# Determine the sequence generated by the following exponential generating functions:

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.

• 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. – Brian Mar 18 at 19:11
• Thank you. Do you think my format is too confusing to get an answers and I should repost it? – radius Mar 18 at 19:13
• I would recommend reformatting your post using an edit. – 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, $$\frac{1}{1-x} = \sum_{k=0}^\infty x^k \implies a_k \equiv 1\ \forall k \in \mathbb{N}.$$