Power series representation of a given integral

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December 10th 2012, 01:27 PM
frank1

Power series representation of a given integral

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?
December 10th 2012, 01:33 PM
prasum

Re: Power series representation of a given integral

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

i guess you know the power series of e^x
December 10th 2012, 02:03 PM
frank1

Re: Power series representation of a given integral

Thanks for yur reply,

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

could you help me?
December 10th 2012, 02:10 PM
Plato

Re: Power series representation of a given integral

Quote:

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.
December 10th 2012, 02:20 PM
frank1

Re: Power series representation of a given integral

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.
December 10th 2012, 02:31 PM
skeeter

Re: Power series representation of a given integral

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

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

finish it ...
December 10th 2012, 02:43 PM
frank1

Re: Power series representation of a given integral

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. (Doh)
December 10th 2012, 02:49 PM
skeeter

Re: Power series representation of a given integral

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

if not, then you need to go research those topics.
December 10th 2012, 03:02 PM
frank1

Re: Power series representation of a given integral

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.
December 10th 2012, 03:24 PM
skeeter

Re: Power series representation of a given integral

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 $\int_2^x \frac{1}{4-t} \, dt$ ?
December 10th 2012, 03:27 PM
frank1

Re: Power series representation of a given integral

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...
December 10th 2012, 03:45 PM
skeeter

Re: Power series representation of a given integral

Quote:

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

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