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Daniel Freedman
I'll give credit to the source of this afterwards, as the solution is on there too.

For all real x, evaluate r=0sin3(3rx)3r \displaystyle \sum_{r=0}^{\infty} \frac{\sin^3{(3^r x)}}{3^r}


never done such a limit with any trig or infact never seen one until now.

some kind of tip?
DeanK22
never done such a limit with any trig or infact never seen one until now.

some kind of tip?


Presumably you've writted out the first few terms and aren't getting anywhere?

A good idea with summations is to try and express the general term in a nicer form that you can sum.
Reply 1322
DeanK22
You are pondering over a question that makes absolutely no sense - 2 is a symbol specifying an amount yet we could have arbitarly chosen another amount [i.e. in our conventional view of numbers we could have gone from 1 object to three objects with no 2 objects represented by a whole number].


I thought it would be something like that! :wink:
Reply 1323
Daniel Freedman
I'll give credit to the source of this afterwards, as the solution is on there too.

For all real x, evaluate r=0sin3(3rx)3r \displaystyle \sum_{r=0}^{\infty} \frac{\sin^3{(3^r x)}}{3^r}


Very nice question. My first though was **** me that looks bitchy

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SimonM
Very nice question. My first though was **** me that looks bitchy

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Telescopes nicely. :smile:

Offtopic: Is University of Leicester good for math? I just got an offer.
SimonM
Very nice question. My first though was **** me that looks bitchy

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how have you taken the sum?
thomaskurian89
Telescopes nicely. :smile:

Offtopic: Is University of Leicester good for math? I just got an offer.


...

... yeah it is good - maths staff (one of them) recieved his degree from there and spoke highly - it is currently an active place for new professors because of ssome kind of scholarship they have in place - apparently some very talented.
DeanK22
how have you taken the sum?


????
DeanK22
how have you taken the limit?


r=0nsin3(3rx)3r=14r=0n3sin(3rx)sin(3r+1x)3r=34sinx14sin(3n+1x)3n \displaystyle \sum_{r=0}^{n}{\frac{\sin^3(3^{r} x)}{3^r}}=\frac{1}{4}\sum_{r=0}^{n} \frac{3\sin(3^r x)-\sin(3^{r+1}x)}{3^r}=\frac{3}{4}\sin x-\frac{1}{4}\frac{\sin(3^{n+1}x)}{3^n}

Now let n tend to infinity, and the second expression vanishes.

Unparseable latex formula:

\bigstar \boxed{ \mbox{Nicely done, Simon [s]smile[/s]}} \bigstar

Reply 1329
Daniel Freedman

Unparseable latex formula:

\bigstar \boxed{ \mbox{Nicely done, Simon [s]smile[/s]}} \bigstar



Thanks, (just got home)

Sauce?
... vm appreciated μon \mu on rather oddly the smiley face looks really scary though :yep:
DeanK22
... vm appreciated μon \mu on rather oddly the smiley face looks really scary though :yep:

Anytime. :top2:

p.s. I prefer μ\mu^- :wink:
... Prove

tan(A)tan(B)tan(C)=tan(A)+tan(B)+tan(C) \displaystyle tan(A)tan(B)tan(C) = tan(A) + tan(B) + tan(C) if A+B+C=π \displaystyle A + B + C = \pi
DeanK22
... Prove

tan(A)tan(B)tan(C)=tan(A)+tan(B)+tan(C) \displaystyle tan(A)tan(B)tan(C) = tan(A) + tan(B) + tan(C) if A+B+C=π \displaystyle A + B + C = \pi


Very similar to a question we've had before.

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Reply 1334
Hi, I couldn't find any thread for past papers, so I'm just gonna post my query here..

Does anyone know where I can get my hands on M3 and M4 past papers? Solomons too would be very helpful, but I have no idea where to get them from. (Btw, don't say school! Is there anywhere else?)
Reply 1335
Merci, Merci beaucoup!
Sue and Bob take turns rolling a 6-sided die. Once either person rolls a 6, the game is over. Sue rolls first. If she doesn’t roll a 6, Bob rolls the die; if he doesn’t roll a 6, Sue rolls again. They continue taking turns until one of them rolls a 6.

Bob rolls a 6 before Sue.

What is the probability Bob rolled the 6 on his second turn?

(from xkcd)
Reply 1337

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SimonM

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Forgive this stupid question, but does it mean 'what is the probability Bob rolled the 6 on his second roll, given that he rolls a 6 before Sue?' If so, I don't see why it's not just:

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If I'm not understanding this right, could you please explain it?
rupertj: "Bob rolls a 6 before Sue."

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