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In the question below, I would have to solve for an upper estimate using Weierstrass Approximation Theorem, however I am not familiar with the theorem, how do I go about solving it?

For any $\epsilon\in [0,1]$, find an upper estimate on the integer $n$ such that there exists an approximation of $f(x)=logx$ on $[1,2]$ by a polynomial $P(x)$ of degree $n$ such that $$\sup_{x\in[-1,1]}|P(x)-logx|\leq \epsilon.$$

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closed as off-topic by Saad, Paras Khosla, Cesareo, José Carlos Santos, Paul Frost Apr 24 at 16:08

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  • $\begingroup$ If you have an analytic function the best n degree approximation is the sum of the n first terms of its Taylor expansion $\endgroup$ – clark Apr 3 at 5:52
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Assuming the fact that the $\log$ you used is in base $e$. Let, $f(x)=\log x$ for all $x\in [1,2]$ then $f(x)$ can be expanded as, $$f(x)=\log x=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3+\dots+\frac{(-1)^{n-1}}{n}(x-1)^n+R_n(x)$$ where $R_n(x)$ is the remainder after $n$ terms.

Set the polynomial $P(x)=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3+\dots+\frac{(-1)^{n-1}}{n}(x-1)^n$, then $P(x)$ is a polynomial of degree $n$. Now choose $n$ in such a way that $\displaystyle \sup_{x\in [1,2]}|R_n(x)|<\epsilon$.

For definition of $R_n(x)$ i recommend this.

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