Taylor series for ln(1+x), Single Variable Calculus
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- Опубликовано: 14 окт 2024
- We find the Taylor series for f(x)=ln(1+x) (the natural log of 1+x) by computing the coefficients with radius and interval of convergence. The Taylor series (or Maclaurin series) is ln(1+x) = Σ ((-1)^(n+1) * (x^n))/n, beginning with n=1. Below are links to similar examples! Please subscribe if you enjoyed this video, thanks!
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#mathematics #math #calculus #sequencesandseries #powerseries #taylorseries #naturallog #iitjammathematics #maclaurinseries
I was so impressed you were writing backwards but then I realized you just flipped the video!
Or did I? (Yes, I did.)
I am so glad I just found this wonderful channel. Thank you for the high-quality videos!
Thank you, and welcome! I'm glad you like my videos.
Thank you for the video. Is there a taylor expansion for ln(-x)? I cannot find it on youtube nor the internet. It is the mirror image of lnx,ie facing to the left. Glad if you could enlighten me. Thank you.
Yes: you need to specify a center and in this case, I imagine you want the center as a = -1. Then use Taylor coefficients (ruclips.net/video/G6lnFAMS0w0/видео.html). You should find your Taylor series to be T(x) = -(x+1) - (x+1)^2/2 - (x+1)^3/3 - (x+1)^4/4 - ...
I'll let you work out what the radius is. :)
@bevinmaultsby thank you so much for the prompt reply. That's the result I got. When I tried to graph it on desmo, even after the 6th terms, it still looked very different from the original. Could you also do Taylor expansion for f(x+dx)? I found it weird that the dx term becomes the power term. And how is it different from f(x-dx)? Hope I am not taking much of your time. Your kindness is greatly appreciated. Thank you.
My graph looked very close, maybe try a few more terms? For your other question, assuming that our Taylor series is centered at x, the coefficients take the form f^(n)(x)/n!, and the series (for a random input z) will be
T(z) = f(x) + f'(x)(z-x) + f''(x)/2! * (z-x)^2 + f'''(x)/3! * (z-x)^3 ...
so when z = x + dx (implying z-x = dx), this becomes
T(x + dx) = f(x) + f'(x)(dx) + f''(x)/2! * (dx)^2 + f'''(x)/3! * (dx)^3 ...
Here dx is either positive or negative. If you try the form z = x-dx, then you will end up with alternating signs.
Thank you so much!!
Everything was super clear but i had one doubt...why was x^n multiplied with that expression at 6:31.....can you pls explain again
The process up to that point was to compute the Taylor Coefficients c_n = f^(n)(0)/n! for the Taylor series. Then once we had the general form of those coefficients, we were ready to set up the Taylor series, which has the form Σ c_n (x-a)^n, where a is the center of the Taylor series. Our series was centered at a=0, so the form became Σ c_n x^n, starting with n=1.
@@bevinmaultsby ohh!..okay okay 👍🏻... thank you
Great explanation ❤
Thank you!
Hello Dr. Bevin Maultsby, I have a question if you kindly help me: what is the expansion of ln(1+x) as x tends to infinity if |x|>1
Since ln(1+x) ~ ln(x) as x goes to infinity, here's a way to find an expansion for ln(x) that converges for x larger than 1/2. We have ln(x) = - ln(1/x) = -ln(1+u), where u = (1-x)/x. Then -ln(1+u) is sum(n>0) (-1)^n/n u^n, so we have ln(x) = sum(n>0) (-1)^n/n (1/x - 1)^n. (This expansion is not a Taylor series as the x is inverted, but it is a neat way to approximate ln(x) for all x larger than 1/2.)
@@bevinmaultsby thank you, really appreciate your help
I didnt get why the sum start from n=1 and not 0
The series expansion is
ln(1+x) = f(0) + f'(0)(x-0) + f''(0)(x-0)^2/2! + ...
For this function, f(0) = ln(1) = 0, so there's no need for that first term.
Great explanation!
Thank you!