I know that the Maclaurin series expansion for $$\frac{1}{\sqrt{1+x^2}}$$ using binonmial coefficients is:

= 1 - $\frac{x^2}{2}$ + $\frac{3x^4}{8}$ - $\frac{5x^6}{16}$ + $\frac{35x^8}{128}$ + O(x^9)

However, when I try to use derivative rules, evaluating at a = 0, I get

f(0) = 1

f '(x) = $\frac{2x}{-2(1+x^2)^(3/2)}$ which = 0 at when a = 0

All other derivatives have x in the numerator, which also give coefficients of 0.

As a result, the Maclaurin series using derivatives would be:

1 + 0 + 0 + 0....

Why is this the case? I would expect that the Maclaurin series using binomial coefficients matches the Maclaurin series obtained by taking derivatives.

  • 2
    $\begingroup$ Check here. $\endgroup$
    – Atbey
    Apr 1 '18 at 0:03
  • $\begingroup$ It might be easier to find series for $(1+x)^{-1/2}$ first, and then plug in $x^2$. $\endgroup$
    – Atbey
    Apr 1 '18 at 0:04
  • $\begingroup$ That's actually how I got the first expansion I showed. What I'm trying to figure out is why using Method 1 (binomial expansion using x^2) gives a different result than using derivatives where the coefficients cn = $\frac{f^n(a)}{n!}$ where a = 0 $\endgroup$ Apr 1 '18 at 0:20
  • $\begingroup$ It does not. Have you checked the link? $f''(x)$ does not vanish at $x=0$. $\endgroup$
    – Atbey
    Apr 1 '18 at 0:24
  • $\begingroup$ Oh, I missed the link. Now I see that every other term disappears, which makes sense with x^2. Thanks! $\endgroup$ Apr 1 '18 at 0:35

Check your derivatives again.

They are not all zero at a=0.

For example when you find your f''(x) using quotient rule, the derivative of the top will generate a non-zero value at a=0.

  • $\begingroup$ Got it - thanks! $\endgroup$ Apr 2 '18 at 2:07

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