a) $f(x)=3e^{3x}$

I have \begin{align}f(0)&= 3 \\ f(1)&=3e^3 \\ f(2)&=3e^6 \end{align} So would my sequence be $a_n=3e^{2n}$?

Or by recurrence $a_n=a_{n-1}(e^3)$?

Or should I find a summation?

b) $f(x)=6e^{5x}-3e^{2x}$

I'm not sure how to start this other than finding values of $f(0)$, $f(1)$, ... etc

But I cannot seem to find a recurrence relation.

c) $f(x)= \frac{1}{1-x}$

For this one, I have the summation from $0$ to infinity of $x^i$ where the sequence is $1+x+x^2+\cdots$

d) $f(x)= \frac{3}{1-2x} +e^x$

I am not sure how to start this one either.

  • $\begingroup$ Welcome to Math.SE! Please format your questions using MathJax. This page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. $\endgroup$ – Brian Mar 18 at 19:11
  • $\begingroup$ Thank you. Do you think my format is too confusing to get an answers and I should repost it? $\endgroup$ – radius Mar 18 at 19:13
  • $\begingroup$ I would recommend reformatting your post using an edit. $\endgroup$ – Brian Mar 18 at 19:14

One way to determine the needed sequence directly is to decompose the function into a Taylor series. Note that $$ 3e^{3x} = 3 \sum_{k=0}^\infty \frac{(3x)^k}{k!} = \sum_{k=0}^\infty \frac{3^{k+1}}{k!} x^k, $$ so the generated sequence is $a_k = 3^{k+1}/k!$.

(b) and (d) can be done similarly, since the Taylor series of the sum is really a sum of a Taylor series, so if $$ h(x) = f(x) + g(x) = \sum_{k=0}^\infty a_k x^k + \sum_{k=0}^\infty b_k x^k = \sum_{k=0}^\infty (a_k+b_k) x^k $$ and the needed sequence is $a_k+b_k$...

Your idea for (c) is correct, but having a summation does not end the problem, you have to rip the sequence out of the summation. In your case, as you write, $$ \frac{1}{1-x} = \sum_{k=0}^\infty x^k \implies a_k \equiv 1\ \forall k \in \mathbb{N}. $$


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