Question

Is there a simple way to express this? \[\Large (A+B)^k\]

Answer 1

\[\Large (A+B)^k\] A and B are matrices and k is positive integer. We can't assume that A and B commute.

Answer 2

how can A+B not commute

Answer 3

They're matrices...

Answer 4

matrices commute under addition

Answer 5

\[\Large (A+B)^2 \ne A^2+2AB+B^2\]

Answer 6

Oh i thought u meant it cant be written (B+A)^k

Answer 7

how about diagonlizing both

Answer 8

Hmm well that's a good idea but technically I'm not really interested in matrices I'm interested in noncommuting operators and thought this question would be a simple analogy to see if we could figure out a binomial theorem to use instead.

So can we do it without diagonalizing them, but still have some kind of binomial theorem form? haha

Answer 9

what happens when you expand
( A + B ) * ( A + B ) using matrix multiplication rules distributivity

Answer 10

is that equal to
(A+B) * A + (A+B)*B

A*A + B*A + A*B + B*B ?

Answer 11

Ok so we can conclude this
Can ::) assune they are diagonalizable?

Answer 12

u can do that perl

Answer 13

My real question is finding the general form of this derivative, I expressed it as this \[\Large \dfrac{d^k}{dx^k}\left( e^{f(x)} \right)=e^{f(x)}(f'(x)\cdot+\frac{d}{dx}\cdot)^k\] So when k=1 we have: \[\large e^{f(x)} (f'(x)+\frac{d}{dx})1=e^{f(x)} (f'(x)*1+\frac{d}{dx}1)=e^{f(x)}f'(x)\]

Answer 14

how to write a general form for (A+B)^k is what we gotta find

Answer 15

Yeah you are right @perl on distributing it out, it is A^2+AB+BA+B^2

Answer 16

ok i see , so we can't assume A, and B commute under multiplication, so it doesn't simplify to A^2 + 2AB + B^2

Answer 17

For comparison, \[e^{f(x)} (f'(x)+\frac{d}{dx})(f'(x)+\frac{d}{dx})1 = e^{f(x)}(f'(x)^2+f''(x)+f'(x)\frac{d}{dx}+\frac{d^2}{dx^2}) \\ e^{f(x)}(f'(x)^2+f''(x))=\frac{d^2}{dx^2}(e^{f(x)})\]

Answer 18

\(e^A=\phi(A)\)

Answer 19

So yeah, the A and B thing is really a lot simpler than all this nonsense, unless you can figure out how to diagonalize either of these, multiplication by f'(x) or the derivative operator.

Answer 20

such that \(\phi(A)\) is a fundamental solution matrices

Answer 21

Hmm well here f(x) is just a regular function of x, not a matrix at all.

Answer 22

hmm :\

Answer 23

Really, the general form of this derivative is what I'm after where f(x) is a function. \[\Large \frac{d^k}{dx^k}(e^{f(x)})\] So if you can help me solve this or the matrix problem would be helping me the same haha.

Answer 24

I think repeated logarithmic differentiation would work pretty well here.

Answer 25

Oh that sounds promising, can you explain what you mean by that?

Answer 26

Do you know what logarithmic differentiation is?

Answer 27

Yes, but I'm not sure how I would apply it to this situation.

Answer 28

Oh another fact, we can assume I already know all the derivatives of f(x), meaning f'(x), f''(x), f'''(x), etc...

Answer 29

Nothing special. Just brute force some logarithmic derivative for a few times and try to find a closed form solution lol.

Answer 30

Yeah brute force how, I'm not really seeing that lol.

Answer 31

the only way i see is expanding :|

Answer 32

Oh like a power series? I'll try that

Answer 33

and we can state special cases:- (it simplify it very good)
A,B diagonalizable
A,B upper triangle (or the other type)
A,B diagonal

Answer 34

i'll take snap for my work

Answer 35

Hmm well I don't really have a clue how we could say multiplication by f'(x) is a diagonal matrix or that the derivative of a function is a upper triangular matrix or something like that... Unless maybe we somehow use a polynomial with an infinite matrix...

Answer 36

hmm no it work with n*n hmm
but i see f(x) and f'(x) as vectors not scalar

Answer 37

By the way, the function f(x) has its derivatives defined as this: \[\Large f^{(k)}(x) = \frac{a}{(x+b)^k}\] for k>0

Answer 38

too much work -.-

Answer 39

Wait show me what you have @Marki it might be helpful since I figured out a way to make an upper triangular matrix out of the derivative operator

Answer 40

i just dumbly expanded it

Answer 41

For reference:
\[
\begin{align*}
\frac{d}{dx}e^{g(x)}=&\,\,e^{g(x)} g'(x)\\
\frac{d^2}{dx^2}e^{g(x)}=&\,\,e^{g(x)} g''(x)+e^{g(x)} g'(x)^2\\
\frac{d^3}{dx^3}e^{g(x)}=&\,\,e^{g(x)} g^{(3)}(x)+e^{g(x)} g'(x)^3+3 e^{g(x)} g'(x) g''(x)\\
\frac{d^4}{dx^4}e^{g(x)}=&\,\,e^{g(x)}
g^{(4)}(x)+3 e^{g(x)} g''(x)^2+e^{g(x)} g'(x)^4+4 e^{g(x)} g^{(3)}(x)
g'(x)\\&+6 e^{g(x)} g'(x)^2 g''(x)\\
\frac{d^5}{dx^5}e^{g(x)}=&\,\,e^{g(x)} g^{(5)}(x)+e^{g(x)} g'(x)^5+5
e^{g(x)} g^{(4)}(x) g'(x)\\
&+10 e^{g(x)} g^{(3)}(x) g''(x)+10 e^{g(x)}
g^{(3)}(x) g'(x)^2+10 e^{g(x)} g'(x)^3 g''(x)\\
&+15 e^{g(x)} g'(x)
g''(x)^2\\
\frac{d^6}{dx^6}e^{g(x)}=&\,\,e^{g(x)} g^{(6)}(x)+10 e^{g(x)} g^{(3)}(x)^2+15 e^{g(x)}
g''(x)^3+e^{g(x)} g'(x)^6\\
&+6 e^{g(x)} g^{(5)}(x) g'(x)+15 e^{g(x)}
g^{(4)}(x) g''(x)+15 e^{g(x)} g^{(4)}(x) g'(x)^2\\
&+20 e^{g(x)} g^{(3)}(x)
g'(x)^3+15 e^{g(x)} g'(x)^4 g''(x)+45 e^{g(x)} g'(x)^2 g''(x)^2\\
&+60
e^{g(x)} g^{(3)}(x) g'(x) g''(x)
\end{align*}
\]

Answer 42

\[\Large \left[\begin{matrix}0 & 1 &0 & 0 \\ 0 & 0 &2 & 0 \\ 0 & 0 &0 & 3 \\ 0 & 0 &0 & 0 \end{matrix}\right] \left[\begin{matrix}1 \\x \\ x^2 \\ x^3\end{matrix}\right]=\left[\begin{matrix}1 \\x \\ x^2 \\ x^3\end{matrix}\right]\] Ok I don't really know how to make the notation correct to saywhat I mean but I have seen this before but never really used it. I think I need to switch these out with the basis and put the coefficients in here, but that's kind of difficult

Answer 43

omg the latex is high

Answer 44

for triangle
(A+B)^k

Answer 45

typo