Can this series be represented by a function

Hello,

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

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

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

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

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

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

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

Thanks

Re: Can this series be represented by a function

Quote:

Originally Posted by

**rainer** Hello,

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

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

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

No, it isn't. The "jth integral of x" is $\displaystyle \frac{x^{j}}{j!}$, not $\displaystyle \frac{x^j}{j}$.

The **exponential** function, $\displaystyle e^x$, has the property you are talking about.

Quote:

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

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

where $\displaystyle f^{(j)}(x)$ 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 $\displaystyle \lim_{j \to \infty}f^{(j)}(x)=1$ or something like that)

Thanks