1. ## topology ans sequences

Can you find any sequence of rationals such that it converges to π/4?

use the fact that the rationals are dense in the reals,
or you can think about the decimal expansion of π/4. How can you get rational numbers out of that, rational numbers that converge to π/4?

Can anybody help me with this immediately??

2. ## 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).

3. ## Re: topology ans sequences

Originally Posted by chath
...
or you can think about the decimal expansion of π/4
If the question was just asking about $\pi$, I'd suggest this rational sequence to you:
$a_0 = 3$
$a_1 = 3.1$
$a_2 = 3.14$
$a_3 = 3.141$
$a_4 = 3.1415$
Etc.

4. ## Re: topology ans sequences

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

5. ## 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???

6. ## Re: topology ans sequences

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 $\frac{\pi}{4}.$

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

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

7. ## Re: topology ans sequences

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

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 $f^{-1}(b_n)$ to be rational.