Hey guys,
I need to prove this claim. Does anybody have an idea how to go about it?
Functions of matrices
Bentley White
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Mason Long
That last property implies [math]\| A^n\| \leq \|A\|^n[/math].
Pretty trivial from there.
Adam Hernandez
I mean the claim that the power series coverges to f(A). Also I should mention I have only had engineering math education. Thanks so far.
Joseph Thompson
Show the sequence of partial sums is Cauchy. This avoids having to find the limit, if you have completeness. Use the standard power series in comparison to bound tails.
Carson Butler
What is the a-priori meaning of f(A)? Typically the power series is used to define sin(A) for example. On the other hand, exp(A) may have another definition that you are trying to tie to the power series.
Joseph Rodriguez
>>sequence of partial sums
which
>>show it cauchy
using frobenius norm?
sorry for the stupid questions
Charles Nguyen
Write out the definition of convergence for both types of series.
Use the triangle inequality, , and the conditions for the first series to show the second one converges too.
Gabriel Davis
effectively trying to solve:
f(A) = \sqrt{1+A}
Brody Cooper
ok I think I get it.
The sum of a norm is larger than a norm of a sum. (triangle inequality)
also the power of a norm is larger or eq. than norm of a power. ()
assuming the condition for the series of x element of the reals^1 hold and assuming two things that can only make it worse (in terms of covergence), convergence with respect to said norm follows.
This is nice thank you. I also would like to show it converges to said \sqrt{I+A}
Alexander Moore
sorry for the bad posting habit.
trying to show that the taylor series converges to this:
[math] \sqrt{I+a} [/math]