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Math Help - What am i missing here?

  1. #1
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    What am i missing here?

    (All integrations are in respect of dx)

    Its known that:
    <br />
e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!} \quad \quad (1)

    Now if we integrate both parts we have that:
    <br />
\int e^x=\int \sum_{n=0}^{\infty} \frac{x^n}{n!} \Rightarrow

    <br />
e^x=\int\frac{x^0}{0!}+\int\frac{x^1}{1!}+...\Righ  tarrow

    and correct me if i am wrong but:
    <br />
\int\frac{x^n}{n!}=\frac{x^{n+1}}{(n+1)!}

    So
    <br />
e^x=\frac{x^1}{1!}+\frac{x^2}{2!}+...\Rightarrow
    <br /> <br />
e^x=\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!}\quad \quad (2)

    And comparing 1 and 2
    <br />
\sum_{n=0}^{\infty} \frac{x^{n}}{(n)!}=\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!}

    What am i missing here?

    Thank you a lot
    Last edited by gdmath; December 5th 2009 at 04:07 AM.
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  2. #2
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    Quote Originally Posted by gdmath View Post
    Its known that:
    <br />
e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!} \quad \quad (1)

    Now if we integrate both parts we have that:
    <br />
\int e^x=\int \sum_{n=0}^{\infty} \frac{x^n}{n!} \Rightarrow

    <br />
e^x=\int\frac{x^0}{0!}+\int\frac{x^1}{1!}+...\Righ  tarrow

    and correct me if i am wrong but:
    <br />
\int\frac{x^n}{n!}=\frac{x^{n+1}}{(n+1)!}

    So
    <br />
e^x=\frac{x^1}{1!}+\frac{x^2}{2!}+...\Rightarrow
    <br /> <br />
e^x=\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!}\quad \quad (2)

    And comparing 1 and 2
    <br />
\sum_{n=0}^{\infty} \frac{x^{n}}{(n)!}=\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!}

    What am i missing here?

    Thank you a lot
    You are missing the fact that \int f(x) dx= F(x)+ C. That is, what you have shown is that
    \sum_{n=0}^{\infty} \frac{x^{n}}{(n)!}=\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!}+ C
    and that's obviously true: let k= n+1 on the right and each term becomes \frac{x^k}{k!}. When n= 0, k= 1 so your equation becomes
    \sum_{n=0}^{\infty} \frac{x^{n}}{(n)!}=\sum_{k=1}^{\infty} \frac{x^{k}}{(k)!}+ C
    which is true with C= 1, the missing "first term" on the right.
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  3. #3
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    Thank you

    Indeed this explains the result.

    However i rely on the fact that e^x is the "neutral" element in calculus (as unit is at multiplication / division).

    So what about if we derive the initial relation???

    I thing we lead to a similar inconvinient result.

    <br />
\sum_{n=0}^{\infty} \frac{x^{n}}{(n)!}=\sum_{n=0}^{\infty} \frac{x^{n-1}}{(n-1)!}

    of course in this last case n! (for negative n) can be defined properly so the last relation to be valid
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  4. #4
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    Quote Originally Posted by gdmath View Post
    Thank you

    Indeed this explains the result.

    However i rely on the fact that e^x is the "neutral" element in calculus (as unit is at multiplication / division).

    So what about if we derive the initial relation???

    I thing we lead to a similar inconvinient result.

    <br />
\sum_{n=0}^{\infty} \frac{x^{n}}{(n)!}=\sum_{n=0}^{\infty} \frac{x^{n-1}}{(n-1)!}

    of course in this last case n! (for negative n) can be defined properly so the last relation to be valid
    \sum_{n=0}^{\infty}\frac{x^n}{n!} = \frac{1}{1} + \frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{6} + ...

    Differentiating that, we get:

    0 + 1 + x + \frac{x^2}{2} + ... = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x

    Why was your result wrong? The derivate of a constant, c, is 0 -- not \frac{x^{n-1}}{(n-1)!}.
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