## 35 Reputation

6 years, 160 days

## @Nikol  the main problem is the co...

the main problem is the complexity of the function f. it is considered very long for 256 by 256. It is the calculation itself that is long.
as an example
, if for example to build plot3d(f,...) then it takes about 20 seconds for 800 points.

## @Carl Love Thanks/ I would lik...

Thanks/ I would like to use in the best scenario nonlinear case/

A*f(B*x,C*y). Scale parameter

## @mmcdara  I use the term copula in...

I use the term copula in the sense of the term pattern.  A template that can be used with scale and offset options.

## @mmcdara Any procedure that gives two n...

@mmcdara

Any procedure that gives two numbers x and y a third z is a function.  It matters what procedure it is.  It must be unambiguous.  What's inside the procedure (x, y) -> z is not important.

## @mmcdara There is 2D spectrum as histog...

@mmcdara

There is 2D spectrum as histogram. 256 x 256 channels. I fit the two-dimensional spectrum with the function f(x, y) := Lowess(mas,fitorder=2,bandwidth=0.07). Now I've got a function f. Next step I try use NonlinearFit(a*f(x, y), mas2, [a],... ). Apparently the function f is considered very slowly.  And my procedure can't issue a solution, it just freeze

## @Nikol  This works well : master ...

This works well :

master := Lowess(<\$ (5 .. 256)>, convert(qw(1, 52 - 45 .. 258), Vector), fitorder = 2, bandwidth = 0.2);

NonlinearFit(B*master(t/a), <(\$10..235)>, wer, t, initialvalues = [a=1], output = [parametervalues]):

## @tomleslie  Thanks for the reply. ...

Thanks for the reply. I'll upload the sheet later. But you're wrong. Lowes can be used in NOnLinearfit. I get a great fit in the one-dimensional case. like B*f(A*x) where f(x):=Lowess()

## @vv Thanks! Why does this happen?...

@vv Thanks!

Why does this happen?

## @epostma  Thank you very match...

Thank you very match

# Is it a bug?

## @vv  Thank you very much...

Thank you very much

## Maple disappoint me/...

Maple disappoints me/

## rsolve({f(n)=0.5*f(n-1)+0.5*f(n+1), f(0)...

rsolve({f(n)=0.5*f(n-1)+0.5*f(n+1), f(0)=1, f(a)=0},f(n));

with parameter is it impossible?

## I think it's just a defect. it's so ...

I think it's just a defect.

it's so obvious problem

## I have rsolve({ f(n+1) = f(n)+c,f(0) = 0...

I have

rsolve({ f(n+1) = f(n)+c,f(0) = 0}, {f});

return {f(n) = c (n + 1) - c}

It's a TRUE

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