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Consider the Taylor series (about $0$) for the function $\dfrac{1}{1+t^2}$. We first consider the function:

$$f(x)=\dfrac{1}{1+x}=(1+x)^{-1}$$

We find that $f^n(x)=(-1)^nn!(1+x)^{-(1+n)}$. The nth order Taylor polynomial $P_n(x)$is thus $\sum_{k=0}^n(-x)^k$ , and the remainder term $R_n(x)$ is $ \dfrac{f^{n+1}(c)x^{n+1}}{n+1!}$= $\dfrac{(-x)^{n+1}}{(1+c)^{n+2}}$, for some $c$ in $(0,x).$

We therefore have, by Taylor's theorem:

$$f(x)=\dfrac{1}{1+x}=\sum_{k=0}^n(-x)^k+\dfrac{(-x)^{n+1}}{(1+c)^{n+2}}$$ The substitution $x=t^2$ results in: $$\dfrac{1}{1+t^2}=\sum_{k=0}^n(-t^2)^k+\dfrac{(-t^2)^{n+1}}{(1+c)^{n+2}}$$ However, the top answer of: Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$?, writes : $$\dfrac{1}{1+t^2}=\sum_{k=0}^n(-t^2)^k+\dfrac{(-t^2)^{n+1}}{1+t^2}$$.

I am struggling to see how can we use the fact that $1<1+c<1+t^2$ ,(the only information we have about c), to reach this result, starting from $\dfrac{(-t^2)^{n+1}}{(1+c)^{n+2}}$.

satan 29
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1 Answers1

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This is because it comes from a high-school identity, independent from Taylor's formula: from the factorisation formula $\;1-x^{n+1}=(1-x)(1+x+x^2+\dots +x^n)$, you can deduce instantly that $$\frac1{1-x}=1+x+x^2+\dots +x^n+\frac{x^{n+1}}{1-x},$$ whence, substituting $x$ with $-t^2$, $$\frac1{1+t^2}=1-t^2+t^4+\dots +(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}.$$

Bernard
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