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Here is an (intuitive) proof that explains the coeffcients in ∞ ∑ k = 0(k + n − 1 n − 1)xk = 1 (1 − x)n. Since 1 1 − x = ∞ ∑ k = 0xk, we just need to figure out how to multiply it n times. To calculate the coefficient of a specific power, say xk, you only need the first k + 1 terms in each series. So we have, (1 + x + x2 + x3 ...
35. As others have noted, a power series is a series ∑∞ n=1anxn ∑ n = 1 ∞ a n x n (or sometimes with x x translated by some x0 x 0, to become (x −x0) (x − x 0)). Normally when one says Taylor series, one means the Taylor series of some particular smooth function f f. (So in mathematical speech, one wouldn't usually say "consider a ...
Assuming your power series is centered at 0 0, As a hint: try expanding it out as a Taylor series expansion, the coefficients have a very nice formula. Look at the series expansion of 1 1−t = 1 + t +t2 + ⋯ 1 1 − t = 1 + t + t 2 + ⋯. and replace t t by x +x4 = x(1 +x3) x + x 4 = x (1 + x 3). Then thin a bit about the ways to get an x11 x 11.
With other types of series, including the very important trigonometric series, one has to be more careful. $\endgroup$ – André Nicolas Commented Jun 29, 2016 at 2:09
The one that fits best here in my opinion is: √1 + x = 1 + x 2 − x2 8 + x3 16 − 5x4 128 + …. Through substitution, we can obtain: 1 √1 − x2 = 1 + x2 2 + 3x4 8 + 5x6 16 + 35x8 128 + …. Then by integration: arcsin(x) = ∫x 0 1 √1 − t2dt = x + x3 6 + 3x5 40 + 5x7 112 + 35x9 1152 + …. Share. Cite.
So. e − x2 = 1 − x2 + x4 2! − x6 3! + ⋯. To find the general series expansion, you just have to remember the series expansion of eA: eA = + ∞ ∑ k = 0Ak k! So again: if A = − x2 you gain finally. e − x2 = + ∞ ∑ k = 0(− x2)k k! = + ∞ ∑ k = 0(− 1)k k! x2k. Share. Cite. answered Jun 29, 2016 at 14:18.
the Taylor series for ln (x) is relatively simple : 1/x , -1/x^2, 1/x^3, -1/x^4, and so on iirc. log (x) = ln (x)/ln (10) via the change-of-base rule, thus the Taylor series for log (x) is just the Taylor series for ln (x) divided by ln (10). – correcthorsebatterystaple. Mar 18 at 14:35. Add a comment. 4 Answers. Sorted by:
Now, ex is a very wel-behaved function, and we have that it actually is equal to its power series at every x. In particular, if you substitute x by 2x in the power series expansion of ex, you will obtain the power sereis expansion for e2x: ∑n=0∞ (2x)n n! =∑n=0∞ 2n ⋅ xn n! Share. Cite.
power-series; taylor-expansion; Share. Cite. Follow edited Nov 6, 2019 at 15:39. José Carlos Santos ...
The power series expansion of 1 z 1 z about z0 = 1 z 0 = 1 is: 1 z =∑n=0∞ (1 − z)n =∑n=0∞ (−1)n(z − 1)n 1 z = ∑ n = 0 ∞ (1 − z) n = ∑ n = 0 ∞ (− 1) n (z − 1) n. Now what I don't get is the logic behind such reasoning. It might be basic but I just don't see how you can get a series expansion simple as that. Can ...