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I am learning Maclaurin Series for the first time and having trouble understanding it.

The thing is, Maclaurin Series has the basic thinking that infinite number of derivatives have coefficients of $f^{(n)}(0)$ that equals $n!C_n$. I get that.

But whenever there is a chain rule, does every $f^{(n)}(0)$ not become 0?

For example, for function $(4+x^2)^{\frac{-1}{2}}$, proper form of binomial function is $\frac{1}{2}(1+\frac{x^2}{4})^{\frac{-1}{2}}$,

the first derivative becomes $\frac{-1}{2}(1+\frac{x^2}{4})^{\frac{-3}{2}}*\frac{x}{2}$ and every following derivatives have $\frac{x}{2}$ at the end, which makes $f^{(n)}(0)=0$ because x being zero makes everything zero.

But the solutions don't seem to mind it and solve problems as if there was not $\frac{x}{2}$

강승태
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2 Answers2

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No

$f'(x)=-\dfrac{x}{{{\left( {{x}^{2}}+4\right) }^{3/2}}}$ and so indeed $f'(0)=0$

But $f''(x)=\dfrac{3 {{x}^{2}}}{{{\left( {{x}^{2}}+4\right) }^{5/2}}}-\dfrac{1}{{{\left( {{x}^{2}}+4\right) }^{3/2}}}$ so $f''(0)=0-\dfrac18$ and something similar happens with all the even derivatives

In fact $${{{\left( {{x}^{2}}+4\right) }^{-1/2}}}= \frac{1}{2}-\frac{1}{16} {{x}^{2}}+\frac{3}{256} {{x}^{4}}-\frac{5}{2048} {{x}^{6}}+\frac{35}{65536} {{x}^{8}}-\cdots$$

Henry
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  • I see what you are saying. The solution did not really mention it. And also the answer does not seem to account for what you are saying, although I agree with you. The question is – 강승태 Aug 11 '18 at 00:56
  • I see what you are saying. The solution did not really mention it. And also the answer does not seem to account for what you are saying, although I agree with you. The answer is $\frac{1}{2}x + \Sigma^\infinity_{n=1}(-1)^n\frac{(1)(3)(5)....(2n-1)}{n!2^{3n+1}}x^{2n-1}$ The original question is $x(x^2+4)^{-1/2}$ The answer should have $(2n)!$ in the denomenator because odd numbered derivatives$f^{(2n-1)}(0)$ are 0, but the answer has n!, which suggests all $f^{n}(0)$ is accounted for. – 강승태 Aug 11 '18 at 01:02
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No, not every derivative has an $x/2$. In your example, $$f^{(2)}(x) = {\frac {3\,{x}^{2}}{32\, \left( 1+{x}^{2}/4 \right) ^{5/2}}}-1/8\, \left( 1+{x}^{2}/4 \right) ^{-3/2} $$ so $f^{(2)}(0) = -1/8$.

Robert Israel
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