3
$\begingroup$

I'm trying to get the Maclauren series of: $ f(x) = {x + 5\over1-x^2}$.

I am sure there is some trick here, the result according to Mathematica is:

$5 + x + 5x^2 + x^3 + 5x^4 + x^5 + 5x^6 + \ ...$

I defined:

$f^n_m = {x^n\over(1-x^2)^m}$

$f^n_m = 0$ for $n < 0 $ (just a definition necessary to make the next formula work for $n = 0$)

I've computed:

$(f^n_m)' = nf^{n-1}_m + 2mf^{n+1}_{m+1}$.

We have $f(x) = f^1_1(x) + 5f^0_1(x)$

I tried going from here, I didn't try induction as I want to see how I could get this result without assuming it first.

If I keep taking derivatives I get more and more terms of the form $f^s_r$.
But didn't find the pattern, I can see in the Mclauren series we evaluate the derivatives at 0 and thus only terms of the form $f^0_r$ contribute.

I tried also to just take some derivatives brute force and looking at the result, but they just keep getting more and more terms and I cannot ignore any of them because eventually as I differentiate they will all contribute at some point to the evaluation at 0.

Any advice will be helpful but I would like to see how I could have arrived at the result, so proving it by induction if it assumes the result won't be of much help.

Thank you very much!

$\endgroup$
4
$\begingroup$

Notice we all know the following

$$ \sum^{\infty} x^n = \frac{1}{1-x} $$

for $|x| < 1 $. Now, applying this

$$ \frac{1}{1 - x^2} = \sum x^{2n} \implies\frac{x+5}{1-x^2} = (x+5) \sum x^{2n} = \sum x^{2n + 1} + \sum 5x^{2n} =$$

$$ = 5 + x + 5x^2 + x^3 + 5x^4 + x^5 + 5x^6 + \ ... $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Haha, thank you, classic case of overthinking, need to be more open minded and try different methods! $\endgroup$ – avdo rian Nov 1 '13 at 8:34
2
$\begingroup$

I think you make the problem more difficult than it is. Perform the division of 1 by (1 - x^2). As a result, you have 1 + x^2 + x^4 + x^6 + ... + x^(2n). Now, multiply by (x+5) and expand. You will get what Mathematica told you. Easy, isn't ?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes it is! definitely taught me that if I should try a wider variety of methods! $\endgroup$ – avdo rian Nov 1 '13 at 8:37

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.