- #1

ChrisVer

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- Homework Statement:
- Show that you can write [itex]e^{\textbf{A}t} = f_1(t) \textbf{1} + f_2(t) \textbf{A} [/itex] and determine [itex]f_{1},~f_{2}[/itex] if [itex]\textbf{A}= \begin{pmatrix} a & b \\ 0 & a \end{pmatrix}[/itex]

- Relevant Equations:
- [itex]e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!} [/itex]

Hi, I think this is a nitpicking question, but oh well let me hear your inputs.

Actually I tried to solve this question straightforwardly, by Taylor expanding the exponential and showing that:

[itex] \textbf{A}^n = \begin{pmatrix} a^n & nba^{n-1} \\ 0 & a^n \end{pmatrix} [/itex]

i.e.

[itex]e^{\textbf{A}t} = \sum_{n=0}^{\infty} \frac{\textbf{A}^n t^n}{n!}= \begin{pmatrix} \sum_{n=0}^{\infty} \frac{a^nt^n}{n!} & bt \sum_{n=0}^{\infty} \frac{na^{n-1} t^{n-1}}{n!} \\ 0 & \sum_{n=0}^{\infty} \frac{a^nt^n}{n!} \end{pmatrix} [/itex]

(forget [itex]t[/itex] for now)

The problem I faced was related to the top-right element of the matrix when summed. As the exponential gets expanded, it gives me a sum running from n=0 to infinity. However that term results to negative values of n when n=0:

[itex]\sum_{n=0}^{\infty} b n \frac{a^{n-1}}{n!} = b \sum_{n=0}^{\infty} \frac{a^{n-1}}{(n-1)!} [/itex]

Which doesn't make much sense. However, I don't know how I could emit the term with [itex]n=0[/itex]. I could try to say that [itex]m = n-1[/itex] meaning it would change to [itex] \sum_{m=-1}^{\infty} \frac{a^{m}}{m!} [/itex], but I have the exact same problem.

(I already know how the matrix is written and that term should be a [itex]be^{a}[/itex])

Actually I tried to solve this question straightforwardly, by Taylor expanding the exponential and showing that:

[itex] \textbf{A}^n = \begin{pmatrix} a^n & nba^{n-1} \\ 0 & a^n \end{pmatrix} [/itex]

i.e.

[itex]e^{\textbf{A}t} = \sum_{n=0}^{\infty} \frac{\textbf{A}^n t^n}{n!}= \begin{pmatrix} \sum_{n=0}^{\infty} \frac{a^nt^n}{n!} & bt \sum_{n=0}^{\infty} \frac{na^{n-1} t^{n-1}}{n!} \\ 0 & \sum_{n=0}^{\infty} \frac{a^nt^n}{n!} \end{pmatrix} [/itex]

(forget [itex]t[/itex] for now)

The problem I faced was related to the top-right element of the matrix when summed. As the exponential gets expanded, it gives me a sum running from n=0 to infinity. However that term results to negative values of n when n=0:

[itex]\sum_{n=0}^{\infty} b n \frac{a^{n-1}}{n!} = b \sum_{n=0}^{\infty} \frac{a^{n-1}}{(n-1)!} [/itex]

Which doesn't make much sense. However, I don't know how I could emit the term with [itex]n=0[/itex]. I could try to say that [itex]m = n-1[/itex] meaning it would change to [itex] \sum_{m=-1}^{\infty} \frac{a^{m}}{m!} [/itex], but I have the exact same problem.

(I already know how the matrix is written and that term should be a [itex]be^{a}[/itex])