# Questions tagged [taylor-expansion]

Questions regarding the Taylor series expansion of univariate and multivariate functions, including coefficients and bounds on remainders. A special case is also known as the Maclaurin series.

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### Formula for calculating $\sum_{n=0}^{m}nr^n$

I want to know the general formula for $\sum_{n=0}^{m}nr^n$ for some constant r and how it is derived. For example, when r = 2, the formula is given by: $\sum_{n=0}^{m}n2^n = 2(m2^m - 2^m +1)$ ...
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1answer
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### Complex Taylor and Laurent expansions

Let $f(z):=\dfrac{1}{2-z-z^2}, z\in\mathbb{C}\setminus\left\{ {1, -2}\right\}$. i) Express $f$ in the form $\dfrac{A}{1-z}+\dfrac{B}{2+z}$. [Answer to this is $\dfrac{1/3}{1-z}+\dfrac{1/3}{2+z}$]. ...
1answer
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### Relationship Between Sine as a Series and Sine in Triangles

I'm taking a single variable calculus course and sine series is defined as $$\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k + 1)!}.$$ Sine is also defined as opposite length over hypotenuse length ...
1answer
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### If $f''(x_0)$ exists then $\lim_{x \to x_0} \frac{f(x_0+h)-2f(x_0)+f(x_0-h)}{h^2} = f''(x_0)$

Prove: if $f''(x_0)$ exists then $\lim\limits_{x \rightarrow x_0} \dfrac{f(x_0+h)-2f(x_0)+f(x_0-h)}{h^2} = f''(x_0)$. I'm not exactly sure how Taylor's theorem fits into all this, but I found the ...
1answer
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### How to prove that: $19.999<e^\pi-\pi<20$?

I would like to know how to prove $$e^\pi-\pi\sim 20.$$ More precisely, I want to show by using only mathematical tools that, $$19.999<e^\pi-\pi<20$$ I have checked with online ...
3answers
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### Proving that $\frac{e^x + e^{-x}}2 \le e^{x^2/2}$

Prove the following inequality: $$\dfrac{e^x + e^{-x}}2 \le e^{\frac{x^2}{2}}$$ This should be solved using Taylor series. I tried expanding the left to the 5th degree and the right site to the ...
3answers
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### Taylor series for different points… how do they look?

I can't understand what it means to do the Taylor series at the point $a$. The best way would be showing me how it looks for different $a$ on a graph. Do I find those graphs on the Internet?