1
$\begingroup$

$f(x)=\sum_{n=-\infty}^\infty e^{nx} \longrightarrow f(x) = \infty\text{ for all }x\text{ values}$

\frac{y f''(x)}{1!}=\sum_{n=-\infty}^\infty \frac {y n^2 e^{nx}}{1!}

$\frac{y^2 f^{(4)}(x)}{2!}=\sum_{n=-\infty}^\infty \frac {y^2 n^4 e^{nx}}{2!}$

$\frac{y^3 f^{(6)}(x)}{3!}=\sum_{n=-\infty}^\infty \frac {y^3 n^6 e^{nx}}{3!}$

$ \vdots $

{g(x,y)=f(x)+\frac{y.f''(x)}{1!}+\frac{y^2 f^{(4)}(x)}{2!}+\cdots=\sum_{n=-\infty}^\infty e^{nx+n^2 y}} \text{ if }y=x/2

$ g(x,x/2)=\sum_{n=-\infty}^\infty e^{\frac{x[(n+1)^2 - 1]}{2}}$

$ g(x,x/2)=e^{-x/2}\cdot\sum_{n=-\infty}^\infty e^{\frac{x(n+1)^2}{2}}=e^{-x/2}\cdot\sum_{n=-\infty}^\infty e^{\frac{xn^2}{2}}$

this result has values for all $x<0$

My question:

How could we get such result although we produced $g(x,x/2)$ from $f(x)$ that doesn't have any defined values for $x<0$?

  • 0
    you are correct, thanks.2012-01-15

2 Answers 2

6

$f(x):=\sum_{n=0}^\infty x\;,$

$g(x,y):=f(y) - f(x)\;,$

$g(x,x)=f(x)-f(x)=0\;.$

How could we get this result although we produced $g(x,x)$ from $f(x)$ that doesn't have any defined values for $x\ne0$? You don't need any fancy exponential series to get this effect; examples of this kind abound. It just goes to show that there's no reason to expect formal manipulations of divergent series to have any meaning, and in particular to expect the resulting series to diverge.

3

If $f(x)$ is infinite for all real $x$, then does it make sense to take a derivative of this function?