Power series representation of a given integral

**frank1** · December 10th 2012, 01:27 PM

Hello guys,

I did a lot of finding the derivative of power series, but I'm stuck on the first exercise of finding the power series representation of a integral.

In this case: $\int^x_0(e^t)$

I dont even know how to begin this.

=/

Any ideas?

**prasum** · December 10th 2012, 01:33 PM

just calculate the integral yu will get e^x -1

i guess you know the power series of e^x

**frank1** · December 10th 2012, 02:03 PM

Thanks for yur reply,

but no, I dont know whats the power series of e^x,

could you help me?

**Plato** · December 10th 2012, 02:10 PM

Originally Posted by frank1

I did a lot of finding the derivative of power series, but I'm stuck on the first exercise of finding the power series representation of a integral.
In this case: $\int^x_0(e^t)$

Please state the entire and exact wording of the question.

**frank1** · December 10th 2012, 02:20 PM

Question) Find the power series representation of $\int^x_0(e^t)dt$ .

I know how to integrate that. But whats the next step? I'm really lost.

**skeeter** · December 10th 2012, 02:31 PM

$e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + ... + \frac{t^n}{n!} + ...$

$\int_0^x e^t \, dt = \left[t + \frac{t^2}{2!} + \frac{t^3}{3!} + ... + \frac{t^{n+1}}{(n+1)!} + ... \right]_0^x$

finish it ...

**frank1** · December 10th 2012, 02:43 PM

I dont undertand why $e^t=1+t+t^2/2!...$

So, basically you integrate the given expression, find the general term of de series of that integrated expression, then make the substitution with given integral limits?

Like, the following is:

$\int^x_2(4-t)dt$

So that integral is: $-ln(4-t)$ ...

aah, =S how the heck will I find this general term expression?

I dont mean to get help in all exercises, I just want to understand the logic here =/

Thanks for the help and patience.

**skeeter** · December 10th 2012, 02:49 PM

do you know anything about power series of functions? ... Taylor series? ... Maclaurin series?

if not, then you need to go research those topics.

**frank1** · December 10th 2012, 03:02 PM

No, Taylor series/McLaurin series are the next topic in my classes/book

I studied the basics of power series, then derivatives of it, and now I'm on integrals.

**skeeter** · December 10th 2012, 03:24 PM

well, it's a bit difficult to integrate using series when you do not know what the general series is for $y = e^x$ .

also, for post #7 , did you mean $\int_2^x (4-t) \, dt$ or $\int_2^x \frac{1}{4-t} \, dt$ ?

**frank1** · December 10th 2012, 03:27 PM

1/(4-t)

Well, but could you explain me the logic you follow in this exercises? Like, "Find the general term, then substitute..."

Just to try to grap something here...

**skeeter** · December 10th 2012, 03:45 PM

1/(4-t)

I thought so ...

you should be familiar with this series representation ...

$\frac{1}{1-x} = 1 + x + x^2 + x^3 + ...$

which converges for values of ${x : |x| < 1}$

using this concept ...

$\frac{1}{4-t} = \frac{1}{4} \cdot \frac{1}{1 - \frac{t}{4}} = \frac{1}{4} \left[1 + \frac{t}{4} + \frac{t^2}{4^2} + \frac{t^3}{4^3} + ... + \frac{t^n}{4^n} + ... \right]$

integrate ...

$-\ln(4-t) = \frac{1}{4} \left[C + t + \frac{t^2}{2 \cdot 4} + \frac{t^3}{3 \cdot 4^2} + \frac{t^4}{4 \cdot 4^3} + ... + \frac{t^{n+1}}{(n+1)4^n} + ... \right]$

at $t = 0$ , $-\ln(4-t) = -\ln(4)$ ...

$-\ln(4-t) = \left[-4\ln(4) + \frac{t}{4} + \frac{t^2}{2 \cdot 4^2} + \frac{t^3}{3 \cdot 4^3} + \frac{t^4}{4 \cdot 4^4} + ... + \frac{t^{n+1}}{(n+1)4^{n+1}} + ... \right]$

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