Author Topic: 4U Maths Question Thread (Read 665307 times) Tweet Share

RuiAce · « **Reply #390 on:** August 13, 2016, 10:46:04 pm »

Quote from: massive on August 13, 2016, 10:24:44 pm

guys how do you do this?

$\text{Nothing stands obvious in the reduction formula so as usual, apply integration by parts}\\ \text{differentiating the whole thing and integrating the }1$
$\begin{align*}I_n&=\int_0^1(1-x^2)^ndx\\ &= \left[x(1-x^2)^n\right]_0^1-\int_0^1x(-2nx)(1-x^2)^{n-1}dx\\ &= 0+2n\int_0^1x^2(1-x^2)^{n-1}dx\end{align*}$
$\text{The tough part in this question lies in how we have to use add/take-away.}\\ \text{Because we need the }x^2\text{ to merge into }(1-x^2)\text{ we must add and subtract 1.}\\ \text{Note that the reason for doing this is just to reintroduce the reduction formula}$
$\textit{Also, watch how carefully I expand it.}$
$\begin{align*}I_n&=2n\int_0^1(x^2-1+1)(1-x^2)^{n-1}dx\\ &= 2n \int_0^1(x^2-1)(1-x^2)^{n-1}dx+2n\int_0^1(1-x^2)^{n-1}dx\\ &= -2n \int_0^1(1-x^2)^ndx + 2n\int_0^1(1-x^2)^{n-1}dx\\ \therefore (2n+1)I_n&=2nI_{n-1}\\ I_n&=\frac{2n}{2n+1}I_{n-1}\end{align*}$
$\text{As you'd expect, we need to apply the reduction formula for the second part.}\\ \textit{However, make sure you don't get lost between }n\textit{ and }2n \\ \begin{align*}I_n&=\frac{2n}{2n+1}I_{n-1}\\ &= \frac{2n}{2n+1}\, \frac{2n-2}{2n-1}I_{n-2}\\ &=\frac{2n}{2n+1}\, \frac{2n-2}{2n-1}\, \frac{2n-4}{2n-3} I_{n-3}\\ &\vdots \\ &= \frac{(2n)(2n-2)(2n-4)\dots (6)(4)(2)}{(2n+1)(2n-1)(2n-3)\dots (7)(5)(3)}I_0 \end{align*}$
$\text{Firstly, observe that }I_0=1\text{ upon evaluating the integral.}\\ \text{Look at what you're trying to prove. You need to get }(2n+1)!\text{ there.}\\ \text{But you're missing terms. In particular, you're missing }(2n)(2n-2)(2n-4)\dots (6)(4)(2)\text{ - what a coincidence!}$
$\text{So we just bring it in.}\\\begin{align*} I_n&=\frac{(2n)^2(2n-2)^2(2n-4)^2\dots(6)^2(4)^2(2)^2}{\left[(2n+1)(2n-1)(2n-3)\dots(7)(5)(3)\right]\left[(2n)(2n-2)(2n-4)\dots(6)(4)(2)\right](1)}\\ &=\frac{\left[(2n)(2n-2)(2n-4)\dots(6)(4)(2)\right]^2}{(2n+1)!}\end{align*}$
$\text{And NOW, observe how for the ugly }(2n)(2n-2)(2n-4)\dots (6)(4)(2)\\ \text{We can factorise out a 2 from EACH and EVERY one of those brackets.}\\ \text{If we factor the 2 out from ALL }n\text{ terms, we get this:}\\ (2n)(2n-2)(2n-4)\dots(6)(4)(2)=2^n\left[n(n-1)(n-2)\dots(3)(2)(1)\right]=2^n n!$
$\text{So putting this back in there:}\\ I_n=\frac{(2^nn!)^2}{(2n+1)!}=\frac{2^{2n}(n!)^2}{(2n+1)!}$

massive · « **Reply #391 on:** August 13, 2016, 11:17:58 pm »

Quote from: RuiAce on August 13, 2016, 10:46:04 pm

$\text{As you'd expect, we need to apply the reduction formula for the second part.}\\ \textit{However, make sure you don't get lost between }n\textit{ and }2n \\ \begin{align*}I_n&=\frac{2n}{2n+1}I_{n-1}\\ &= \frac{2n}{2n+1}\, \frac{2n-2}{2n-1}I_{n-2}\\ &=\frac{2n}{2n+1}\, \frac{2n-2}{2n-1}\, \frac{2n-4}{2n-3} I_{n-3}\\ &\vdots \\ &= \frac{(2n)(2n-2)(2n-4)\dots (6)(4)(2)}{(2n+1)(2n-1)(2n-3)\dots (7)(5)(3)}I_0 \end{align*}$

How do you get that?

RuiAce · « **Reply #392 on:** August 13, 2016, 11:23:52 pm »

Quote from: massive on August 13, 2016, 11:17:58 pm

How do you get that?

$\text{By repeatedly using the reduction formula.}\\ \begin{align*}I_n &= \frac{2n}{2n+1}I_{n-1}\\ I_{n-1}&=\frac{2(n-1)}{2(n-1)+1}I_{n-2}=\frac{2n-2}{2n-1}=I_{n-2}\\ \therefore I_n&=\frac{2n}{2n+1}\,\frac{2n-2}{2n-1}I_{n-2}\\ I_{n-2}&=\frac{2(n-2)}{2(n-2)+1}I_{n-3}=\frac{2n-4}{2n-3}I_{n-3}\\ \therefore I_n &= \frac{2n}{2n+1}\,\frac{2n-2}{2n-1}\,\frac{2n-4}{2n-3}I_{n-3}\end{align*}\\ \text{Then evaluate it down to }I_0\text{ because the formula doesn't go beyond }n=0$

massive · « **Reply #393 on:** August 14, 2016, 07:51:46 am »

Yo guys how do you do part iii. Rui I tried doing the method you did by taking it to I₀ but i ended up with:
1/(I_n) = (n+2)(n+1), am i doing something wrong?

RuiAce · « **Reply #394 on:** August 14, 2016, 08:49:24 am »

Quote from: massive on August 14, 2016, 07:51:46 am

Yo guys how do you do part iii. Rui I tried doing the method you did by taking it to I₀ but i ended up with:
1/(I_n) = (n+2)(n+1), am i doing something wrong?

I subbed in values for n on WolframAlpha. That question has a mistake; it is (n+1)(n+2)/2 as you said.

massive · « **Reply #395 on:** August 14, 2016, 01:43:23 pm »

hey can someone get this? i got the answer except with a 1/2 infront :S

RuiAce · « **Reply #396 on:** August 14, 2016, 02:06:55 pm »

Quote from: massive on August 14, 2016, 01:43:23 pm

hey can someone get this? i got the answer except with a 1/2 infront :S

$\text{What a trick question. Don't need integration by parts here, let alone reduction.}\\ \text{Let }u=1-x \implies -du=dx$
$\begin{align*}\int_0^1x(1-x)^ndx&=\int_1^0(1-u)u^n(-du)\\ &=\int_0^1 (u^n-u^{n+1})du\\ &= \left[\frac{u^{n+1}}{n+1}-\frac{u^{n+2}}{n+2}\right]_0^1\\ &= \frac{1}{n+1}-\frac{1}{n+2}\\ &= \frac{1}{(n+1)(n+2)}\\ &= \frac{n!}{n!(n+1)(n+2)}\\ &= \frac{n!}{(n+2)!}\end{align*}$
$\text{If you did integration by parts, however, you should've gotten this}\\ \frac{1}{n+1}\int_0^1 (1-x)^{n+1}dx$

massive · « **Reply #397 on:** August 14, 2016, 02:29:39 pm »

Thanks Rui!

Also how do do you do the q attached?

RuiAce · « **Reply #398 on:** August 14, 2016, 03:22:39 pm »

Quote from: massive on August 14, 2016, 02:29:39 pm

Thanks Rui!

Also how do do you do the q attached?

$\text{There's so many ways to do this question and I don't even know }\\ \text{what's technically the fastest way. So here's just one possible method.}$
$\begin{align*}\tan \frac{\theta}{2}&=\tan \left(\frac{\pi}{2}-\frac{1}{2}\tan^{-1}\frac{3}{4}\right)\\ &= \cot \left(\frac{1}{2}\tan^{-1}\frac{3}{4}\right)\end{align*}\\ \text{So how do we find that? Well.}$
$\text{Let }x=\frac{1}{2}\tan^{-1}\frac{3}{4} \text{ for the sake of neatness.}\\ \text{Using our }\textbf{double angle formula}$
$\begin{align*}\frac{2\tan x}{1-\tan^2x}&=\tan (2x)\\ &= \tan \left(\tan^{-1}\frac{3}{4}\right)\\ &= \frac{3}{4}\end{align*}$
$\text{So now we're actually doing nothing but solving a quadratic.}\\ \begin{align*}\frac{2\tan x}{1-\tan^2 x}&=\frac{3}{4}\\ 8\tan x&=3-3\tan^2x\\ 3\tan^2x+8\tan x-3&=0\\ (3\tan x-1)(\tan x+3)&=0\\ \tan x = \frac{1}{3} \text{ o}&\text{r } \tan x = -3\\ \cot x = 3 \text{ o}&\text{r }\cot x=-\frac{1}{3}\end{align*}$
$\text{Remember that }x=\frac{1}{2}\tan^{-1}\frac{3}{4}\\ \text{This will be a first quadrant angle.}\\ \text{Tan and Cot are positive in the first quadrant! Therefore we reject the negative solution.}$
$\begin{align*}\cot x &= 3\\ \cot \left(\frac{1}{2}\tan^{-1}\frac{3}{4}\right)&=3\\ \tan \frac{\theta}{2} &=3 \end{align*}$

birdwing341 · « **Reply #399 on:** August 14, 2016, 09:49:54 pm »

Hello!
Anyone got some top tips for how to deal with harder projectile motion questions that you don't know how to do?
Any help would be much appreciated!!
Thanks in advance

RuiAce · « **Reply #400 on:** August 14, 2016, 09:56:13 pm »

Quote from: birdwing341 on August 14, 2016, 09:49:54 pm

Hello!
Anyone got some top tips for how to deal with harder projectile motion questions that you don't know how to do?
Any help would be much appreciated!!
Thanks in advance

It is extremely rare that a projectile question should appear in the 4U exam. As far as my memory goes, the only time it was asked was sometime during 2003-2006 where they combined it with resisted motion.

Despite this, projectile motions are a simple matter of:
Early parts - Using relevant definitions as appropriate (e.g. y=0 for a time of flight) and just dealing with messy algebra.
Late parts - Expect to use anything you know. One can never predict what they will need on the day. (However, fact is after a truly sufficient amount of past papers, you'd be trained to adjust to the new ways of thought and thus be better equipped to handle the peculiar questions.)

The biggest tips with projectile motion are basically to know your fundamentals and then expect the unexpected. It's much easier to answer a question than to give generic tips, especially for questions on topics such as this.

massive · « **Reply #401 on:** August 15, 2016, 09:59:45 am »

hey guys is ln(x/2) the same as ln( (tanx/2 - 1) / (tanx/2) )? Also if it is true, how? Thanks!!

RuiAce · « **Reply #402 on:** August 15, 2016, 10:04:10 am »

Quote from: massive on August 15, 2016, 09:59:45 am

hey guys is ln(x/2) the same as ln( (tanx/2 - 1) / (tanx/2) )? Also if it is true, how? Thanks!!

Nope.

Change the formatting a bit and still nope.

massive · « **Reply #403 on:** August 15, 2016, 10:29:15 am »

oh then how do you do the q attached. my ans was ln( (tanx/2 - 1) / (tanx/2) ) but the answer is ln(x/2). Thanks.

RuiAce · « **Reply #404 on:** August 15, 2016, 10:46:52 am »

Quote from: massive on August 15, 2016, 10:29:15 am

oh then how do you do the q attached. my ans was ln( (tanx/2 - 1) / (tanx/2) ) but the answer is ln(x/2). Thanks.

$\text{Let }t=\tan \frac{\theta}{2}\text{ so that }\theta = 2 \tan^{-1}t\\ \text{Then, }\sin \theta = \frac{2t}{1+t^2}, \, \cos \theta = \frac{1-t^2}{1+t^2}\text{ and }d\theta = \frac{2}{1+t^2}dt$
$\begin{align*}\int \frac{d\theta}{1-\cos \theta - \sin \theta}&=\int \frac{\frac{2}{1+t^2}dt}{1-\frac{1-t^2}{1+t^2}-\frac{2t}{1+t^2}}\\ &= \int \frac{2}{1+t^2-1+t^2-2t}dt\\ &= \int \frac{1}{t(t-1)}dt\\ &= \int\left(\frac{1}{t-1}-\frac{1}{t}\right)dt\\ &= \ln |t-1|-\ln |t|+C\\&= \ln \left|\frac{\tan \frac{\theta}{2} -1}{\tan \frac{\theta}{2}}\right| \end{align*}\\ \text{As was what you said.}$
$\textbf{The given answer is undeniably wrong. This can be shown working backwards.}\\ \frac{d}{dx} \ln \frac{x}{2} = \frac{1}{x}\\ \textbf{which looks nothing like the original integral.}$

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