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please help me to do this problem...

Let $f: \mathbb R\to \mathbb R$ be a continuous function such that $f(q)=\sin q$ for $q\in\mathbb Q$ (rational numbers). Find the value of f(π/4).

ccc
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  • Please read: http://meta.math.stackexchange.com/q/1803 –  Oct 16 '12 at 16:33
  • What values could $f$ take on the irational numbers? What would make sense? And if you got this, how do you prove it? – Stefan Oct 16 '12 at 16:33
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    I think you are in an exam. Hence I will refrain from answering any more of your question for the next 2 hours. – Rudy the Reindeer Oct 16 '12 at 16:35
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    You would make it easier to other people to help you, if you wrote your thoughts about the problem. Otherwise some people might post answers employing some facts you have not learned yet (and thus useless for you). What do you know about continuity? If you know something about dense sets, you might have a look here: Continuous functions between metric spaces are equal if they are equal on a dense set. Other possibility: What do you know about continuity and convergent sequences? – Martin Sleziak Oct 16 '12 at 16:39
  • to Martin,,I referred your link about cts function between metric spaces are equal....but that says about two functions right???here there's only one and I know Q is dense in R and the thing is I cant find any function that is convergent to π/4. – ccc Oct 17 '12 at 02:05

2 Answers2

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I'd do it as follows:

$f$ is continuous which means that you can swap limits and function: $\lim_{n \to \infty} f(x_n) = f( \lim_{n \to \infty} x_n) = f(x)$. So to find $f( \pi / 4)$ you pick a sequence $x_n \to \pi / 4$ and investigate what $f(x_n)$ converges to.

I looked it up (because I didn't know any such sequence off the top of my head), and for example, you could take the formula found by Leibniz:

$$ \frac{\pi}{4} = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k - 1}$$

Rudy the Reindeer
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  • yes exactly...I agree with this idea.but still I couldn't find any sequence such that xn→π/4.can anybody help me?? – ccc Oct 17 '12 at 00:14
  • @cccjay I added a formula to my answer, hope this helps! – Rudy the Reindeer Oct 17 '12 at 06:46
  • thanks alot MattN...I'll check that...:) – ccc Oct 17 '12 at 07:30
  • another problem!!!!What is the use of giving f(q)=sin(q) there??? – ccc Oct 17 '12 at 07:41
  • @cccjay That's the function definition! Otherwise, what function are you evaluating at $\pi / 4$? – Rudy the Reindeer Oct 17 '12 at 07:45
  • @ Matt N,you mean that I should find the limit of the function sin(∑(−1)^k+1/2k−1) as n goes to infinity? – ccc Oct 19 '12 at 00:52
  • sorry there's a mistake,,,,limit should be found as k goes to infinity.. – ccc Oct 19 '12 at 01:31
  • I found that the series ∑(−1)^k+1/2k−1 converges to zero using the alternating series test.but is that sufficient for us to say that the "limit of the sin of this series" converges to zero??? – ccc Oct 19 '12 at 01:34
  • @cccjay I think when you wrote $n$ was correct: you want $y = \lim_{n \to \infty} \sum_{k=1}^n \frac{(-1)^{k+1}}{2k - 1}$. If $f$ is the sine function, shouldn't $f(y) = f(\pi/4) \neq 0 $? – Rudy the Reindeer Oct 19 '12 at 05:28
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The problem OP seems to have is with finding a sequence of rationals converging to $\pi/4$.

If you know that the rationals are dense in the reals, you can just state that fact as a proof of the existence of such a sequence of rationals.

If you don't want to (or can't) use that, then think about the decimal expansion of $\pi/4$. How can you get rational numbers out of that, rational numbers that converge to $\pi/4$?

Gerry Myerson
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