# Can this series be represented by a function

• Oct 10th 2011, 12:17 PM
rainer
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)

Thanks
• Oct 11th 2011, 08:58 AM
HallsofIvy
Re: Can this series be represented by a function
Quote:

Originally Posted by rainer
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.

Quote:

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.

Quote:

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

Thanks