The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Year 13 Maths Help Thread

Scroll to see replies

Original post by Ano123
Any ideas?


Coincidentally enough I was learning about reduction formulae when you posted this and I got:

In=∫012πcos⁡nx dx=n−1n⋅In−2I_n=\displaystyle \int _0^{\frac{1}{2} \pi } \cos^{n} x \ dx = \frac{n-1}{n} \cdot I_{n-2}

Which would mean that

I100=∏i=1100(2i−1)2(50!)⋅π2\displaystyle I_{100}=\frac{\prod_{i=1}^{100} (2i-1)}{2(50!)} \cdot \frac{\pi}{2}

To be honest I have no idea if I am correct or how to get it in the wanted form, so good luck.
Original post by RDKGames
Coincidentally enough I was learning about reduction formulae when you posted this and I got:

In=∫012πcos⁡nx dx=n−1n⋅In−2I_n=\displaystyle \int _0^{\frac{1}{2} \pi } \cos^{n} x \ dx = \frac{n-1}{n} \cdot I_{n-2}

Which would mean that

I100=Πi=1100(2i−1)2(50!)⋅π2\displaystyle I_{100}=\frac{\Pi_{i=1}^{100} (2i-1)}{2(50!)} \cdot \frac{\pi}{2}

To be honest I have no idea if I am correct or how to get it in the wanted form, so good luck.


Your answer is not quite right. The answer is πΓ(100)2100(50Γ2(50)) \displaystyle \frac{ \pi \Gamma (100)}{2^{100} ( 50\Gamma ^2(50))} if it gives you a clue haha.
(edited 7 years ago)
Original post by B_9710
Your answer is not quite right. The answer is πΓ(100)2100(50Γ2(50)) \displaystyle \frac{ \pi \Gamma (100)}{2^{100} ( 50\Gamma ^2(50))} if it gives you a clue haha.


Well, that's neat. I never came across that gamma function and notation before and I've no clue how to work with it nor how to get to that form.
Original post by RDKGames
That would be the limit, yes. I think it would make more sense to rewrite it as:

∫abx−2dx=1a,0<a<b,b→∞\int^b_a x^{-2} dx=\frac{1}{a}, 0<a<b, b\rightarrow \infty

but I'm not sure. Something about limits should be in there anyway.


Should be ∫a∞x−2 dx=lim⁡b→∞∫abx−2 dx\displaystyle \int_a^{\infty} x^{-2} \, \mathrm{d}x = \lim_{b \to \infty} \int_a^b x^{-2} \, \mathrm{d}x
(edited 7 years ago)
Original post by RDKGames
Well, that's neat. I never came across that gamma function and notation before and I've no clue how to work with it nor how to get to that form.


Basically you have to know how to get ∏n=0N(2n+1) \displaystyle \prod _{n=0}^{N} (2n+1) and ∏n=0N2n \displaystyle \prod _{n=0}^{N} 2n in terms of N! N! . And because Γ(x)=(x−1)! \Gamma (x) = (x-1)! I just used it for banter.
Hi, whats a good textbook for OCR A2 Maths? I've heard the CGP guides aren't very good.
Reply 446
Avatar for Palette
Palette
OP
14
Is it unheard of for the second part of a two-part STEP question to be easier than the first part? That's what's going on with me with the question I'm currently on (can do the second part with ease but can't do the first part).
Original post by Palette
Is it unheard of for the second part of a two-part STEP question to be easier than the first part? That's what's going on with me with the question I'm currently on (can do the second part with ease but can't do the first part).


No, it's quite common. Especially if the second part is just checking some cool special case or something, whilst the first part is more general.
Original post by DarkEnergy
Hi, whats a good textbook for OCR A2 Maths? I've heard the CGP guides aren't very good.


Not familiar with OCR but.. I'd have hoped that the textbook was enough :tongue: with maths it's all about practice, a good textbook should offer this as well as the learning material.. all you need (I hope the OCR textbooks do this as well as Edexcel does). You can probably find extension material online.
Original post by SeanFM
Not familiar with OCR but.. I'd have hoped that the textbook was enough :tongue: with maths it's all about practice, a good textbook should offer this as well as the learning material.. all you need (I hope the OCR textbooks do this as well as Edexcel does). You can probably find extension material online.


I don't know if I'm being stupid (probably am) but I don't know what textbook people keep referring to, we were never told about any textbook in class or anything, only the CGP revision guides. Thanks for the reply.
Original post by DarkEnergy
I don't know if I'm being stupid (probably am) but I don't know what textbook people keep referring to, we were never told about any textbook in class or anything, only the CGP revision guides. Thanks for the reply.


I've just googled and I'm just as confused :tongue: there doesn't seem to be a set textbook or even individual ones.. I see a few versions of ones that combine C3 and C4... interesting. Probably can't go too far wrong with any of those.

Is this with MEI or Non-MEI?
Original post by SeanFM

Is this with MEI or Non-MEI?


Non-MEI
Im doing some questions on algebra & functions (first few C3 topics) but i dont have any answers :3
Is anyone willing to mark them for me? - let me know and ill send a PM with the questions and my answers.
Original post by kiiten
Im doing some questions on algebra & functions (first few C3 topics) but i dont have any answers :3
Is anyone willing to mark them for me? - let me know and ill send a PM with the questions and my answers.


Sure.
Original post by RDKGames
Sure.


Sorry someone already sent me a PM :/. But thank you anyway :smile:

Posted from TSR Mobile
Why is the integral of x/x^2 +1 =1/2 ln|x^2 +1| not sure what c4 method to use to get and understand this results
Reply 456
Avatar for solC
solC
14
Original post by youreanutter
Why is the integral of x/x^2 +1 =1/2 ln|x^2 +1| not sure what c4 method to use to get and understand this results


Notice how the numerator is similar to the derivative of the denominator, and go from there.
Original post by youreanutter
Why is the integral of x/x^2 +1 =1/2 ln|x^2 +1| not sure what c4 method to use to get and understand this results


int(x/x^2+1) = 1/2int(2x/x^2+1) inside the integral we now have derivative over function so the integral becomes ln of the denominator ln(x^2+1) and the result is (1/2)ln(x^2+1) + C
Original post by DarkEnergy
Non-MEI


:hmmmm: strange that there is no set textbook.. sorry.
Original post by SeanFM
:hmmmm: strange that there is no set textbook.. sorry.


No problem, cheers for looking anyways. Reckon this will be alright? link

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you