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Let's consider a map $L$ between complex matrices: $L(A) = kA + B$ for some real value $k$ and matrix $B$. I am trying to calculate $e^{tL}(A)$ for some real value $t>0$. Here is my attempt:

Since $L^n(A) = k^nA + \big(\sum_{i=0}^{n-1}k^i\big)B$, we have:

$e^{tL}(A) = A + tL(A) + \frac{t^2}{2!}L^2(A) + \frac{t^3}{3!}L^3(A) + \dotsc $

$= A + t(kA + B) + \frac{t^2}{2!}\big(k^2A + (k+1)B\big) + \frac{t^3}{3!}\big(k^3A + (k^2+k+1)B\big) + \dotsc $

$= \big(1 + k + \frac{(kt)^2}{2!} + \dotsc\big)A + \big(t + \frac{t^2}{2!}(k+1) + \frac{t^3}{3!}(k^2+k+1) + \dotsc \big)B$

$= e^{kt}A + \big(t + \frac{t^2}{2!}(k+1) + \frac{t^3}{3!}(k^2+k+1) + \dotsc \big)B$.

I am not sure how to calculate the coefficient for the matrix $B$, which is $\sum_{i=1}^\infty \frac{t^{i}}{{i}!} (\sum_{j=0}^{i-1} k^j)$.

Does it converge/diverge? If it diverges, is it possible to obtain a condition on $k$ such that it converges?

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1 Answer 1

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We have that by geometric series with $k\neq 1$

$$\sum_{j=0}^{i-1} k^j=\frac{k^i-1}{k-1}$$

and then

$$\sum_{i=1}^\infty \frac{t^{i}}{{i}!} \left(\sum_{j=0}^{i-1} k^j\right)=\sum_{i=1}^\infty \frac{t^{i}}{{i}!} \frac{k^i-1}{k-1}=\frac{1}{k-1}\sum_{i=1}^\infty \frac{(tk)^{i}}{{i}!}-\frac{1}{k-1}\sum_{i=1}^\infty \frac{t^{i}}{{i}!}=$$

$$=\frac{e^{tk}-1}{k-1}-\frac{e^t-1}{k-1}=\boxed {\frac{e^{tk}-e^t}{k-1}}$$


For the case $k=1$

$$\sum_{j=0}^{i-1} k^j=\sum_{j=0}^{i-1} 1=i$$

and then

$$\sum_{i=1}^\infty \frac{t^{i}}{{i}!} \left(\sum_{j=0}^{i-1} k^j\right)=\sum_{i=1}^\infty \frac{t^{i}}{{i-1}!}=t\sum_{i=1}^\infty \frac{t^{i-1}}{{i-1}!}=\boxed {te^t}$$

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  • $\begingroup$ And for $k=1$ it is $te^t$. $\endgroup$
    – user
    Jul 3, 2023 at 20:32
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    $\begingroup$ @user Thanks, I add also a derivation for that! $\endgroup$
    – user
    Jul 3, 2023 at 20:57

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