Thread: vexingly simple proof (sum periodic functns)

I have a feeling this proof is supposed to be more complex, is this legitimate?
Let f(x) and g(x) be continuous, periodic functions defined for -infinity<x<+infinity. If limx->+infinity (f(x)-g(x))=0 prove that f(x):=g(x).

Proof:
Let h(x)=f(x)-g(x).

-g(x) is continuous as it is the product of continuous functions t(x)=-1 and g(x). h(x) is continuous as it is the sum of continuous functions f(x) and -g(x).

Therefore, (f(x)-g(x))=limx->+infinity (f(x)-g(x))=0 for all x>N where N is the N defined in "if limx->+infinity (f(x)-g(x))=0 <=> for any epsilon>0, if x>N then abs((f(x)-g(x))-0)<epsilon."

Now, f(x)-g(x)=0 for all x>N.

If f(x) has period p then f(x+p)-g(x+p)=0 for all x+p>x>N.

Now, f(x)-g(x)=0=f(x+p)-g(x+p) for all x+p>x>N.

Hence f(x)-f(x+p)-g(x)=-g(x+p) for all x+p>x>N.

But f(x)=f(x+p) as p is a period of f(x); therefore, -g(x)=-g(x+p) i.e. g(x)=g(x+p) for all x+p>x>N.

All x+p>x>N includes all (x+(m/q)b)+p>x>N where (m/q) is any rational number and b is the base period of g(x).Therefore, p must be a period of the periodic function g(x).

Now, for any integer k such that (x-kp)<N i.e. k>((x-N)/p), f(x)=f(x-kp) and g(x)=g(x-kp) [as p is a period of both f(x) and g(x)].

Therefore, 0=f(x)-g(x)=f(x-kp)-g(x-kp) for all x>N, i.e. for all (x-kp)<N.

Therefore, f(x-kp)-g(x-kp)=0 for all x>N, i.e. for all (x-kp)<N; and f(x)-g(x)=0 for all x>N.

Note that the same logic follows if k is replaced by some integer z such that (x-z)<<N i.e. z>>((x-N)/p) [just to reaffirm that this proof holds for x approaching -infinity aswell].

Therefore, f(x)=g(x) for all x.

Finally, as f(x) and g(x) have the same domain, -infinity<x<+infinity, and as f(x) and g(x) also have the same values at each x in their domain, i.e. f(x)=g(x) for all x in -infinity<x<+infinity, it must be true that f(x):=g(x). /QED

Is my proof faulty because I didn't prove f(N)=g(N) so I didn't really prove f(x)=g(x) for ALL x? Could I fix this by defining k as any integer such that k>or=((x-N)/p)?

Thanks for reading

Originally Posted by 234578

I have a feeling this proof is supposed to be more complex, is this legitimate?
Let f(x) and g(x) be continuous, periodic functions defined for -infinity<x<+infinity. If limx->+infinity (f(x)-g(x))=0 prove that f(x):=g(x).

Proof:
Let h(x)=f(x)-g(x).

-g(x) is continuous as it is the product of continuous functions t(x)=-1 and g(x). h(x) is continuous as it is the sum of continuous functions f(x) and -g(x).

Therefore, (f(x)-g(x))=limx->+infinity (f(x)-g(x))=0 for all x>N where N is the N defined in "if limx->+infinity (f(x)-g(x))=0 <=> for any epsilon>0, if x>N then abs((f(x)-g(x))-0)<epsilon."

Now, for all x>N.

How so?

If f(x) has period p then f(x+p)-g(x+p)=0 for all x+p>x>N.

Now, f(x)-g(x)=0=f(x+p)-g(x+p) for all x+p>x>N.

Hence f(x)-f(x+p)-g(x)=-g(x+p) for all x+p>x>N.

But f(x)=f(x+p) as p is a period of f(x); therefore, -g(x)=-g(x+p) i.e. g(x)=g(x+p) for all x+p>x>N.

All x+p>x>N includes all (x+(m/q)b)+p>x>N where (m/q) is any rational number and b is the base period of g(x).Therefore, p must be a period of the periodic function g(x).

Now, for any integer k such that (x-kp)<N i.e. k>((x-N)/p), f(x)=f(x-kp) and g(x)=g(x-kp) [as p is a period of both f(x) and g(x)].

Therefore, 0=f(x)-g(x)=f(x-kp)-g(x-kp) for all x>N, i.e. for all (x-kp)<N.

Therefore, f(x-kp)-g(x-kp)=0 for all x>N, i.e. for all (x-kp)<N; and f(x)-g(x)=0 for all x>N.

Note that the same logic follows if k is replaced by some integer z such that (x-z)<<N i.e. z>>((x-N)/p) [just to reaffirm that this proof holds for x approaching -infinity aswell].

Therefore, f(x)=g(x) for all x.

Finally, as f(x) and g(x) have the same domain, -infinity<x<+infinity, and as f(x) and g(x) also have the same values at each x in their domain, i.e. f(x)=g(x) for all x in -infinity<x<+infinity, it must be true that f(x):=g(x). /QED

Is my proof faulty because I didn't prove f(N)=g(N) so I didn't really prove f(x)=g(x) for ALL x? Could I fix this by defining k as any integer such that k>or=((x-N)/p)?

Thanks for reading

Originally Posted by 234578

I have a feeling this proof is supposed to be more complex, is this legitimate?
Let f(x) and g(x) be continuous, periodic functions defined for -infinity<x<+infinity. If limx->+infinity (f(x)-g(x))=0 prove that f(x):=g(x).

Here is a short proof (and I'm like Failure, I don't understand yours).

Let be the periods of respectively. For any , we have (where ), hence . Therefore, we have (taking as in the previous formula). But we also have using the definition of . As a consequence, comparing the limits, we have .

Thus is a period for as well. Then the conclusion is straightforward: for all , hence .

NB: I did not even use the continuity of ...

Thanks Just to clarify, does the symbol mean the limit of f(x)-g(x) as x approaches infinity is zero? Or does it mean as x approaches k, in which case I was wrong to think I understood this proof--why is k in the limit sign? (sorry if I sound ignorant)

Originally Posted by Laurent

Here is a short proof (and I'm like Failure, I don't understand yours).

Let be the periods of respectively. For any , we have (where ), hence . Therefore, we have (taking as in the previous formula). But we also have using the definition of . As a consequence, comparing the limits, we have .

Thus is a period for as well. Then the conclusion is straightforward: for all , hence .

NB: I did not even use the continuity of ...

Originally Posted by 234578

Thanks Just to clarify, does the symbol mean the limit of f(x)-g(x) as x approaches infinity is zero? Or does it mean as x approaches k, in which case I was wrong to think I understood this proof--why is k in the limit sign? (sorry if I sound ignorant)

This is the limit of the sequence indexed by . Another notation is .

In the case of my last limit (i.e. ), no appears in the "sequence"; therefore it is a constant sequence, and its limit is its common value, . The idea is that by periodicity , which now depends on . And since , we may apply the hypothesis: we have (the limit when is equal to the limit along any sequence tending to : if the limit exists).