The Proof is Trivial!

Problem 381***

Let A be an nxn real matrix with the modulus of each entry strictly less than 1/n. Is

I-A

invertable in all cases? If not, when is it invertable.

Reply 2381

Dalek1099

21

Original post by keromedic

I only just noticed the problem and thought "oh, I can finally solve a problem on here". Too late though :tongue:

Spoiler

I solved it another way by substituting the identity for cos2x which makes it sinx(cos^2x-sin^2x) and integrating sinxcos^2x as -cos^3x/3 which left me with -sin^3x=-sinx(1-cos^2x)=-sinx+sinxcos^2x and the integral of this is cosx-cos^3x/3 giving me an integral of cosx-2cos^3x/3 +c in total.

Reply 2382

Jooooshy

17

Original post by Mladenov

As Nikolai Nikolov, the leader of our mathematics team, says: Why simple, when it could be complicated?!

Spoiler

So weird, he tutors at my college (Univ)!

Reply 2383

henpen

Problem 382**

Let

a_1 \in (1,2)

and

a_{k+1}=a_k+\frac{k}{a_k}

. Prove that there are at most two terms in the sequence that add to make an integer.

(edited 10 years ago)

Reply 2384

Arieisit

3

Problem 383**

Ms. Janis Smith takes out an endowment policy with an insurance company which involves making a fixed payment of

\$P

each year. At the end of

n

years, Janis expects to receive a payout of a sum of money which is equal to her total payments together with interest added at the rate of

\alpha \%

per annum of the total sum in the fund.

Show, by mathematical induction or otherwise, that the total sum in the fund at the end of the

Unparseable latex formula:

n^t^h

year is

\$\frac{PR(R^n -1)}{R - 1}

where

R=(1+\frac{\alpha}{100})

Reply 2385

henpen

Problem 384***

It seems I incorrectly starred this one. Here's a better rewording:

Prove that if

a,b,\frac{a^2+b^2}{1+ab} \in \mathbb{Z}

, then

\frac{a^2+b^2}{1+ab}

is a perfect square.

(edited 10 years ago)

Reply 2386

james22

16

Original post by henpen

Problem 384**

Prove that for some

a,b

there exists an

n

such that

\frac{a^2+b^2}{1+ab}, a, b, n \in \mathbb{Z} \Rightarrow \frac{a^2+b^2}{1+ab}=n^2.

Do yoyu mean for all a,b?

Reply 2387

username937760

16

Original post by henpen

Problem 384**

Prove that for some

a,b

there exists an

n

such that

\frac{a^2+b^2}{1+ab}, a, b, n \in \mathbb{Z} \Rightarrow \frac{a^2+b^2}{1+ab}=n^2.

What are you trying to ask? if it is just to find integers a,b,n satisfying this equation then it is absolutely trivial; a=b=n=1 is an easy solution.

If, on the other hand, you are asking the form that question is usually asked in (Which is to prove that if

\frac{a^2+b^2}{1+ab}

is an integer for integers a,b then it is a perfect square) then you should rewrite it to make things explicit and add another star to the difficulty as said question was infamously difficult for the IMO at its time of conception.

Reply 2388

MAyman12

11

Original post by henpen

Problem 384**

Prove that for some

a,b

there exists an

n

such that

\frac{a^2+b^2}{1+ab}, a, b, n \in \mathbb{Z} \Rightarrow \frac{a^2+b^2}{1+ab}=n^2.

I believe this was the hardest IMO question ever. In year 1988 I think.

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Reply 2389

Flauta

9

Okay, so I'm new to this thread :hello:

This was in a book I'm reading and I thought it was a really beautiful result, I'm sorry if anyone has asked a question like this before. This might be possible with just *, I haven't tried it.

Problem 385***

0<k<1[br]Find in terms of k \displaystyle\int^{2\pi}_0 \frac{1}{1-2k\cos \theta +k^2}\ d\theta[br]

(edited 10 years ago)

Reply 2390

Felix Felicis

13

Original post by Flauta

Okay, so I'm new to this thread :hello:

This was in a book I'm reading and I thought it was a really beautiful result, I'm sorry if anyone has asked a question like this before. This might be possible with just *, I haven't tried it.

Welcome to the thread! :cool:

Problem 385***

0<k<1[br]Find in terms of k \displaystyle\int^{2\pi}_0 \frac{1}{1-2k\cos \theta +k^2}\ d\theta[br]

Solution 385

\displaystyle\begin{aligned} \int_0^{2\pi} \frac{dx}{k^2 - 2k\cos{x} + 1} \ dx & \overset{t = \tan{\frac{x}{2}}}= \int_{-\infty}^{\infty} \frac{2dt}{(k+1)^2 t^2 + (k-1)^2} \\ \\ & = \frac{2\arctan{\left(\frac{k+1}{k-1} t\right)}}{k^2 - 1}\bigg|_{-\infty}^{\infty} \end{aligned}

Now, note that

0 < k < 1 \implies k - 1 < 0

so we will use the fact that

\arctan\left(\frac{k+1}{k-1} t\right) = -\arctan \left(\frac{k+1}{1-k} t \right)

as

\arctan{x}

is an odd function so that the

\frac{k+1}{k-1}

does not affect the sign of the argument of the arctan. Thus the integral is given by:

\dfrac{2}{1-k^2} \left(\arctan{\left(\frac{k+1}{1-k} t \right)} \bigg|_{-\infty}^{\infty}\right) = \dfrac{2\pi}{1-k^2}

(edited 10 years ago)

Reply 2391

Flauta

9

Original post by Felix Felicis

Welcome to the thread! :cool:

x

Thank you!

I didn't know it could be done that way, correct answer though :smile:

Reply 2392

Flauta

9

Original post by Arieisit

Problem 383**

Ms. Janis Smith takes out an endowment policy with an insurance company which involves making a fixed payment of

\$P

each year. At the end of

n

years, Janis expects to receive a payout of a sum of money which is equal to her total payments together with interest added at the rate of

\alpha \%

per annum of the total sum in the fund.

Show, by mathematical induction or otherwise, that the total sum in the fund at the end of the

Unparseable latex formula:

n^t^h

year is

\$\frac{PR(R^n -1)}{R - 1}

where

R=(1+\frac{\alpha}{100})

I THINK I've got it?

Solution 383

[br]S_n=\frac{PR(R^n-1)}{R-1}[br]S_{n+1}=R(\frac{PR(R^n-1)}{R-1}+P)[br]S_{n+1}=PR(\frac{R(R^n-1)}{R-1}+1)[br]S_{n+1}=PR(\frac{R^{n+1}-R+R-1}{R-1})[br]S_{n+1}=PR(\frac{R^{n+1}-1}{R-1})[br]n+1=k[br]S_{k}=\frac{PR(R^k-1)}{R-1}[br][br]True for n=1 because[br][br]S_1=\frac{PR(R-1)}{R-1}[br]S_1=PR[br]

Is that correct?

(edited 10 years ago)

Reply 2393

Arieisit

3

Original post by Flauta

I THINK I've got it?

Solution 383

[br]S_n=\frac{PR(R^n-1)}{R-1}[br]S_{n+1}=R(\frac{PR(R^n-1)}{R-1}+P)[br]S_{n+1}=PR(\frac{R(R^n-1)}{R-1}+1)[br]S_{n+1}=PR(\frac{R^{n+1}-R+R-1}{R-1})[br]S_{n+1}=PR(\frac{R^{n+1}-1}{R-1})[br]n+1=k[br]S_{k}=\frac{PR(R^k-1)}{R-1}[br][br]True for n=1 because[br][br]S_1=\frac{PR(R-1)}{R-1}[br]S_1=PR[br]

Is that correct?

You are on the right path :yy:

but the proof is not complete.

I could finish it off for you but I wouldn't rob you of the joy of doing it yourself :biggrin:

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Reply 2394

ukdragon37

14

Original post by Arieisit

Problem 383**

Ms. Janis Smith takes out an endowment policy with an insurance company which involves making a fixed payment of

\$P

each year. At the end of

n

years, Janis expects to receive a payout of a sum of money which is equal to her total payments together with interest added at the rate of

\alpha \%

per annum of the total sum in the fund.

Show, by mathematical induction or otherwise, that the total sum in the fund at the end of the

Unparseable latex formula:

n^t^h

year is

\$\frac{PR(R^n -1)}{R - 1}

where

R=(1+\frac{\alpha}{100})

When I was in school I had fun spending a few days investigating this class of problems. :smile:

Solution 383**

Spoiler

Reply 2395

Arieisit

3

Original post by ukdragon37

When I was in school I had fun spending a few days investigating this class of problems. :smile:

Solution 383**

Spoiler

This is the way I did it as well. It's more "fun" than the induction :biggrin:

Posted from TSR Mobile

Reply 2396

Flauta

9

Original post by Arieisit

This is the way I did it as well. It's more "fun" than the induction :biggrin:

Posted from TSR Mobile

I disagree, induction is pretty damn exciting, like an expedition to try to get it back into the original form :tongue:

Did I miss out a step or something? Reading over it and can't see where my mistake is.

EDIT: Oh I see it now, nevermind. I was a bit too brief. Thanks :smile:

(edited 10 years ago)

Reply 2397

Flauta

9

Okay so I need more LaTeX practice, and I still think problems like this are cool. We had to numerically integrate this in maths today, was dreadfully inaccurate aha. There're various ways of doing this.

Problem 386**/***

[br]Find \displaystyle\int^{2\pi}_0 \frac{1}{5-3\sin x} dx[br]

Hints

Spoiler

(edited 10 years ago)

Reply 2398

henpen

Original post by Flauta

Okay so I need more LaTeX practice, and I still think problems like this are cool. We had to numerically integrate this in maths today, was dreadfully inaccurate aha. There're various ways of doing this.

Problem 386**/***