
3. Solution: (x^i-1 )/(i-1) + C
No matter what method you use, maths stays consistent once again. Never fails to blow my mind.
My man's is out here wearing a supreme jacket. Damn homie I can't believe you been flexing on us like that since you got big
You can tell that red pen and black pen are friends. They finish eachother's equations.
Exellent, so simple, isn´t it. A few years ago I watched a video where a group of men honored a salsa singer. They shouted at the end : "The world needs salsa". Let me tell you man, you really love maths. I´d love my students and colleagues to be like you. What we could shout as one has to be: "The world needs maths". It really needs!! Above all those countries like mine the ones that nowadays continue to worshiping politicians. Go ahead guy!!! You have too much to give yet. Congratulations.
Your videos gave me the opportiunity to show my class in Israel the beauty of Mathematics.Thank you!
josh writeman what a wonderful comment! Thank you for making my day (Friday! Woohooo!)
Your enthusiasm is infectious, thanks!
@486 s (8:08) - CHEERS for actually doing it the way that makes clear the underlying math rules being employed and not just that hackneyed old "FOIL it out" stuff.
Wohoo,u have 41^2 videos in total,btw great video...
wow, that is a LOT of videos.
At first I read this as 2^41 and I was like WTF?!
He seemed so excited at the end this shit is great.
I've come up with something similar in the past (basically the same truck but for different integrals) and I just want to warn that Re(z1)*Re(z2) /= Re(z1*z2) it's pretty easy to see why, but even easier to assume it's true without even realizing it. So this method always works for integrals like this, but doesn't for products unless you get really tricky. If you want to see what I mean, try to use this a trick to solve the two following integrals.
e^x*sin(x)
e^x*cos(x)*sin(x)
I already knew Euler's formula from differential equation use, but I never would have thought of using it like that before now...
This is the solution to the drag equation, chemical reactions, logistic growth and many other limiting equations when constants A and B are both equal to i. ln(f(theta)) is the derivative of tan(theta) which when inverted becomes theta = atan(x) which has (1+ x^2)^-1 as a derivative. All limiting value problems can be factored as [(x-a)(x-b)]^-1 * dx = kt * dt.
I love your enthusiasm
So actually, if you don't expand the integrand to its trigonometric form, you get a complex constant term. But if you do, then you get two constants, one for the real and one for the imaginary part, amazing. By the way, how do you feel about the word "imaginary" for describing complex numbers? I personally don't like using it. It leads to confusion and then students feel they don't really exist nor represent something. Complex numbers are as real as real numbers!!! They're just in another axis!!
Kiritsu well you could call them squarerootofminusone numbers, or complex numbers, i guess it's just a name. Technically complex numbers are ordered pair of reals where the product is defined in a particular way, (a,b)*(c,d) = (ac-bd,ad+bc);
So it's better to call them imaginary, to have that feel they might be family
I agree. I do not like calling them imaginary numbers, either. Particularly when you start getting into the higher levels of physics and engineering, complex numbers become a VERY real thing that describes VERY real phenomena shockingly accurately. Numbers exist in more than a one dimensional number line! When you get into field analysis, you have to deal with *three* dimensional numbers! But they're still very real numbers! Just complicated. And way above my head right now. :)
Take for instance electrical impedance. It has a real part: resistance. And an "imaginary" part: reactance. Reactance is the "resistance" that capacitors and inductors exhibit in the face of an AC signal. In the DC world, all you have is resistance. Or rather, an electrical impedance of R+0j ohms. In the AC world? Pop a capacitor or an inductor in series with that resistor and suddenly that 0j is no longer 0j. It's something else entirely. PLUS you get a phase shift on top of the AC signal that depends upon the angle of the impedance vector, to boot.
Calyo Delphi We have learned that in psychics about impendance, but still most of students in my class think that imaginary numbers aren't real.
Aaaagh, that's gotta be frustrating... :(
Luka Popovic To be fair, Imaginary numbers aren’t Real.
SO GOOD !!!
Muhammad Wibowo wah... ketemu orang indo nih...
Muhammad Agil Ghifari mantaap bang, jarang2 ada orang indo buka ginian hehehe
You know shit gets real, if blackpenredpen takes out a blue pen
That's pretty delightful!
Your university allows you to use a lecture theatre, surely they would allow you to use a clip on microphone. Awesome channel btw.
Here, here!
Absolutely brilliant! Your cheerful attitude plus the way you make a mind-blowing problem sound so simple is what makes me want to watch MORE of your videos! So good!
I am Spanish. I love this channel. THANKS and more videos!!
I'm so curious now about what other nuggets of curious investigation are yet to come. Can't wait for the next video!
Every time I do questions with imaginary numbers, I am feeling like the usage of imaginary number and Euler’s formula is similar to the “hyperspace” concept in Star Wars. Hyperspace is another universe that have their own rules which don’t make sense in the original world, but by entering it, you can reach the destination much faster.
found this channel later but worth it
Discovery in mathematics is nothing short of miraculous
Your unbridled eagerness to share your epiphany far and wide here on youtube speaks to a passion for teaching that I'm sure would make my late mom proud.
Also I just get so squeeful watching geeky people like me get giddy and show off the geeky things they do for fun.
My first guess would be (x^(i + 1))/(i + 1), applying the formula for the integral of a power of a variable.
I keep remembering the graduate student named Kwak from South Korea during the eighties he talked just like you, and we liked him and went to bar with him and laughed a lot. I didn't find anything wrong with your explanation an proof, you just don't a shorter ways to solve these problems.
Why is this so fun and satisfying?
The problem of closed symbolic integration was solved in 1968. en.wikipedia.org/wiki/Risch_algorithm
If a closed solution exists then the algorithm will find it, and if it doesn't then the algorithm will produce a solution that is the sum of a closed form (that is maximal in some sense) and an infinite series.
So why can't I stop watching these videos?
In real domain we would integrate it by parts du=dx, v=cos(ln(x)) , u=x , dv=-1/xsin(lnx)dx
To calculate both of these integrals in real domain two integrations by parts will be enough
because we do not need to calculate them separately
We will get system of equation after integrating by parts twice and we will be able to calculate both integrals from that system of equations
This man is a treasure
6:55 Phrasing.
Your videos are ALWAYS GREAT!! I like the way you think to solve the problems.
thank you!!!!!!!!!!!!
Can you do a series of videos for integral, derivative, etc, of the hyperbolic functions and their relationship with i? Or, if possible, do a video with quarternions?
never smashed the mfking like button so fast
**Series of A(i): A(1)=√2, A(2)=√(2-√2), A(3)=√(2-√(2+√2)), A(4)=√(2-√(2+√(2+√2))), ... determine the type of convergence. Hi from Belarus :))
I think you are a genius. The way of your teaching is fabulous
4:45 no meed for a substitution, just do 2 integration by parts and then transfer a term from the RHS to the LHS and divide by two
I clicked the video just to see the domain of integration, because i wanted to try to solve it trought complex integration method. I didn't expect you would calculate the primitive. Nice!
You are artist of maths! what a talent!
Wish my profs had this much enthusiasm when teaching.
I like Your Teaching Technique
I love your passion
He should have factored out the negative in at the end and rearranged the sine and cosine to be in the same order as the real term, this would make it easier to remember and it would show more symmetry
"so good!" is now officially a blackpenredpen meme.
Now my phobia of maths is getting converted into love for it... All thanks to you.
His définition of x^i as cos(ln(x))+i*sin(ln(x)) only works if x is strictly positive, as x is a real number
You can define complex logarithms as a multivalued function and then choose the principal branch though
I integrated by Parts and got that but I’m kind of sad now because the way you did it is really cool and not boring IBP. Haha great video
SO GOOD! This vid literally fulfill my nerdism.
Tanks for the short Oreo's video.
He should have his own channel.
Complex numbers are powerful. That magic trick was very cool.
Isnt it mind bogling that complex powered by a compled is a real no.
The fact that that was your own making earned a sub
The solution and the set up is amazing, but in this problem, I think the most important thing is to justify that you can apply the power rule as you did to x^i. That's not trivial to the point where we can just skip over it.
I loved this so much, your enthusiasm and the maths that I was able to follow, thank you.
Those crazy functions!
I never took complex analysis (looking into the Reimann Hypothesis), and tried to do this myself another way, and am sure I messed up with all the substitutions when I wound up with something that looked more like a derivative. It's nice come back to this video and remind myself why integrals and derivatives of complex powers behave so nicely
Once, I wanted to 'spin' a telescope mirror with epoxy resin.
But I needed to calculate the focal length of the mirror based upon something I could measure, like the change of 'height' of the edge of the liquid as it rose up the container due to centripital force. I could place a marker on the edge of the circular container and spin the liquid until the edge reached the marker.. But how high above the flat resting level of the liquid to place the marker?
After an hour or two and a little bit of differentiation by rule, I solve the problem. I think I was even happier than this guy is here :)
One of the greatest achievements of my mathematical life (I'm an engineer really, not a mathematician).
Of course subsequently I lost the piece of paper on which I did the work, and have never been able to recreate it :)
that is a WILD function !
BPRP - you are not only a great teacher, but you also have a very beautiful heart. From your response to the hate that was spewed in the comments section, I could think of only two other people who would have acted in the same way and they are both very great people; Mahatma Gandhi and Martin Luther King Jr. I not only love your channel and the math you teach here, I respect you as a great person too. And that respect has gone up a couple of notches.
Thank you so much mohan153doshi! I am for sure nowhere as great as Gandhi and MLK. I can only try my best. Once again, thank you very much.
YAMW. Continue to teach us the beauty of math. I just love all your videos.
:) :) :) He's a little Meehhmmhhrr!!!! :) :) :)
@blackpenredpen : aww meehhmmhhrr...
Though I would say not quite to sell short - but again, that's modesty and I commend that. So I am not asking you to arrogate anything but I'd offer my opinion and that's that I'd say how "great" someone is when it comes to morals or ethical virtue shouldn't be measured quite so much by absolute _quantity_ of achievements in how that _SOCIETY_ says so, because we as humans have _differing capacities_ and the ultimate judge - God - and no I'm not asking you to believe in one (and moreover I certainly do not believe in _dogmatic religion_ 's ideas about God), merely sharing my perspective; if you want God can be considered as an imaginary ideal being or ideal point than a real one that stands for perfect moral justice and thus does not have to literally exist to be useful just as the real number line need not literally exist in the physical universe but is useful and is the representation of an ideal geometric/arithmetical continuum, an idea you should surely understand - would only judge us relative to our capacity, especially when Sie has all by Hir decree said that our capacities and allotments in life are not to be equal. With that, it would be grotesquely unfair to judge the way that our SOCIETY does for that would be to consign to condemnation people simply for something that was beyond their ability to control.
Maybe then the comparison is fundamentally not fair. Not because one can be ranked but because they can't. We cannot and should not be expected of beyond our capacity; and what matters most is what we do with that which we are given. Humankind is like the great army and like any such army it must have those in all roles and neither can be considered to excel the other for were an army to be composed solely of that role considered as the one awarded the status of greatest excellence it would fall into utter depravity and ruin and thus be entirely bereft of the ability to function.
That's a great video, it helps me a lot with my calculus teaching
Simple derivation if you use 1-I=sqrt(2)*e^(-i•pi/4)
really youre a great teacher ..............
Great talk. Most books on complex functions discuss these tricks.
Yayy another blackpenredpen complex number video
Yay!!
4:52 A N O T H E R O N E
ProdZ hahahahaha!!
The answers are by no mean complete. x belongs to R; x^i. But x belongs to R+ when we use log(x).
I want you as a professor you get so excited
Now that I think of it, that answer can also be presented as (x/2)*x^i + (x/2) *x^-i + C1 + C2, or as x times the average between x^i and x^-i plus the constants.
Love your videos so much :D