Is there a proof for this?

Heirot

Hello,

let's say we have two functions of two variables: f(x,y) and g(x,y). Say we know that the sum / integral over all y's of f^n * g does not depend on x for every natural number n (and zero). Does that mean that f and g both don't depend on x?

Thanks

HallsofIvy

Not necessarily. Let g(x,y)= 0 for all x and y. Then f^n*g (I assume you mean composition) is f^n(0) for all x and y so the sum/integral is a constant no matter what f is. If f^n*g is ordinary multiplication of functions, then f^n*g= 0 for all x and y and again, the integral is a constant no matter what f is.

Office_Shredder

If it's composition, then g better be a vector

Heirot

Sorry for not being clear - f(x,y) and g(x,y) are scalar functions and * is ordinary multiplication. f^n is then f multiplied n times by itself. Now, if we don't assume the trivial null solution, f(x,y)=0 or g(x,y)=0, does my statement hold?

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Heirot	Is there a proof for this? Tweet Hello, let's say we have two functions of two variables: f(x,y) and g(x,y). Say we know that the sum / integral over all y's of f^n * g does not depend on x for every natural number n (and zero). Does that mean that f and g both don't depend on x? Thanks

HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	Not necessarily. Let g(x,y)= 0 for all x and y. Then f^ng (I assume you mean composition) is f^n(0) for all x and y so the sum/integral is a constant no matter what f is. If f^ng is ordinary multiplication of functions, then f^n*g= 0 for all x and y and again, the integral is a constant no matter what f is.

Office_Shredder Blog Entries: 1 Recognitions: Homework Help	If it's composition, then g better be a vector

Heirot	Is there a proof for this? Sorry for not being clear - f(x,y) and g(x,y) are scalar functions and * is ordinary multiplication. f^n is then f multiplied n times by itself. Now, if we don't assume the trivial null solution, f(x,y)=0 or g(x,y)=0, does my statement hold?