You could try it in the 1d case and know soon enough if it works.

du/dx = u, u(0) = 1.

aka GO/NOGO

- Cuchulainn
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You could try it in the 1d case and know soon enough if it works.

du/dx = u, u(0) = 1.

aka GO/NOGO

du/dx = u, u(0) = 1.

aka GO/NOGO

I'd rather not go down the ODE solving route

- Cuchulainn
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The reason being?I'd rather not go down the ODE solving route

What about a home-grown solver?

Isn't it going to be slower than an approximant? I will be calculating these derivatives in an inner loop.

- katastrofa
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It's not a Padé approximant.Take the Pade (0,1) (or was it (1,0) ?) [$]e^x[$] approximate by [$]1+x[$] on [-1,1]I thought so...No.

1: Compute maximum error using 101 calculus

2. Compute derivative

3. GOTO 1

Gets worser and worser.

exp x = exp(x/2) / exp(-x/2) for Padé.

- Cuchulainn
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ItIt's not a Padé approximant.Take the Pade (0,1) (or was it (1,0) ?) [$]e^x[$] approximate by [$]1+x[$] on [-1,1]I thought so...

1: Compute maximum error using 101 calculus

2. Compute derivative

3. GOTO 1

Gets worser and worser.

exp x = exp(x/2) / exp(-x/2) for Padé.

https://en.wikipedia.org/wiki/Pad%C3%A9_table

Not sure what your above identity is telling me.

- katastrofa
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There's no such a thing as Padé(.,0). It would kill the whole idea of the approximant. My identity was a hint how to calculate the Padé approximant of exp. Nevermind, I need to go and do something mathematically sound now.

Last edited by katastrofa on July 25th, 2018, 11:09 am, edited 1 time in total.

- Cuchulainn
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Read the Wiki link.

The identity is cute but far away from rational approximation (IMO). But I might be wrong.Or you.

*Feller was an ebullient man, who would rather be wrong than undecided. *

The identity is cute but far away from rational approximation (IMO). But I might be wrong.Or you.

- katastrofa
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We essentially don't agree whether the Taylor series is a Padé approximant. I think the motivation for the latter is to achieve(/extend the radius of) or accelerate the convergence of the first, hence it has the form of a quotient of two Taylor series expansions:

[$]\sum_{i=0}^M a_i x^i / ( 1 + \sum_{j=1}^N a_j x^j )[$]

Thus the Pade approximant of exp x should be derived from exp(x/2) / exp(-x/2) = T(x/2) / T(-x/2).

I think ISayMoo's could use Padé approximant to calculate exp(A), but needs to assure that the norm of A is low (e.g. by applying the Gershgorin - a.k.a. some Belarusian guy who's name I forgot - circle theorem, which we have once intensely discussed in the forum with Lovenatalya).

BTW, why do you use the word "cute" so often when we talk?

[$]\sum_{i=0}^M a_i x^i / ( 1 + \sum_{j=1}^N a_j x^j )[$]

Thus the Pade approximant of exp x should be derived from exp(x/2) / exp(-x/2) = T(x/2) / T(-x/2).

I think ISayMoo's could use Padé approximant to calculate exp(A), but needs to assure that the norm of A is low (e.g. by applying the Gershgorin - a.k.a. some Belarusian guy who's name I forgot - circle theorem, which we have once intensely discussed in the forum with Lovenatalya).

BTW, why do you use the word "cute" so often when we talk?

Last edited by katastrofa on July 26th, 2018, 12:32 pm, edited 1 time in total.

- Cuchulainn
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This is the first time I ever used that word Cute foto.

*Thus the Pade approximant of exp x should be derived from exp(x/2) / exp(-x/2) = T(x/2) / T(-x/2).*

What's 'T'?

What's 'T'?

Last edited by Cuchulainn on July 25th, 2018, 6:18 pm, edited 2 times in total.

- katastrofa
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It's *relatively* often then.

T is for Taylor expansion.

T is for Taylor expansion.

- Cuchulainn
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What about somefing on the lines of BCH?Simple question which stumps me: I have a complex square matrix H. There are some nice methods for calculating exp(H). What about calculating the derivative of exp(H) over elements of H? To be precise: let M = exp(H). I want to calculate dM_{jk} / dH_{mn} numerically, accurately and (relatively) quickly.

https://math.stackexchange.com/question ... -of-matrix

A successful implementation would improve the code (e.g. maintainability and readability) for backpropagation etc.

How many terms would I have to keep? I suppose I can bound it myself, but I'm lazy.

Higham's method looks v. appealing to me now.

Higham's method looks v. appealing to me now.

- katastrofa
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Is what you call Higham's method simply a complex step derivative, which has been used by statisticians for sensitivity analysis since complex was added in the C99 standard? Who's Higham?

- Cuchulainn
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https://www.genealogy.math.ndsu.nodak.e ... p?id=96012

*complex step derivative, which has been used by statisticians for sensitivity analysis*

Interesting. So, statisticians discovered it?

BTW Fortran has complex type forever. C99 is playing catch-up.

Interesting. So, statisticians discovered it?

BTW Fortran has complex type forever. C99 is playing catch-up.

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