#Calculus Proof

38 messages · Page 1 of 1 (latest)

quick sparrow
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I tried using the Weighted Mean Value Theorem but I'm not able to prove the single signed property of g(x). Are there any alternate proofs the the question?

left acornBOT
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wide tendon
wide tendon
quick sparrow
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It could be intersecting the x-axis at some point in the interval and the integral could be 0 too

wide tendon
quick sparrow
wide tendon
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you know that the integral of f(x)g(x) is 0 for all g

quick sparrow
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Yes

wide tendon
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what about when ||g=f||

quick sparrow
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Then integral of f(x)^2 would be 0

wide tendon
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Yes

quick sparrow
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If there exists a point in [a, b] where f(x) is not 0, then f(x)^2 would be greater than 0 at that point.

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Which means f(x) has to be 0 at any point in the interval?

wide tendon
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well yeah f(x)^2=0 => f(x)=0

quick sparrow
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But what if g is not equal to f?

wide tendon
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You know that it’s true for any continuous g

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That includes f

quick sparrow
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We're taking a specific case to prove it and generalizing it to the entire interval.

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Is there a generic solution to it?

wide tendon
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Well there is nothing else to prove, you know that the integral of f(x)g(x) is 0 for any continuous function g, if it’s true for any function g then it’s true for g=f so you have that the integral of f(x)^2 is 0, with what I said before this implies that f(x)^2=0 for all x ie f(x)=0 for all x

quick sparrow
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Thanks @wide tendon

wide tendon
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Np

quick sparrow
# wide tendon Np

If it is alright with you, could I contact you in private for a couple more problems of a similar structure?

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I'm having a bit of a trouble wrapping my head around those

wide tendon
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And I’ll comment if I have time

quick sparrow
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+close

lunar agateBOT
# quick sparrow +close
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# lunar agate

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