# Diff EQ, i'm confused on plotting this solution! (series)

1. Feb 8, 2006

### mr_coffee

Hello everyone. This problem has 4 parts. I got part (a) which is suppose to be the hardest part but i'm confused on how to
(b) Plot Qn(t) for n = 1...4 observe wether the iterates seem to be converging. Well this is the series solution I got for the differential equation:

Here is the orginal problem and how i got the answer to part (a).

They also say:
Plot |Q(t)-Qn(t)| for n = 1...4. For each of Q1(t),...,Q4(t), estimate the interval in which it is a reasonably good approximation to the actual solution. A few people asked the professor and he said, solve the Differential equation and hats the answeR? urnt?

2. Feb 8, 2006

### benorin

Well, without plotting it I can tell you that (only because I did a very similar problem in reverse a few hours ago)

$$\sum_{k=0}^{\infty} \frac{(-1)^{k+1}t^{k+1}}{(k+1)!2^{k-1}} = 4\sum_{k=0}^{\infty} \frac{1}{(k+1)k!}\left( -\frac{t}{2}\right) ^{k+1} = -2\sum_{k=0}^{\infty}\int_{x=0}^{t} \frac{1}{k!}\left( -\frac{x}{2}\right) ^{k} dx = -2\int_{x=0}^{t} \sum_{k=0}^{\infty} \frac{1}{k!}\left( -\frac{x}{2}\right) ^{k} dx = -2\int_{x=0}^{t} e^{-\frac{x}{2}}dx = 4\left( e^{-\frac{t}{2}}-1\right)$$

but your sum started at k=1 and not k=0, so subtract off the k=0 term from both sides, namely

$$\mbox{your sum } = 4\left( e^{-\frac{t}{2}}-1\right) -(-2t) = 4e^{-\frac{t}{2}}+2t-4$$

and so (if I haven't gone astray in my calculation)

$$Q_n(t)\rightarrow 4e^{-\frac{t}{2}}+2t-4 \mbox{ as }n\rightarrow\infty$$
which maple confirms.

Last edited: Feb 8, 2006
3. Feb 9, 2006

### mr_coffee

ahh i c, that does look just like mine!
So I can say it converges to $$Q_n(t)\rightarrow 4e^{-\frac{t}{2}}+2t-4 \mbox{ as }n\rightarrow\infty$$