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The Proof is Trivial!

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thank you !
surely thats just saying that cosy=cosx=1 , but that doesn't cover all the possible solutions, thats just a specific case i think -
Original post by GCSEHELPPLS
surely thats just saying that cosy=cosx=1 , but that doesn't cover all the possible solutions, thats just a specific case i think -


Yeah forget it, I think you're right :smile:
Original post by Indeterminate
I'm going for some food, and I'll leave you guys with some food for thought :smile:

Problem 514***

Find a sequence sn,n≥1s_n, n \geq 1 such that

∑n=1∞sn \displaystyle \sum_{n=1}^{\infty} s_n

converges but

∑n=1∞(sn+1) \displaystyle \sum_{n=1}^{\infty} (s_n + 1)

diverges.

Find also a sequence such that

∑n=1∞(sn+1) \displaystyle \sum_{n=1}^{\infty} (s_n + 1)

converges but

∑n=1∞sn \displaystyle \sum_{n=1}^{\infty} s_n

diverges.


Is this question stated correctly? Unless I've missed something you can have s_n = 0 for all n for the first part and s_n = -1 for all n for the second part.
Original post by metaltron
Is this question stated correctly? Unless I've missed something you can have s_n = 0 for all n for the first part and s_n = -1 for all n for the second part.


I've checked my source and it turns out I missed a condition that it must be a sequence of positive real numbers.

I've edited it in now :smile:
Original post by Indeterminate
I've checked my source and it turns out I missed a condition that it must be a sequence of positive real numbers.

I've edited it in now :smile:


Hmm even then the sum of (s_n + 1) is unbounded. The sum from 1 to k will be greater than k, so it can't converge.
(edited 8 years ago)
Original post by metaltron
Hmm even then the sum of (s_n + 1) is unbounded. The sum from 1 to n will be greater than n, so it can't converge.


That's what has me worried too. I didn't think much about it before posting but I think something's fishy.

I think I better get rid of it and post something else instead.


Posted from TSR Mobile
Original post by Indeterminate
That's what has me worried too. I didn't think much about it before posting but I think something's fishy.

I think I better get rid of it and post something else instead.


Posted from TSR Mobile


Yeah good idea! :biggrin:
Original post by Indeterminate
That's what has me worried too. I didn't think much about it before posting but I think something's fishy.

I think I better get rid of it and post something else instead.


Posted from TSR Mobile


Converges with respect to a metric which is not the normal one perhaps?
I'll take problem 514 then...

Problem 514

Consider a 5x5 board. There is a lamp in every square, initially all off. When we switch on a lamp, it switches on all lamps orthogonally next to it (5 lamps are switched on every move, unless at a side or corner). After some number of moves, exactly 1 lamp is on. Find all squares in which this lamp may be in.
Original post by metaltron
Converges with respect to a metric which is not the normal one perhaps?


It's possible that this question needs some specific context. I'll see what I can find out.

In the meantime

Problem 515***

Prove that if

f:[0,1]→Rf: [0,1] \rightarrow \mathbb{R}

is an increasing function then f is Lebesgue measurable.
Reply 3311
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SParm
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Original post by Indeterminate
It's possible that this question needs some specific context. I'll see what I can find out.

In the meantime

Problem 515***

Prove that if

f:[0,1]→Rf: [0,1] \rightarrow \mathbb{R}

is an increasing function then f is Lebesgue measurable.


Spoiler

Problem 516
Considering that you hang out at the math section of TSR, it should be an elementary task for you guys to prove the following:

csc⁡(2x)−cot⁡(2x)=tan⁡(x)\csc(2x)-\cot(2x) = \tan(x)

... so that's not going to be your job. Assume that you have proved this beforehand (since the equation is true anyway) and you are ready to move on to the next task. Your task now is to
use the equation above to prove that there is a constant k that fulfills the following equation:

k∣1+cos⁡(2x)∣=∣sec⁡(x)∣\displaystyle \frac{k}{\sqrt{|1+\cos(2x)|}} = |\sec(x)|

(There are ways to prove the equation that does not involve the previously 'proven' equation, those are not acceptable as a solution to this problem.)

---

On a side note, I am confused how to mark the difficulty level of this question in comparison to the A-Levels since I am doing the IB. How similar is A-Levels to IB Mathematics HL?
(edited 8 years ago)
For the benefit of any A-level students :smile:

Problem 517

Find

\displaystyle \int x\ln\left(x^2 + a^2 + \sqrt{x^2 - a^2}\right)\ dx
(edited 8 years ago)
An easy one (compared to my previous ones)

Problem 518

a) A pirate ship has 2015 treasure chests, all of which are closed. Each chest contains some amount of gold. To distribute the gold, the pirates will do the following: The captain first decides how many chests he will keep and tells this number to the rest of the pirates. He opens all the chests and selects which ones he will keep. The captain wants to make sure he gets at least half the total gold. What is the minimum number the captain should say to ensure that he gets at least half the gold?

b) A pirate ship has 2015 treasure chests, all of which are closed. Each chest contains some amount of gold and some amount of silver. To distribute the gold and silver, the pirates will do the same as above. The captain wants to make sure he gets at least half the total gold AND half the total silver. What is the minimum number the captain should say to ensure that he gets at least half the gold and half the silver?
(edited 8 years ago)
By the way, can we start starring problem depending on the level of the knowledge they require? The details are in the OP, I believe. :smile:
Original post by udvuvvdu
Problem 516


I like the way you set the question out, neat and nice! The one thing I don't really like is that you only allowed certain ways to solve the problem, this isn't really the best thing do to (imo) as there are a variety of ways to do things and seeing multiple solutions especially short elegant ones that surpass the standard ones are valuable and enjoyable. :smile:

To answer your side note, the general consensus is that A-Level Further Mathematics (which everybody on this thread should posses indepth knowledge of) > IB Mathematical HL > A-Level Mathematics.

So for this thread, anything requiring IB-only level knowledge is most definitely * only.

Solution 516

csc⁡2x−cot⁡2x=1sin⁡2x−cos⁡2xsin⁡2x=2sin⁡2x2sin⁡xcos⁡x=sin⁡xcos⁡x=tan⁡x\displaystyle \csc 2x - \cot 2x = \frac{1}{\sin 2x} - \frac{\cos 2x}{\sin 2x} = \frac{2 \sin^2 x}{2 \sin x \cos x} = \frac{\sin x}{\cos x} = \tan x

I'm not quite sure what your second part is asking, perhaps you could tell me where I'm going wrong in my solution:

k∣1+cos⁡2x∣=k∣2cos⁡2x∣=k2∣cos⁡x∣=k∣sec⁡x∣2≠∣sec⁡x∣\displaystyle \frac{k}{\sqrt{|1+\cos 2x|}} = \frac{k}{\sqrt{|2 \cos^2 x|}} = \frac{k}{\sqrt{2}|\cos x|} = \frac{k |\sec x|}{\sqrt{2}} \neq |\sec x| unless k=2k = \sqrt{2}.
(edited 8 years ago)
[QUOTE=Zacken;60096521

Solution 516

csc⁡2x−cot⁡2x=1sin⁡2x−cos⁡2xsin⁡2x=2cos⁡2x2sin⁡xcos⁡x=cos⁡xsin⁡x=cot⁡x\displaystyle \csc 2x - \cot 2x = \frac{1}{\sin 2x} - \frac{\cos 2x}{\sin 2x} = \frac{\boxed{2\cos^2 x}}{2 \sin x \cos x} = \frac{\cos x}{\sin x} = \cot x



I boxed where did you went wrong, and if you need further hints:

Spoiler

I only set one solution because it is too elementary to prove the second part of the problem using trigonometry or otherwise, which you just did.

Actually, in the second part you need to show that there is a constant value 'k' that fulfills the equation as said by 'where k is a constant'. Which you just proved saying 'unless k=2k = \sqrt{2}' since 2\sqrt{2} is a constant and not a function of x.

Sorry for making it hard to understand, I'll rephrase the question now. I'll accept your solution since it proves the equation although it does not use the previous equation. I am however still open to solutions if there's anyone else or you who can solve the problem using the previous equation :smile:
(edited 8 years ago)
Original post by udvuvvdu
x


My first part contained an elementary mistake, a byproduct of doing maths in latex-only at 3 in the morning. :rofl:
Original post by Indeterminate
For the benefit of any A-level students :smile:

Problem 517

Find

\displaystyle \int x\ln\left(x^2 + a^2 + \sqrt{x^2 - a^2}\right)\ dx


Solution 517
Is there an easier way to do this? :colondollar:

Let t=x2−a2  ⟹  dt=2x dxt= x^2 - a^2 \implies \mathrm{d}t = 2x \, \mathrm{d}x so that we get I=12∫ln⁡(t+t+2a2) dt\displaystyle I=\frac{1}{2}\int \ln(t + \sqrt{t}+2a^2) \, \mathrm{d}t and then using part with u=ln⁡f(t)u = \ln f(t) and v=tv=t, yielding:
I=tln⁡(t+t+2a2)−∫1+12t−12t+t+2a2t dt\displaystyle I=t \ln(t+\sqrt{t}+2a^2)-\int\frac{1+\frac{1}{2}t^{-\frac{1}{2}}}{t+\sqrt{t}+2a^2} t \, \mathrm{d}t

Now use t=w2t = w^2 so that I=tln⁡(t+t+2a2)−∫w2+12ww2+w+2a22w dw\displaystyle I=t \ln(t+\sqrt{t}+2a^2)-\int\frac{w^2+\frac{1}{2}w}{w^2+w+2a^2} 2w \, \mathrm{d}w

=tln⁡(t+t+2a2)−∫(2w−1)dw−∫w−2a2w+a2w2+w+2a2 dw\displaystyle =t \ln(t+\sqrt{t}+2a^2)-\int (2w-1) dw-\int\frac{w-2a^2w+a^2}{w^2+w+2a^2} \, \mathrm{d}w

=tln⁡(t+t+2a2)−w2+w−∫12(2w+1)(1−2a2)w2+w+2a2dw−∫−12+2a2w2+w+2a2 dw\displaystyle = t \ln(t+\sqrt{t}+2a^2)-w^2+w-\int\frac{\frac{1}{2}(2w+1)(1-2a^2)}{w^2+w+2a^2} dw-\int\frac{-\frac{1}{2}+2a^2}{w^2+w+2a^2} \, \mathrm{d}w

So that we get I=tln⁡(t+t+2a2)−w2+w−∫12(1−2a2)ydy−∫−12+2a2(w+12)2+2a2−14 dw\displaystyle I = t \ln(t+\sqrt{t}+2a^2)-w^2+w-\int\frac{\frac{1}{2}(1-2a^2)}{y} dy-\int\frac{-\frac{1}{2}+2a^2}{(w+\frac{1}{2})^2+2a^2-\frac{1}{4}} \, \mathrm{d}w

where y=w2+w+a2\displaystyle y = w^2+w+a^2

To get:

I=tln⁡(t+t+2a2)−w2+w−12(1−2a2)ln⁡∣y∣−−12+2a22a2−14arctan⁡(w+122a2−14)+c[br]\displaystyle I = t \ln(t+\sqrt{t}+2a^2)-w^2+w-\frac{1}{2}(1-2a^2)\ln|y|-\frac{-\frac{1}{2}+2a^2}{\sqrt{2a^2-\frac{1}{4}}}\arctan\left(\frac{w+\frac{1}{2}}{\sqrt{2a^2-\frac{1}{4}}}\right)+c[br]

and then plug in the relevant expressions to get it in terms of x again.
(edited 8 years ago)

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