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The Proof is Trivial!

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Original post by Occams Chainsaw
Felix didn't use induction. I saw his answer and realised what my friend was trying to get me to do...
We just started FP1 so he was trying to play with it!

(2n1)=n2[br]n=1:1=1check![br][br](2k1)=k2[br][br]k+1:2(k+1)1=(k+1)2[br]2(k+1)1=2(k+1)1[br]k2+(2(k+1)1)=(k+1)2[br]2(k+1)1=(k+1)2(2n - 1) = n^2[br]n = 1: 1=1 check![br][br](2k - 1) = k^2 [br][br]k+1: 2(k + 1) - 1 = (k + 1)^2[br]2(k + 1) - 1 = 2(k + 1) - 1[br]k^2 + (2(k + 1) - 1) = (k + 1)^2 [br]2(k + 1) - 1 = (k + 1)^2
Not even sure if that makes sense right now. Brain = fried.


I know it's not induction. That's why I was questioning it.

I honestly cannot see what you are trying to do....That top line isn't an identity.
Original post by bananarama2
I know it's not induction. That's why I was questioning it.

I honestly cannot see what you are trying to do....That top line isn't an identity.


I think I'll come fix it in the morning after some sleep and save further embarrassment.

I think I just proved that 1+3+5+7+9...+(2n1)=n2 1 + 3 + 5 + 7 + 9... + (2n-1)=n^2 , didn't I?
Original post by Occams Chainsaw
I think I'll come fix it in the morning after some sleep and save further embarrassment.

I think I just proved that 1+3+5+7+9...+(2n1)=n2 1 + 3 + 5 + 7 + 9... + (2n-1)=n^2 , didn't I?


Aha :biggrin: I think that's wise.

It's not what is written, but I think it may be have been latexed badly :tongue:
Original post by bananarama2
Aha :biggrin: I think that's wise.

It's not what is written, but I think it may be have been latexed badly :tongue:


I've not used latex before :sexface:
I guess that ones been done a few times, right? .... :sexface:

Anyway, I hope to redeem myself after some kip!... :sexface: It's exactly difficult lol. I'm just plugging in k+1! :biggrin:
Original post by Occams Chainsaw
I've not used latex before :sexface:
I guess that ones been done a few times, right? .... :sexface:

Anyway, I hope to redeem myself after some kip!... :sexface: It's exactly difficult lol. I'm just plugging in k+1! :biggrin:


Spoiler-How it should be done

Any of you made up these questions yourself?

Posted from TSR Mobile
Original post by bananarama2

Spoiler-How it should be done



Yeah, I'm not really sure how use the latex yet. I'll look through the guide. I definitely didn't know how to display sigma. It's quite a nice tool, isn't it?!
What I was trying to prove was what you interpreted. I now realise (after a rejuvinating 10 hours sleep and a Sunday roast) that I should have said I was trying to prove that the sum odd numbers is a square number!

A nice little pattern I noticed is this:
if I have an n*n box - or a perfect square
xxxx
xxxx
xxxx
xxxx

in this case where n = 4, x = 16

where x = a = b = c = d and we split it into (the only way I can describe it is 'mini'-squares' so the sum of the x-coordinate value for a will always equal the sum for a's y coordinate etc...)

abcd
abcc
abbb
aaaa

So we have produced 7a + 5b + 3b + 1a = 16x
or, 16 (a square number) = 7 + 5 + 3 + 1...

Is that very obvious? It seems that way now I have written it down but I when I figured it out I felt pretty pleased with myself lol
(edited 10 years ago)
Original post by jack.hadamard
Problem 345 **/***

Evaluate limx00tanxtansint dt0sinxln(1+sint) dt\displaystyle \lim_{x \to 0} \frac{\displaystyle \int_{0}^{\tan x} \tan\sin t\ dt}{\displaystyle \int_{0}^{\sin x} \ln(1 + \sin t)\ dt}.


An alternative solution to the one given:

Using well known aproximations and ignoring higher order terms (the limits go to 0 so we can assume higher roder terms don't matter) we have

tan(sin(t))=tan(t)=t\tan(\sin(t))=\tan(t)=t

and

ln(1+sin(t))=ln(1+t)=tln(1+\sin(t))=ln(1+t)=t

we can then evaluate the integrals explicitly to get the original limit to be equal to the limit of

1cos(x)2\frac{1}{\cos(x)^2}

Which is 1.

I wonder if there is a better way of doing ths though? Both L'H and series expansions feel like cheating. A bit like solving an inequality by just using calculus.
(edited 10 years ago)
Original post by james22
An alternative solution to the one given:

Using well known aproximations and ignoring higher order terms (the limits go to 0 so we can assume higher roder terms don't matter) we have

tan(sin(t))=tan(t)=t\tan(\sin(t))=\tan(t)=t

and

ln(1+sin(t))=ln(1+t)=tln(1+\sin(t))=ln(1+t)=t

we can then evaluate the integrals explicitly to get the original limit to be equal to the limit of

1cos(x)2\frac{1}{\cos(x)^2}

Which is 0.

I wonder if there is a better way of doing ths though? Both L'H and series expansions feel like cheating. A bit like solving an inequality by just using calculus.


1/1 is 0?
Reply 2269
Original post by Occams Chainsaw
Yeah, I'm not really sure how use the latex yet. I'll look through the guide. I definitely didn't know how to display sigma. It's quite a nice tool, isn't it?!
What I was trying to prove was what you interpreted. I now realise (after a rejuvinating 10 hours sleep and a Sunday roast) that I should have said I was trying to prove that the sum odd numbers is a square number!

A nice little pattern I noticed is this:
if I have an n*n box - or a perfect square
xxxx
xxxx
xxxx
xxxx

in this case where n = 4, x = 16

where x = a = b = c = d and we split it into (the only way I can describe it is 'mini'-squares' so the sum of the x-coordinate value for a will always equal the sum for a's y coordinate etc...)

abcd
abcc
abbb
aaaa

So we have produced 7a + 5b + 3b + 1a = 16x
or, 16 (a square number) = 7 + 5 + 3 + 1...

Is that very obvious? It seems that way now I have written it down but I when I figured it out I felt pretty pleased with myself lol


(n+1)^2=n^2+2n+1; In essence, to step from the nth square number to the n+1th, you must add the nth odd number. You can see this must be so as you say, the nth square number is a square, n*n. To get to the (n+1)th, you need to add a column of n, a row of n, and then one more (ie the 3 a's going downwards, the 3 a's at the bottom, then the one in the bottom left). Fairly clear, from a certain viewpoint. Otherwise, it's not entirely obvious that the sum of odd numbers should be square.
This may have been on here before so if it has please tell me and I'll delete it.

Problem 346 *
Unparseable latex formula:

\displaystyle \int^b_a \arccos \left(\frac{x}{\sqrt{\left(a+b \right)x - ab\right)}}\right)\ dx

Reply 2271
Some number theory!

Problem 347*

Prove that gcd(2a1,2b1)=2gcd(a,b)1 gcd(2^a - 1, 2^b - 1) = 2^{gcd(a,b)}- 1 .

Problem 348**

Prove that x2+1y25 \displaystyle \frac{x^2+1}{y^2-5} is never an integer for x,yZ x, y \in \mathbb{Z} .
Reply 2272
Original post by MW24595

Problem 348**

Prove that x2+1y25 \displaystyle \frac{x^2+1}{y^2-5} is never an integer for x,yZ x, y \in \mathbb{Z} .


This isn't true:

22+1025=1 \dfrac{2^2+1}{0^2-5}=-1
x2+1225=x21 \dfrac{x^2+1}{2^2-5}=-x^2-1

If we change integer to positive integer:

Spoiler

Solution 347

WLOG bab\geq a, let δ\delta be s.t. δ2a1\delta |2^a-1 and δ2b1\delta |2^b-1.

Define an+1=min(an,bnan)a_{n+1}=\min (a_n, b_n-a_n) and bn+1=max(an,bnan)b_{n+1}=\max (a_n,b_n-a_n) with a1=aa_1=a and b1=bb_1=b. Note that an0a_n\to 0 and bngcd(a,b)b_n\to \gcd (a,b). For any n:n:

δ2bn1δ2an1δ2bn2anδ2bnan1δ2bn+11δ2an+11δ2bn+12an+1    δ2gcd(a,b)1\begin{aligned} &\delta |2^{b_n}-1\wedge \delta| 2^{a_n}-1\Rightarrow \delta | 2^{b_{n}}-2^{a_n}\Rightarrow \delta | 2^{b_n-a_n}-1\\& \Rightarrow \delta |2^{b_{n+1}}-1\wedge \delta| 2^{a_{n+1}}-1\Rightarrow \delta | 2^{b_{n+1}}-2^{a_{n+1}}\Rightarrow \cdots \\& \;\;\cdots \\&\quad \Rightarrow \delta |2^{\gcd (a,b)}-1\end{aligned}

Obviously 2gcd(a,b)12^{\gcd{(a,b)}}-1 divides both initial numbers and hence δmax=2gcd(a,b)1\delta_{\text{max}}= 2^{\gcd{(a,b)}}-1
Problem 349 / **

Let h:R0R+h : \mathbb{R}^{\geq 0} \to \mathbb{R}^{+} be such that h(x)>2xh'(x) > 2x for all xx. Prove that

k=1nh(1+Fk2)>4FnFn+1\displaystyle \sum_{k=1}^{n} h(1 + F_k^2) > 4F_nF_{n+1}

where FnF_n is the nn-th Fibonacci.


Problem 350 / **

The Tribonacci numbers, TnT_n, are defined as

T0=0,T1=T2=1T_0 = 0, T_1 = T_2 = 1 and Tn+3=Tn+2+Tn+1+TnT_{n+3} = T_{n+2} + T_{n+1} + T_{n} for n0n \geq 0.

Find k=1nTk2\displaystyle \sum_{k=1}^{n} T_k^2.
Solution 349

I will use the identity

k=1nFk2=FnFn+1\displaystyle\sum_{k=1}^n F_k^2=F_nF_{n+1}

There are several ways to prove this, most straight forward is to use the explicit formula for the fibonacci numbers and it is just an exercise in summing geometric series.

Using standard results from analysis we have that h is increasing and that

h(x)>x2h(x)>x^2

This is the best inequality we can have, as h can be made as close to x2x^2 as we like.

Using both of these we have

i=1nh(1+(Fk)2)>i=1n(1+(Fk)2)2=i=1n(1+2(Fk)2+(Fk)4)[br][br]i=1n4(Fk)2=4FnFn+1\displaystyle\sum_{i=1}^n h(1+(F_k)^2) > \displaystyle\sum_{i=1}^n (1+(F_k)^2)^2 = \displaystyle\sum_{i=1}^n (1+2(F_k)^2+(F_k)^4)[br][br]\geq * \displaystyle\sum_{i=1}^n 4(F_k)^2 = 4F_nF_{n+1}

You may not be happy with the starred inequality, but this is true because x2+1>2xx^2+1>2x is immediate from (x1)2>0(x-1)^2>0
Original post by Pterodactyl
This may have been on here before so if it has please tell me and I'll delete it.

Problem 346 *
Unparseable latex formula:

\displaystyle \int^b_a \arccos \left(\frac{x}{\sqrt{\left(a+b \right)x - ab\right)}}\right)\ dx

Surprisingly tricky (I was barking down the wrong tree at 2am with inverse trig identities and far too many substitutions) but this has a very nice geometric link. (apologies if this is a repeat question, I haven't followed this thread much). Apologies also for the concise workings.
Use IBP and set x - \frac{ab}{a+b} = v and arccos(...) = u
Our uv term (arcos1-arccos1) can be safely set as 0 if I am to make any geometrical sense of this.
After a bit of algebra bashing we obtain;
Unparseable latex formula:

\displaystyle \int^b_a \arccos \left(\frac{x}{\sqrt{\left(a+b \right)x - ab\right)}}\right)\ dx = \int^b_a \frac{ax+bx-2ab}{2(a+b)(\sqrt{ax+bx-ab-x^2})}dx


We now work on the RHS.
We set y=2xabba y = \frac{2x-a-b}{b-a}
11(b2a2)(y+1)4(a+b)(1y2)dy\displaystyle \int^1_{-1} \frac{(b^2-a^2)(y+1)}{4(a+b)(\sqrt{1-y^2})} dy
Taking out the constants from the integral, we see our integral reduces down to
b2a24(a+b)1111y2dy\displaystyle \frac{b^2-a^2}{4(a+b)} \int^1_{-1} \frac{1}{\sqrt{1-y^2}} dy
As y/(1-y^2) is obviously 0 over the interval as it is an odd function.
This is now a well known integral;
1111y2dy=arcsin(1)arcsin(1)=π\displaystyle \int^1_{-1} \frac{1}{\sqrt{1-y^2}} dy = \arcsin(1) - \arcsin(-1) = \pi
And thus
I=π(b2a2)4(a+b) I = \pi \frac{(b^2-a^2)}{4(a+b)}

Now (for the nice part) a sketch shows this describes a semicircle which intersects the x axis at a and b, and whereby the integrand describes the angle (for each x).

I'll take a look at the tribonacci when I have some more time at a computer.
Original post by jack.hadamard

Problem 350 / **

The Tribonacci numbers, TnT_n, are defined as

T0=0,T1=T2=1T_0 = 0, T_1 = T_2 = 1 and Tn+3=Tn+2+Tn+1+TnT_{n+3} = T_{n+2} + T_{n+1} + T_{n} for n0n \geq 0.

Find k=1nTk2\displaystyle \sum_{k=1}^{n} T_k^2.


This is harder than I first thought. I have now failed with 3 methods, all failing because the reccurance relation has an odd number of terms (I think they would work with all ones with an even number of terms), the sums just don't all cancel as 1 is always left out.

I do have a method that will certainly work, but it involves solving teh recurrance relation to get it into an explict form, squaring that and then summing all the geometric series. I don't want to do this ans I doubt it was the intention.
(edited 10 years ago)
Original post by Llewellyn
Surprisingly tricky (I was barking down the wrong tree at 2am with inverse trig identities and far too many substitutions) but this has a very nice geometric link. (apologies if this is a repeat question, I haven't followed this thread much). Apologies also for the concise workings.
Use IBP and set x - \frac{ab}{a+b} = v and arccos(...) = u
Our uv term (arcos1-arccos1) can be safely set as 0 if I am to make any geometrical sense of this.
After a bit of algebra bashing we obtain;
Unparseable latex formula:

\displaystyle \int^b_a \arccos \left(\frac{x}{\sqrt{\left(a+b \right)x - ab\right)}}\right)\ dx = \int^b_a \frac{ax+bx-2ab}{2(a+b)(\sqrt{ax+bx-ab-x^2})}dx


We now work on the RHS.
We set y=2xabba y = \frac{2x-a-b}{b-a}
11(b2a2)(y+1)4(a+b)(1y2)dy\displaystyle \int^1_{-1} \frac{(b^2-a^2)(y+1)}{4(a+b)(\sqrt{1-y^2})} dy
Taking out the constants from the integral, we see our integral reduces down to
b2a24(a+b)1111y2dy\displaystyle \frac{b^2-a^2}{4(a+b)} \int^1_{-1} \frac{1}{\sqrt{1-y^2}} dy
As y/(1-y^2) is obviously 0 over the interval as it is an odd function.
This is now a well known integral;
1111y2dy=arcsin(1)arcsin(1)=π\displaystyle \int^1_{-1} \frac{1}{\sqrt{1-y^2}} dy = \arcsin(1) - \arcsin(-1) = \pi
And thus
I=π(b2a2)4(a+b) I = \pi \frac{(b^2-a^2)}{4(a+b)}

Now (for the nice part) a sketch shows this describes a semicircle which intersects the x axis at a and b, and whereby the integrand describes the angle (for each x).

I'll take a look at the tribonacci when I have some more time at a computer.


I used a different second substitution so missed out on the nice geometric link. :frown:

Problem 351 **
0(xx32+x524x7246+...)(1+x222+x42242+x6224262+...) dx\displaystyle \int^\infty_0 \left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot4}-\frac{x^7}{2\cdot4\cdot6}+... \right)\left(1+\frac{x^2}{2^2}+ \frac{x^4}{2^2\cdot4^2}+\frac{x^6}{2^2\cdot4^2\cdot6^2}+... \right)\ dx

Problem 352 */**
n=0(1)n3n+1\displaystyle \sum^\infty_{n=0} \dfrac{\left(-1\right)^n}{3n+1}
(edited 10 years ago)
Original post by Pterodactyl

Problem 351 **
0(xx32+x524x7246+...)(1+x22+x42242+x6224262+...) dx\displaystyle \int^\infty_0 \left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot4}-\frac{x^7}{2\cdot4\cdot6}+... \right)\left(1+\frac{x^2}{2}+ \frac{x^4}{2^2\cdot4^2}+\frac{x^6}{2^2\cdot4^2\cdot6^2}+... \right)\ dx


Pardon me if I'm wrong, but should the second term of your second bracket be 122x2\frac{1}{2^2} x^2?

Problem 353*/**
For a positive integer nn, let SnS_n be the total sum of the intervals of xx such that sin4nxsinx\sin 4nx \geq \sin x in 0xπ20\leq x\leq \frac{\pi}{2}.

Find limnSn\displaystyle \lim_{n\to\infty} S_n.

Problem 354*/**

Evaluate limn(0πsin2nxsinx dxk=1n1k)\displaystyle\lim_{n\to\infty} \left( \int_0^{\pi}\frac{\sin^2 nx}{\sin x} \ dx - \sum_{k=1}^{n}\frac{1}{k}\right).
(edited 10 years ago)

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