0
$\begingroup$

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.

$\endgroup$
5
  • 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
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Got it - thanks! $\endgroup$ Apr 2 '18 at 2:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.