Develop the Taylor series for the function $$f(x)={\frac{1-x}{\sqrt{1+x}}}$$ for $$ a=0$$ I have tried to differentiate it, knowing that the Taylor series have the next form: $F(x)={\frac{f^{(n)}(a)}{n!}(x-a)}$ But it's far too complicated after 2/3 derivatives
Develop the Taylor series for the function $f(x)={\frac{1-x}{\sqrt{1+x}}}$
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calculus
sequences-and-series
taylor-expansion
2 Answers
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The generalized binomial theorem says that $$ (1+x)^{-1/2}=\sum_{n\ge0}\binom{-1/2}{n}x^n \tag{1} $$ (the radius of convergence is discussed later on).
Then $$ \frac{1-x}{\sqrt{1+x}}= (1-x)\sum_{n\ge0}\binom{-1/2}{n}x^n $$ Now you can distribute and collect terms: \begin{align} \frac{1-x}{\sqrt{1+x}} &=(1-x)\sum_{n\ge0}\binom{-1/2}{n}x^n \\[6px] &=\sum_{n\ge0}\binom{-1/2}{n}x^n-\sum_{n\ge0}\binom{-1/2}{n}x^{n+1} \\[6px] &=1+\sum_{n\ge1}\binom{-1/2}{n}x^n-\sum_{n\ge1}\binom{-1/2}{n-1}x^{n} \\[6px] &=1+\sum_{n\ge1}\left(\binom{-1/2}{n}-\binom{-1/2}{n-1}\right)x^n \end{align} The radius of convergence of a power series doesn't change when it's multiplied by a polynomial (verify it), so it's sufficient to look at the radius of convergence of $(1)$. With the ratio test, $$ \left|\frac{\dbinom{-1/2}{n+1}x^{n+1}}{\dbinom{-1/2}{n}x^{n}}\right|= \frac{n+1}{n+1/2}|x| $$ because $$ \binom{k}{n}=\frac{k(k-1)\dotsm(k-n+1)}{n!} $$ so $$ \frac{\dbinom{k}{n+1}}{\dbinom{k}{n}}= \frac{k(k-1)\dotsm(k-n+1)}{n!} \frac{(n+1)!}{k(k-1)\dotsm(k-n)}= \frac{n+1}{k-n} $$ Since the limit at $\infty$ of the ratio is $|x|$, the ratio of convergence is $1$.
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0The binomial theorem do say $(a+b)^k=\sum_{k\ge0}\binom{k}{n}a^{k-n}b^n$ so it's not like yours.. Can you explain? I mean, your $k=-1/2$ but $x$ do not have $n-(-1/2)$ – 2017-01-13
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1@NeacsuMihai $a=1$, $b=x$. Anyway, the “generalized” is about $k$ not being integer and is $(1+x)^k=\sum_{n\ge0}\binom{k}{n}x^n$. – 2017-01-13
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0ok, I did not read that properly, sry. But, let's say that I want to find its Radius of convergence, how do I do that? – 2017-01-13
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0@NeacsuMihai The radius of convergence is $1$. – 2017-01-13
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0Why? And sorry for so many questions.. But I rly want to unerstand this type of exercises.. – 2017-01-13
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0The equality is only true for $|x|<1$. Of course you can do the Taylor series using this but just the notation is not the best. – 2017-01-13
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0@NeacsuMihai The radius of convergence of the binomial series is $1$, as shown above. – 2017-01-13
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0@Masacroso Is it better now? – 2017-01-13
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A different method: let two functions $f,g\in C^\infty(D,\Bbb R)$, i.e. they are infinitely differentiable in some domain $D$. I will represent the n-th derivative of some one-variable function $h$ as $\partial^n h$.
You can check with induction that
$$\partial^n(f\cdot g)=\sum_{k=0}^n\binom{n}{k}(\partial^k f)(\partial^{n-k} g)\tag{1}$$
If you define $f(x):=1-x$ and $g(x):=(1+x)^{-1/2}$ then (1) reduces to
$$\partial^n\left(\frac{1-x}{\sqrt{1+x}}\right)=(1-x)\partial^n (1+x)^{-1/2}-n\cdot\partial^{n-1}(1+x)^{-1/2}$$
and $$\partial^k(1+x)^{-1/2}=(-1/2)^{\underline k}(1+x)^{-1/2-k}$$
where $a^{\underline k}$ represent a falling factorial.