Property Function exp

juaninf

How can prove this

[tex]\exp(At)\exp(-At_0)=\exp(A(t-t_0))[/tex]?

using [tex]\displaystyle\sum_{i=0}^n{(1/k!)A^kt^k}[/tex]

and this properties
in t=0
[tex]
[\exp(At)]_{t} = I
[/tex]


[tex]exp(At)exp(-At)=I[/tex]
[tex]\frac{dexp(At)}{dt}=Aexp(At)=exp(At)A[/tex]

ross_tang

For your first equation, please refer to this question in Voofie/Mathematics:

How to prove e^At e^(-At_0) = e^A(t-t_0)?

I think you typed wrong in this formula:

[tex]exp(At)_t=0 = I[/tex]

0 is not equal to I. And your what's your meaning of [tex]exp(At)_t[/tex]?

For this one [tex]exp(At)exp(-At)=I[/tex], you can use my result to prove easily. For the last one, you should try to use the power series expansion and differentiate term by term. You will get the answer easily too.

Petr Mugver

To conclude, i suppose you mean

[tex][\exp(At)]_{t=0} = I[/tex]

Well, it's pretty simple:

[tex][\exp(At)]_{t=0} = \left[\sum_{k=0}^\infty\frac{A^kt^k}{k!}\right]_{t=0}=I+0+0+\cdots=I[/tex]

juaninf

fix question my question is
How prove this,
[tex]
\exp(At)\exp(-At_0)=\exp(A(t-t_0))
[/tex]
using as above properties

juaninf

Thank,I am reading this web
http://www.voofie.com/content/152/ho...-at_0-eat-t_0/

but i dont understand how change sumatoria infinite to finite, Where i can read this?

ross_tang

I have answered your question. Please have a look.

How to prove e^At e^(-At_0) = e^A(t-t_0)?