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Math Help - vexingly simple proof (sum periodic functns)

  1. #1
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    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
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by 234578 View Post
    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){\color{red}=}0 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
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  3. #3
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    Quote Originally Posted by 234578 View Post
    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 T_f,T_g be the periods of f,g respectively. For any x, we have f(x)-g(x+k T_f)=f(x+k T_f)-g(x+k T_f)\to_k 0 (where k\in\mathbb{N}), hence g(x+k T_f)\to_k f(x). Therefore, we have g((x+T_g)+kT_f)\to_k f(x+T_g) (taking x+T_g as x in the previous formula). But we also have g((x+T_g)+kT_f)=g(x+kT_f)\to_k f(x) using the definition of T_g. As a consequence, comparing the limits, we have f(x+T_g)=f(x).

    Thus T_g is a period for f as well. Then the conclusion is straightforward: for all x, f(x)-g(x)= f(x+kT_g)-g(x+kT_g)\to_k 0 hence f(x)=g(x).

    NB: I did not even use the continuity of f,g...
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  4. #4
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    Thanks Just to clarify, does the symbol f(x)-g(x)\to_k 0 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)

    Quote Originally Posted by Laurent View Post
    Here is a short proof (and I'm like Failure, I don't understand yours).

    Let T_f,T_g be the periods of f,g respectively. For any x, we have f(x)-g(x+k T_f)=f(x+k T_f)-g(x+k T_f)\to_k 0 (where k\in\mathbb{N}), hence g(x+k T_f)\to_k f(x). Therefore, we have g((x+T_g)+kT_f)\to_k f(x+T_g) (taking x+T_g as x in the previous formula). But we also have g((x+T_g)+kT_f)=g(x+kT_f)\to_k f(x) using the definition of T_g. As a consequence, comparing the limits, we have f(x+T_g)=f(x).

    Thus T_g is a period for f as well. Then the conclusion is straightforward: for all x, f(x)-g(x)= f(x+kT_g)-g(x+kT_g)\to_k 0 hence f(x)=g(x).

    NB: I did not even use the continuity of f,g...
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  5. #5
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    Quote Originally Posted by 234578 View Post
    Thanks Just to clarify, does the symbol f(x)-g(x)\to_k 0 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 k. Another notation is a_k \to_{k\to\infty,\,k\in\mathbb{N}}\ell.

    In the case of my last limit (i.e. f(x)-g(x)\to_k 0), no k appears in the "sequence"; therefore it is a constant sequence, and its limit is its common value, f(x)-g(x). The idea is that by periodicity f(x)-g(x)=f(x+kT_g)-g(x+kT_g), which now depends on k. And since x+kT_g\to_k+\infty, we may apply the hypothesis: we have f(x+kT_g)-g(x+kT_g)\to_k 0 (the limit when x\to\infty is equal to the limit along any sequence (x_k)_k tending to +\infty: \lim_{x\to\infty}\phi(x)=\lim_k \phi(x_k) if the limit \lim_{x\to\infty}\phi(x) exists).
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