The Proof is Trivial!

I'm not quite sure what you mean, but

\displaystyle \int_{-1}^1 \frac{dx}{\sqrt{x}}

has finite area, despite having an asymptote.

Edit: Please be clearer, I don't see how the above is not a counterexample to your claim.

probably because you didnt read his last sentence...

Reply 2581

IceKidd

Problem 424 */**

Find the integral of ln(cosx) between x=pi/2 and x=0

(edited 10 years ago)

Reply 2582

Arcane1729

12

Problem 424 */**

The sum of the base-10 logarithms of the divisors of 10^N is ~~729~~ 792 for some natural number N
.What is the value of N ?

Edit: There was a mistake

(edited 10 years ago)

Reply 2583

Flauta

9

Original post by IceKidd

Problem 424 */**

Find the integral of ln(cosx) between x=pi/2 and x=0

This is going to look dreadful, although I have put in some effort putting in arrows and stuff

Solution 424

Building on what we've seen previously when the boundaries are pi/2 and 0 and the interchangeability therein of sin(x) and cos(x)..

I = \displaystyle \int^{\frac{\pi}{2}}_0 \ln \cos (x) dx = \displaystyle \int^{\frac{\pi}{2}}_0 \ln \sin (x) dx \\ \sin (x) \equiv 2 \sin (\frac{x}{2}) \cos(\frac{x}{2}) \\ \Rightarrow I = \frac{\pi}{2} \ln 2\ + \displaystyle \int^{\frac{\pi}{2}}_0 \ln \cos (\frac{x}{2}) + \ln \sin(\frac{x}{2}) dx \\ I = \frac{\pi}{2} \ln 2\ + 2\displaystyle \int^{\frac{\pi}{4}}_0 \ln \cos (x) + \ln \sin(x) dx \\ \displaystyle \int^{\frac{\pi}{4}}_0 \ln \sin (x) dx = \displaystyle \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \ln \cos (x) dx \\ I = \frac{\pi}{2} \ln 2\ + 2I \Rightarrow I = -\frac{\pi}{2} \ln 2

(edited 10 years ago)

Reply 2584

Felix Felicis

13

Original post by nahomyemane778

Problem 424 */**

The sum of the base-10 logarithms of the divisors of 10^N is 729 for some natural number N
.What is the value of N ?

I think you mean

792

.

Reply 2585

Llewellyn

14

Thanks for that felix, I was wondering where I went wrong with N = 1/3 (-2+(19684-81 (59055)^(1/2))^(1/3)+(19684+81 (59055)^(1/2))^(1/3)) :lol:

Reply 2586

IceKidd

Original post by Llewellyn

Thanks for that felix, I was wondering where I went wrong with N = 1/3 (-2+(19684-81 (59055)^(1/2))^(1/3)+(19684+81 (59055)^(1/2))^(1/3)) :lol:

Hows durham? Are u top of ur class

Reply 2587

Arcane1729

12

Original post by Felix Felicis

I think you mean

792

.

How on earth did you possibly know? Have you see this question before?
Nice to see good old feynman back for christmas

(edited 10 years ago)

Reply 2588

Felix Felicis

13

Original post by Llewellyn

Thanks for that felix, I was wondering where I went wrong with N = 1/3 (-2+(19684-81 (59055)^(1/2))^(1/3)+(19684+81 (59055)^(1/2))^(1/3)) :lol:

Original post by nahomyemane778

How on earth did you possibly know? Have you see this question before?
Nice to see good old feynman back for christmas

Nope - I got to the stage where I had an equation with no integer solutions for the value you gave (

\frac{1}{2} N(N+1)^2 = 729

), checked my working over and realised there's a mistake in the question and 792 just so happened to give a nice value for N. :cool:

Posted from TSR Mobile

Reply 2589

Arcane1729

12

Original post by Felix Felicis

Nope - I got to the stage where I had an equation with no integer solutions for the value you gave (

\frac{1}{2} N(N+1)^2 = 729

), checked my working over and realised there's a mistake in the question and 792 just so happened to give a nice value for N. :cool:

Posted from TSR Mobile

Ahh ok then- well done- I've amended the OP so you can post the solution now.

Reply 2590

Zakee

N = 11 using the Bertrand-Chebyshev theorem.

Reply 2591

Arcane1729

12

Original post by Zakee

N = 11 using the Bertrand-Chebyshev theorem.

err... what? :s-smilie:

How does that solve the problem?

Proof please.

Reply 2592

Zakee

Original post by nahomyemane778

err... what? :s-smilie:

How does that solve the problem?

Proof please.

Proof:

Unparseable latex formula:

\frac{4^n}{2n } \le \binom{2n}{n} = \left(\prod_{p \le \sqrt{2n}} p^{R(p,n)}\right) \left(\prod_{\sqrt{2n} < p \le \frac{2n}{3}} p^{R(p,n)}\right)[br]\lt (2n)^{\sqrt{2n}} \prod_{1 < p \leq \frac{2n}{3} } p = (2n)^{\sqrt{2n}} \Big( \frac{2n}{3}\Big)\# \le (2n)^{\sqrt{2n}} 4^{2n/3}.\

By concavity of the right-hand side as a function of n, the last inequality is necessarily verified on an interval. Since it holds true for

n = 11

and not

n = 12

, upper bound is the equating value which is:

n = 11

The proof is trivial

(edited 10 years ago)

Reply 2593

Arcane1729

12

Original post by Zakee

Proof:

Unparseable latex formula:

\frac{4^n}{2n } \le \binom{2n}{n} = \left(\prod_{p \le \sqrt{2n}} p^{R(p,n)}\right) \left(\prod_{\sqrt{2n} < p \le \frac{2n}{3}} p^{R(p,n)}\right)[br] < (2n)^{\sqrt{2n}} \prod_{1 < p \leq \frac{2n}{3} } p = (2n)^{\sqrt{2n}} \Big( \frac{2n}{3}\Big)\# \le (2n)^{\sqrt{2n}} 4^{2n/3}.\

By concavity of the right-hand side as a function of n, the last inequality is necessarily verified on an interval. Since it holds true for

n = 11

and not

n = 12

, upper bound is the equating value which is:

n = 11

The proof is trivial

Well that's some pretty heavy machinery you've used- I dont understand at all to be honest. Is it university level?

Reply 2594

Zakee

Original post by nahomyemane778

Well that's some pretty heavy machinery you've used- I dont understand at all to be honest. Is it university level?

I'm currently in year 13, and I think it may be University level, I'm not sure. I heard from someone that it comes up in FP3 (A-level), but I haven't studied FP3 yet. I'm just extremely interested in Pure Mathematics and have a knack for problem-solving.

(edited 10 years ago)

Reply 2595

Flauta

9

Original post by Zakee

Proof

The proof is trivial

What is that? :redface:

(edited 10 years ago)

Reply 2596

username219321

14

Original post by Zakee

I'm currently in year 13, and I think it may be University level, I'm not sure. I heard from someone that it comes up in FP3 (A-level), but I haven't studied FP3 yet.