Is there a proof for this?

by Heirot
Tags: proof
Heirot is offline
Mar25-09, 11:30 AM
P: 151

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?

Phys.Org News Partner Science news on
SensaBubble: It's a bubble, but not as we know it (w/ video)
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Microbes provide insights into evolution of human language
HallsofIvy is offline
Mar25-09, 12:44 PM
Sci Advisor
PF Gold
P: 38,900
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 is offline
Mar25-09, 01:40 PM
P: 4,499
If it's composition, then g better be a vector

Heirot is offline
Mar25-09, 04:32 PM
P: 151

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?

Register to reply

Related Discussions
Proof: Compare two integral(Please look at my surgested proof) Calculus & Beyond Homework 11
help with an proof Precalculus Mathematics Homework 10
Proof Calculus & Beyond Homework 5
Proof: One more irrationality proof :) Introductory Physics Homework 5
A proof is a proof---says Canadian Prime Minister General Math 0