Series expansion by integrating

*Sweet 16*

If we define operator T be an integration from 0 to x we find:

Te^x = e^x - 1

1 = e^x - Te^x

= (1 - T)e^x

inverting:

e^x = 1/(1-T) . 1

= (1 + T + T² + T³ ...) . 1

= 1 + x + x²/2! + x³/3! ...

:pac:

Also

cosx = 1/(1+T²) . 1

sinx = 1/(1+T²) . x

*Sweet 16*

*Sweet 16* wrote: »

e^x = 1/(1-T) . 1

I don't think this is valid, but, I could be wrong. The right side of the equation is an operator while the left side is a function of x. I don't think that can be so, or am I mistaken?

Kryx

= (1 - T)e^x

Not sure you can do that, grouping operator and function!

The 1 in the bracket would have to become its equivillant in terms of T in the respected constraints. Can't be bothered doing it now.

Think thats right

Fremen

Is there a question here or did you just feel like dropping some maths?

This kind of thing pops up in deriving numerical schemes for the heat diffusion equation. You can take formal power series of operators rigorously, though all I know is that "It works". Can't say I've ever seen a mathematical rationale for the idea.

I guess "1" just becomes the identity operator rather than a function here.