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Math Help - Can this series be represented by a function

  1. #1
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    Can this series be represented by a function

    Hello,

    The infinite series \ln(1-x)=-\sum_{j=1}^{\infty }\frac{x^j}{j}\: \: \:; \: \: \: -1< x\leq 1

    might also be expressed -\sum_{j=1}^{\infty }F^{(j)}(x)\: \: \:; \: \: \: -1< x\leq 1

    where F^{(j)}(x) is the jth integral of x, and F^{(1)}(x)=x

    Again, this series can be expressed in the function \ln(1-x)

    Now, my question is does there exist a function that expresses the series -\sum_{j=1}^{\infty }f^{(j)}(x)\: \: \:; \: \: \: -1< x\leq 1

    where f^{(j)}(x) is the jth *derivative* of x?

    (...and I guess \lim_{j \to \infty}f^{(j)}(x)=1 or something like that)


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  2. #2
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    Re: Can this series be represented by a function

    Quote Originally Posted by rainer View Post
    Hello,

    The infinite series \ln(1-x)=-\sum_{j=1}^{\infty }\frac{x^j}{j}\: \: \:; \: \: \: -1< x\leq 1

    might also be expressed -\sum_{j=1}^{\infty }F^{(j)}(x)\: \: \:; \: \: \: -1< x\leq 1

    where F^{(j)}(x) is the jth integral of x, and F^{(1)}(x)=x
    No, it isn't. The "jth integral of x" is \frac{x^{j}}{j!}, not \frac{x^j}{j}.
    The exponential function, e^x, has the property you are talking about.

    Again, this series can be expressed in the function \ln(1-x)

    Now, my question is does there exist a function that expresses the series -\sum_{j=1}^{\infty }f^{(j)}(x)\: \: \:; \: \: \: -1< x\leq 1

    where f^{(j)}(x) is the jth *derivative* of x?
    ??? the first derivative of x is 1 and the "jth derivative of x" for j> 1 is 0. So such a series would be x+ 1+ 0+ 0+ 0...= x+ 1.

    (...and I guess \lim_{j \to \infty}f^{(j)}(x)=1 or something like that)


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