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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 exists. One may note that most of the most famous Taylor series are a Maclaurin series, probably since they look nicer. For example,
calculus - Find Taylor polynomial of an integral function 28 Nov 2020 · Computing Taylor polynomial of a infinitely differentiable function that is a composite of non-differentiable functions 0 Find the Taylor series for an arbitrary polynomial
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 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 ...
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 property that lim_{x \to 0} [f(x) - p_2(x)]/[f(x) - q_2(x)] = 0 for all degree 2 polynomials q_2(x) different than p_2(x); it is in this sense that the degree 2 MacLaurin polynomial of f(x) is the …
computer science - Why do we need Taylor polynomials? 1 Oct 2012 · 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 function, why not just use it and get an exact value instead of using it's Taylor polynomial and only getting an approximation?
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)! This means that we can differentiate a function as easily as we can differentiate a polynomial, and we can compare functions by comparing their series expansions.
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 the first order expansion, $\sin(x)=x-\dfrac{x^3}{3!}+r_3(x)$ is the third order expansion, $\sin(x)=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+r_5(x)$ is the fifth order ...
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 ...
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 an easier way to solve for the 5th taylor polynomial of this function below if …