I'm missing something about finding the first few coefficients of a function represented as a power series. I can get the series correct, I just can't for the life of me figure out what to use to get the first few coefficients. I'm given that
$$f(x) =\frac 8{(1-6x)^2}$$
I'm able to find the radius of convergence as $R = 1/6$ and I end up with the $C_n$ as $6^{2n}$ and also $X^{2n}$ for $X^n$. I'm supposed to get the first five coefficients, which I have
$C_0 = 8$ for $(8\cdot6^{2\cdot0})$ (this one is correct)
$C_1 = 288$ for $(8\cdot6^{2\cdot1})$
$C_2 = 1728$ for $(8\cdot6^{2\cdot2})$
$C_3 = 373248$ for$(8\cdot6^{2\cdot3})$
$C_4 = 13436928$ for$( 8\cdot6^{2\cdot4})$
$C_1$ to $C_4$ are all wrong. This is plugging in $n = 0$ through $n = 4$ directly into the Cn of the series. I also tried plugging into the original function, which was also incorrect. Where are the additional coefficients found if not by plugging in directly? I can't seem to find anything on that. X isn't a part of my answer, so I can't do $8\cdot6^{2n}\cdot{x^{2n}}$ and have it submit successfully. My math course uses a website called Webwork and it won't allow X as a part of the answer.
Sorry for my poor formatting, I don't know how to format for math notation.
Thank you in advance! -Frank
$\endgroup$3 Answers
$\begingroup$I think you have made an error in finding the general term.
The geometric series $\displaystyle\sum_{n=0}^{\infty}8 \cdot 6^{2n} \cdot x^{2n}$ evaluates to $\dfrac{8}{1-(6x)^2}$ which is not $\dfrac{8}{(1-6x)^2}$.
Instead, start with the following well known series: $\dfrac{1}{1-y} = \displaystyle\sum_{n = 0}^{\infty}y^n$.
Then, differentiate term by term to get another series $\dfrac{1}{(1-y)^2} = \displaystyle\sum_{n = 1}^{\infty}ny^{n-1} = \sum_{n = 0}^{\infty}(n+1)y^n$.
Finally, replace $y$ with something in terms of $x$ and multiply both sides by a constant to get a series for $\dfrac{8}{(1-6x)^2}$.
$\endgroup$ $\begingroup$Don't need to overthink it, just use the definition and start cranking out derivatives since you know $C_k = f^{(k)}(0)/k!$:
- $C_0 = 8$ since$f(0) = 8$.
- $f'(x) = 96/(1 - 6x)^{3}$ so $C_1 = 96$.
- $f''(x) = 1728/(1 - 6x)^4$ and $C_2 = f''(0)/2!$
and so forth.
$\endgroup$ $\begingroup$If your function is $f(x) = 8/(1-6x)^2$, not $f(x) = 8/(1-{6x}^2)$, your solution is incorrect. But please observe that $\left(\dfrac{1}{1-x}\right)'=\dfrac{-1}{(1-x)^2}$ and use this fact, with proper modification, to find the coefficients.
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