- #1
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Hey, I am not trying to prove anything here, just merely asking a question to something I tryed:
I found a relation to the Taylor series and how the integration by parts expand.
for example:
[tex]\int e^x dx = x e^x - \int xe^x dx[/tex]
[tex]\int xe^x dx = \frac{x^2}{2}e^x - \int \frac{x^2}{2}e^x dx[/tex]
[tex]\int \frac{x^2}{2}e^x dx = \frac{x^3}{6}e^x - \int \frac{x^3}{6}e^x dx[/tex]
We see that this goes on and on to this:
[tex]\int e^x dx = e^x[/tex]
[tex]e^x = xe^x - \frac{x^2}{2}e^x + \frac{x^3}{6}e^x - \frac{x^4}{24}e^x...\frac{x^n}{n!}e^x[/tex]
[tex]e^x=e^x \cdot \sum^{\infty}_{k=1}\frac{x^k}{k!}(-1)^{k-1}[/tex]
[tex]\sum^{\infty}_{k=1}\frac{x^k}{k!}(-1)^{k-1} = 1[/tex]
And that's definitely NOT true! What is wrong?
I found a relation to the Taylor series and how the integration by parts expand.
for example:
[tex]\int e^x dx = x e^x - \int xe^x dx[/tex]
[tex]\int xe^x dx = \frac{x^2}{2}e^x - \int \frac{x^2}{2}e^x dx[/tex]
[tex]\int \frac{x^2}{2}e^x dx = \frac{x^3}{6}e^x - \int \frac{x^3}{6}e^x dx[/tex]
We see that this goes on and on to this:
[tex]\int e^x dx = e^x[/tex]
[tex]e^x = xe^x - \frac{x^2}{2}e^x + \frac{x^3}{6}e^x - \frac{x^4}{24}e^x...\frac{x^n}{n!}e^x[/tex]
[tex]e^x=e^x \cdot \sum^{\infty}_{k=1}\frac{x^k}{k!}(-1)^{k-1}[/tex]
[tex]\sum^{\infty}_{k=1}\frac{x^k}{k!}(-1)^{k-1} = 1[/tex]
And that's definitely NOT true! What is wrong?