anonymous
  • anonymous
Here's a fun integral question I made for a competition a few years ago. I was inspired by this much easier question: Warmup: Let f(x) be a continuous real-valued function on [0,1]. Calculate: \[\lim_{n\to\infty}\int_0^1x^nf(x)dx\] Now for the real question: Let f(x) be a continuous real-valued function on [0,1]. Calculate \[\lim_{n\to\infty} n\int_0^1x^nf(x)dx\]
Mathematics
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anonymous
  • anonymous
Here's a fun integral question I made for a competition a few years ago. I was inspired by this much easier question: Warmup: Let f(x) be a continuous real-valued function on [0,1]. Calculate: \[\lim_{n\to\infty}\int_0^1x^nf(x)dx\] Now for the real question: Let f(x) be a continuous real-valued function on [0,1]. Calculate \[\lim_{n\to\infty} n\int_0^1x^nf(x)dx\]
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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JamesJ
  • JamesJ
The first one is clearly zero. The second one .... one sec
JamesJ
  • JamesJ
f(1)
anonymous
  • anonymous
Didn't take too long. I meant to say "calculate with proof" but I suppose that is trivial once you know the answers.

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JamesJ
  • JamesJ
Yes, I know. But I can write the proofs for these. These integrals are actually a kind of 'primitive' form of the Laplace transform.
JamesJ
  • JamesJ
we usually have e^(-st) as the function we're integrating against. If we write x = e^-s, then that's how we get back to these integrals.
anonymous
  • anonymous
Makes sense, never real thought about that connection too much. Very clever! Want a harder question (that should take longer than 10 seconds for you to solve)?
JamesJ
  • JamesJ
Talking of writing proofs though, writing a good proof for that first integral isn't to obvious, I don't think.
JamesJ
  • JamesJ
It seems to me you need to write it as \[ \int_0^{1-\delta} + \int_{1-\delta}^1 \] Showing the first piece goes to zero is straightforward as the function \( g(x) = x^n f(x) \) converges uniformly to the zero function on \( [0,1-\delta] \). Then you just need to deal with the the other bit.
JamesJ
  • JamesJ
Anyway. What's the other question?
anonymous
  • anonymous
I don't think all of that machinery is too necessary. Since the function is continuous on a closed interval, it must be bounded. Then some manipulations followed by a squeeze theorem will give you the result. Anyhoo, I'll type up the next question (an inequality) real quick
JamesJ
  • JamesJ
Yes, you don't need the uniform convergence. But I'd still break it up into the two pieces.
JamesJ
  • JamesJ
...which isn't to say there isn't another way to skin that cat. Just the way I'd do it.

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