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Math Help - Power series representation of a given integral

  1. #1
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    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?
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  2. #2
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    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
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    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?
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  4. #4
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    Re: Power series representation of a given integral

    Quote Originally Posted by frank1 View Post
    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.
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  5. #5
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    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.
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  6. #6
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    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 ...
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  7. #7
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    Question 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.
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  8. #8
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    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.
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  9. #9
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    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.
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  10. #10
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    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 ?
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  11. #11
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    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...
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  12. #12
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    Re: Power series representation of a given integral

    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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