Re: topology ans sequences

use the fact that the rationals are dense in the reals to find the answer for this,,,

Let f:R→R be a continuous function such that f(q)=sinq for q∈Q (rational numbers). Find the value of f(π/4).

Re: topology ans sequences

Quote:

Originally Posted by

**chath** ...

or you can think about the decimal expansion of π/4

If the question was just asking about $\displaystyle \pi$, I'd suggest this rational sequence to you:

$\displaystyle a_0 = 3$

$\displaystyle a_1 = 3.1$

$\displaystyle a_2 = 3.14$

$\displaystyle a_3 = 3.141$

$\displaystyle a_4 = 3.1415$

Etc.

Re: topology ans sequences

no not about pi.i need a rational function that converges to pi/4

Re: topology ans sequences

I have this idea...to find f(π/4) we need to pick a sequence xn→π/4 and investigate what f(xn) converges to....will that be ok for this???

Re: topology ans sequences

Quote:

Originally Posted by

**chath** no not about pi.i need a rational function that converges to pi/4

The example I gave was meant to be suggestive of how you could construct a rational sequence for $\displaystyle \frac{\pi}{4}.$

Think about it a minute. Each of those, each term in the sequence, is rational, and their limit goes to $\displaystyle \pi$.

So can you think of a different-but-related sequence of rationals whose limit goes to $\displaystyle \frac{\pi}{4} \ ?$

Re: topology ans sequences

Quote:

Originally Posted by

**chath** I have this idea...to find f(π/4) we need to pick a sequence xn→π/4 and investigate what f(xn) converges to....will that be ok for this???

it sounds like you're thinking along these lines:

Find a sequence $\displaystyle b_n$ converging to $\displaystyle \frac{\sqrt{2}}{2}$, and then $\displaystyle f^{-1}(b_n)$ will converge to $\displaystyle \pi.$

If so, that isn't going to get you very far, because there's a complication with the function not having an inverse (though, by restricting domains, that can be delt with), and, vastly more intractable, you're going to need these $\displaystyle f^{-1}(b_n)$ to be rational.