An infinite series reduction is it true? I made a reduction of the following serie, but I wonder if this is okey:
$$w(x,t)= \sum^{\infty}_{k=0} \frac{1}{(2k)!}\frac{d^kg(t)}{dt^k} x^{2k} \space ,t >0 $$
where: $$g(t)=e^{-1/t^2} \space ,t>0$$
So $g(\frac{1}{\sqrt{t}})=e^{-t}$
So I can say that :
$$w(x,\frac{1}{\sqrt{t}}) = \sum^{\infty}_{k=0} \frac{1}{(2k)!}\frac{d^kg(\frac{1}{\sqrt{t}})}{dt^k} x^{2k}=
 \sum^{\infty}_{k=0} \frac{(-1)^ke^{-t} x^{2k}}{(2k)!} = e^{-t}\sum^{\infty}_{k=0} \frac{(-1)^k x^{2k}}{(2k)!}$$
but:
$$cos(x)=\sum^{\infty}_{k=0} \frac{(-1)^k x^{2k}}{(2k)!}$$
Hence:
$$w(x,\frac{1}{\sqrt{t}})=e^{-t}cos(x)$$
so we conclude that:
$$w(x,t)=e^{-1/t^2}cos(x)=g(t)cos(x)$$
this is very weird for me, but I think I didn't make any mistake? did I?
 A: Yes you did make a mistake.
Just because $$\frac{d}{dt}h(t)=f(t)$$
Doesn't mean we have that: $$\frac{d}{dt}h(\frac{1}{\sqrt{t}})=f(\frac{1}{\sqrt{t}})$$
If your going to make a transformation in the variable your differentiating with respect to, you also have to apply that same transformation to the variable under the differential operator. For example this would be true:
$$\frac{d}{d\frac{1}{\sqrt{t}}}h(\frac{1}{\sqrt{t}})=f(\frac{1}{\sqrt{t}})$$

Likewise you can't say that if:
$$w(x,t)= \sum^{\infty}_{k=0} \frac{1}{(2k)!}\frac{d^kg(t)}{dt^k} x^{2k}$$
Then we have: $$w(x,\frac{1}{\sqrt{t}})= \sum^{\infty}_{k=0} \frac{1}{(2k)!}\frac{d^kg(\frac{1}{\sqrt{t}})}{dt^k} x^{2k}$$
When you made the transformation $t\to\frac{1}{\sqrt{t}}$ the same transformation needed to be made under the differential operator, which if made would have given the real expression of $w(x,\frac{1}{\sqrt{t}})$ as:
$$w(x,\frac{1}{\sqrt{t}})= \sum^{\infty}_{k=0} \frac{1}{(2k)!}\frac{d^kg(\frac{1}{\sqrt{t}})}{d\frac{1}{\sqrt{t}}^k} x^{2k}$$
