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Mmmh, I've found another way that might work in some situations: use the exponential of the matrix:

$$e^M=\sum_{k\geq 0} \frac{1}{k!}M^k$$

then use derivatives to recover the powers of the matrix:

$$\left(\frac{\partial}{\partial k} \right)^k e^M = M^k$$

For example:

sage:  B = matrix(SR, 2, [1,2,3,4])
...    var('t')
...    f = exp(t*B)[1,1]
sage:  print n(f.derivative(t,4)(t=0))
634.000000000000
sage:  print n((B^4)[1,1])
634.000000000000

In order words, the function f = exp(t*B)[1,1] has as derivatives the coeffient [1,1] of M^k. Maybe that works for you. This could be useful in combination with LazyPowerSeries, as seen in the sage book in french.

Mmmh, I've found another way that might work in some situations: use the exponential of the matrix:

$$e^M=\sum_{k\geq 0} \frac{1}{k!}M^k$$

then use derivatives to recover the powers of the matrix:

$$\left(\frac{\partial}{\partial k} \right)^k e^M = M^k$$

For example:

sage:  B = matrix(SR, 2, [1,2,3,4])
...    var('t')
...    f = exp(t*B)[1,1]
sage:  print n(f.derivative(t,4)(t=0))
634.000000000000
sage:  print n((B^4)[1,1])
634.000000000000

In order words, the k-th order derivative of the function f = exp(t*B)[1,1] has as derivatives is $\frac{1}{k!}$ of the coeffient [1,1] coefficient of M^k. Maybe that works for you. solves your problem. This could be useful in combination with LazyPowerSeries, as seen in the sage book in french.

Mmmh, I've found another way that might work in some situations: use the exponential of the matrix:

$$e^M=\sum_{k\geq 0} \frac{1}{k!}M^k$$

then use derivatives to recover the powers of the matrix:

$$\left(\frac{\partial}{\partial k} \right)^k e^M = M^k$$

For example:

sage:  B = matrix(SR, 2, [1,2,3,4])
...    var('t')
...    f = exp(t*B)[1,1]
sage:  print n(f.derivative(t,4)(t=0))
634.000000000000
sage:  print n((B^4)[1,1])
634.000000000000

In order words, the k-th order derivative of the function f = exp(t*B)[1,1] at zero is $\frac{1}{k!}$ of the [1,1] coefficient of M^k. B^k. Maybe that solves your problem. This could be useful in combination with LazyPowerSeries, as seen in the sage book in french.

Mmmh, I've found another way that might work in some situations: use the exponential of the matrix:

$$e^M=\sum_{k\geq 0} \frac{1}{k!}M^k$$

then use derivatives to recover the powers of the matrix:

$$\left(\frac{\partial}{\partial k} \right)^k e^M = M^k$$

For example:

sage:  B = matrix(SR, 2, [1,2,3,4])
...    var('t')
...    f = exp(t*B)[1,1]
sage:  print n(f.derivative(t,4)(t=0))
634.000000000000
sage:  print n((B^4)[1,1])
634.000000000000

In order words, the k-th order derivative of the function f = exp(t*B)[1,1] at zero is the [1,1] coefficient of B^k. Maybe that solves your problem.

This could be useful in combination with LazyPowerSeries, as seen in the sage book in french.

Mmmh, I've found another way that might work in some situations: use the exponential of the matrix:

$$e^M=\sum_{k\geq 0} \frac{1}{k!}M^k$$

then use derivatives to recover the powers of the matrix:

$$\left(\frac{\partial}{\partial k} \right)^k e^M = M^k$$

For example:

sage:  B = matrix(SR, 2, [1,2,3,4])
...    var('t')
...    f = exp(t*B)[1,1]
sage:  print n(f.derivative(t,4)(t=0))
634.000000000000
sage:  print n((B^4)[1,1])
634.000000000000

In order words, the k-th order derivative of the function f = exp(t*B)[1,1] at zero is the [1,1] coefficient of B^k. Maybe that solves your problem.

This could be useful in combination with LazyPowerSeries, as seen in the sage book in french.

Once you've got the function $f = (e^M)_{ij}$, you could get the function k -> k-th [k-th order derivative of $f$ $f$] = (i,j) [(i,j) entry of $M^k$ $M^k$] by Fourier transform, product with $\psi^k$, and inverse Fourier transform. But those integrals may be non-trivial...