Power series of tan^-1(x), with radius & interval of convergence, calculus 2 tutorial
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- เผยแพร่เมื่อ 7 ก.พ. 2025
- We will find the power series expansion of tan^-1(x), i.e. the arctan(x). We will do so by integrating the power series of 1/(1+x^2). Remember the radius of convergence stays the same when we integrate or differentiate a power series. HOWEVER, we must do more work to check the convergence at the endpoints of the interval of convergence.,
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ima plug in -x^2 into you lil bro
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I just realized that the arctg(1) = pi/4 is the leibneiz series, omg math is so amazing
Multiply both arctan(1) and the resulting power series by 4 and you'll get yourself a PIE.
But who guarantees that the alernating series Σ [(-1)^n][x^(2n+1)]/(2n+1) converges to arctan(1) and arctan(-1) for x = +1, -1 in order to assume that arctan(x) can be described by that series on the boundaries of the interval?
Just a small correction. At 20:20 you say b(n+1)+oo the denominators are equal. If the series is alternating a.k.a. (- 1)^n, you only need b(n)->0 for it to converge.
I think youtube's (-crossout-) mechanism might have got you? may wanna edit that.
also, b(n+1) is less than b(n) because infinity is not considered, i.e. the formula for e is (1+1/n)^n, but for n=infinity the answer is (1+0)^infinity=1, which is obviously wrong.
so you check along the way, and lo and behold: [1/9=1/(2*4+1)]
I didnt know how to edit it out, a simple space seems to do the trick, thanks. As far as the limit is concerned, of course you set n->+oo. In this case you go like this b(n+1)/b(n)=(2n+1)/(2n+3)=(1+1/(2n))/(1+3/(2n))=(1+0)/(1+0)=1 as n->+oo. That is why the series sum(1/(2n+1)) does not converge. The limit with e that you mentioned is an undefined form that is why you must try to figure it out in another way.
actually if you use ratio test you will see that this infinite series always converge (x belong to real set )but does it converge to give arctanx ? i am saying that at -1 and 1 it is right to say series converges but how do we know that it converges to produce arctanx again?
Thank you your teaching id amazing.
Checking the power series converges at the endpoints is easy, it is more tricky to check that it converges to arctan. Abel's theorem is what one would use here.
Great work
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I hate math and still ended up smiling after the first 30 secs of the video hehe
Ur an idiot if u hate mathematics
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Also, you can get a series good for |x|>1 if you realize that the series using 1/x as the argument gives the complementary angle.
How about an error formula for if we use a partial sum of the series (like say use k terms from the series , then get a formula for the max error in the approximation in terms of k ) . ?
Very well done, thank you!
I am quite great at math.but sometimes I come up blank looking at this dudes math calculus
nice video
6:48 can't you just throw away the absolute value since x^2 is always positive?
No, because we need to solve for the radius of convergence, which is in the form of |x-a|
blackpenredpen oh, I should have watched it till the end then :)
yup,
thank so much! best friend
Series converge for x equals to 1. but why it converges to pi/4? what is the proof?
Why is 4 * tan^-1(x) so horrible in estimating pi?
so.... I'm good at algebra, trig, number theory, e, vectors, derivatives, (because of you, integrals), and I know a few identities of pi, but all I can assume here is (by the looks of it) is that for x=1, this sum gives pi? or some pi-based term? I think...
Ryan Roberson its actually pi/4
that's the one! but it seems a bit off: if there is pi/4 and 6/pi, and pi^2/2... where do these arbitrary numbers keep coming from? is there some formula that makes an infinite series for (pi/j)^k?
I know it is pi/4 ,because if you look at what arctan does, if you plug in opposite/adjacent of a right triangle into the arctan function you will get the angle. And if you plug in 1, that means
opposite/adjacent = 1 which means both have the same lenght so the angle must be 45 degrees = pi/4
i am not sure if such formulas exist, i believe these numbers will just pop up if you plug in the right number into the right function
Interestingly, this means that pi/4 = 1 - 1/3 + 1/5 -1/7... . This means that you can calculate pi just by doing 4 - 4/3 + 4/5 - 4/7...
@@shehannanayakkara4162 yes but it converges SUPER slowly. There are better, more efficient ways to calculate pi by hand. StandupMaths has some good videos on calculating pi
I have a question,
the radius of convergence is 1, that means if you plug in a number bigger than one or smaller that -1 the series diverges, but I know that there are values for arctan(x) where x is bigger than one.. how do you calculate those values?
Ah! In those cases.. you will have to change the center and use the Taylor formula to find the power series. This is just one quick way that I can think of as of now.
WHY CAN'T YOU BE MY TA?!?!!
0 is not really a number because it does not share the elemental properties of numbers.0 does not have a ratio to all other numbers like numbers do.It is a error in math to treat zero like it is a number so you should not let x = 0,you must use a positive number to get meaningful results.
The radius of convergence is 1 so the series outputs valid answers for all inputs [-1,1], not just positive.
If you're REALLY gonna argue about zero, consider that numbers are completely abstract and made up. Arguing that 0 "isn't really a number" is the same as arguing that "Optimus Prime is a ghost busters," whether it is true or not has no meaning and doesn't affect anything since it's all made up.
thanks for the knowledge.
You're welcome1
Hey BPRP, my friend was challenged to prove that (1+x)e^x = sum of (1+k)x^k/k! WITHOUT using differentiation. Do you have any ideas?
Am I allowed to start off with the fact that e^x=sum of x^k/k! ?
If so, this is just about arithmetic of series.
(1+x)e^x
=e^x+x*e^x
=sum of x^k/k! + x* ( sum of x^k/k!)
and you will have to fix the power, index, etc
you can see the last part of this video for fixing index. th-cam.com/video/X8c64zq8Lno/w-d-xo.html
And maybe I will make a quick video tmr before my classes.
maestro ,!
Great series of tutorials, but this time you spent so much time on the totally obvious part and missed to present a solution for abs(x)>1
By power series, I think he means taylor series. Correct me iff I m wrong but for abs(x)>1 arctanx has Laurent series, so he wont try to dive into that.
MrGeorge1896 I know I spent a lot of time but it was for my calc2 students who are studying series for the first time.
which is why perhaps u re-did the alternating series test instead of using the result that the second infinite series is the negation of the first, and since the first converges, so does the second.
What university do you go to?
great
Solve tan inverse x using tylers theorm
lies i see a blue pen
Like c1+c2+...=original c (s.o.c's)
It mesns pi is rational number
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dude... do you realize you're saying "isn't it" in completely inapplicable sentences...?
Juliano Ciaramello come on man
Rip on someones English when they're providing free, excellent math help. People like you are never satisfied. Get over yourself or watch someone else.
come on, man
I'm neutral here, but in terms of context, I just learn to filter things like that, my brain translates that to "right?" rhetorical question, as it has the same meaning as used by bprp. but one thing you don't do is insult someone for saying their sayings (unless that saying encourages crime... isn't it?)
I'm not trying to insult him! I'm just wondering if he knows what he's saying! he's using the phrase really oddly is all!
I watch this channel all the time! love this guy!