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computer science - Why do we need Taylor polynomials? but In order to write out a Taylor polynomial for a function, which we will use to approximate said function at a given value, we must first have the function itself. BUT, if we already have the …
calculus - Finding the 5th order taylor polynomial of a function ... 20 Jan 2017 · I am currently stuck on this 5th order taylor polynomial question for almost an hour now, it's quite difficult if you brute force it but there was a hint given today by my tutor that there is …
calculus - Taylor's polynomial uniqueness proof - why are these … 12 Sep 2016 · So I'm viewing a short proof on the uniqueness of Taylor polynomials. Uniqueness of Taylor polynomial: Let ...
Taylor series expansion of sin(x) - Mathematics Stack Exchange 11 Dec 2017 · Now a Taylor expansion is written up to a remainder term, with as many terms as you like. The word order is used and equals the highest degree. So you can say $\sin(x)=x+r_1(x)$ is …
Taylor Series for $\\log(x)$ - Mathematics Stack Exchange 29 Nov 2013 · The Taylor series centered at $1$ can be easily derived with the geometric series $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$
How are the Taylor Series derived? - Mathematics Stack Exchange The Taylor series is an extremely powerful representation because it shows that every function can be represented as an infinite polynomial (with a few disclaimers, such as interval of convergence)! …
How do Taylor polynomials work to approximate functions? $\begingroup$ Continuing along @KevinCarlson 's line of thought, if we let p_2(x) be the second MacLaurin polynomial of the function f(x), then p_2(x) is the unique degree 2 polynomial with the …
What is the difference between the Taylor and Maclaurin series? 31 Jan 2017 · If a function is equal to it's Taylor series locally, it is said to be an analytic function, and it has a lot of interesting properties. However, not all functions are equal to their Taylor series, if a …
Taylor series of $\\ln(1+x)$? - Mathematics Stack Exchange Note that $$\frac{1}{1+x}=\sum_{n \ge 0} (-1)^nx^n$$ Integrating both sides gives you \begin{align} \ln(1+x) &=\sum_{n \ge 0}\frac{(-1)^nx^{n+1}}{n+1}\\ &=x-\frac{x^2 ...
Taylor Polynomial of $e^{-x^2}$ - Mathematics Stack Exchange Because that is the degree 4 Taylor polynomial. You gave the degree 8 Taylor polynomial. Share. Cite ...