Getting to Cambridge: STEP by STEP!

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The lad himself. Big up to us self-teaching bros, we'll get through it in the end. How did you find your Jan exams?

(If STEP does go south, which it won't, It'd be awesome to see you at Warwick mate)

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Reply 141

Zacken

Original post by aymanzayedmannan

...

Do you have a favourite reduction formulae question? :biggrin:

Reply 142

Ko 20

Original post by Zacken

Who said I got high physics grades?

Well if you got a offer from cambridge and they think your going to get a A*/A you must be doing well

Reply 143

Zacken

Original post by Princepieman

Haven't seen you around in a while, what's new? Congrats on the purple. Self-teaching lads ftw.

Jan exams were mainly okay, messed M3, floundered one question in M2 - missed a few marks in S2, etc...

You've firmed Warwick, then?

Reply 144

Zacken

Original post by Ko 20

Well if you got a offer from cambridge and they think your going to get a A*/A you must be doing well

Fair. I wouldn't really take revision tips from me. I haven't taken notes in years. Just watch online lectures on youtube and do past papers is how I roll, I can't see that being optimal for a lot of people though.

Reply 145

username738914

Original post by Zacken

Haha, it's been a while man. Just quit my part time job on Sunday so it's nose to brimstone (a.k.a no-lifing) from now until exam season. Apart from that, not much else! Thanks, I feel somewhat powerful now :colone:

Good to hear - you've smashed them no doubt. Ach, a few marks won't mean much, as long as you gave it your best shot.

I would, if Bristol actually got back to me (coming up on 5 months of waiting now)!!!

Reply 146

Ayman!

Original post by Zacken

Do you have a favourite reduction formulae question? :biggrin:

It's difficult choosing one in particular, but I really love it when I can do the algebraic ones like these :colondollar:

Unparseable latex formula:

\displaystyle \[I_{n} =\int_{0}^{1}x^{n}\sqrt{(1-x^{2})}\mathrm{d}x

Show that,

Unparseable latex formula:

\displaystyle (n+2)I_{n}=(n-1)I_{n-2}\]

Not STEP standard, but it's decent for FP3 level!

Edit: I attached the screenshots before but they didn't seem to upload on the preview so I latex'ed the integral later. You can ignore them :tongue:

(edited 8 years ago)

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Screen Shot 2016-02-09 at 23.53.16.png137.5KB

Reply 147

Zacken

Original post by aymanzayedmannan

...

\displaystyle I_n = \int_0^1 x^n \sqrt{1-x^2} \, \mathrm{d}x = \int_0^1 x^{n-1} \cdot x \sqrt{1-x^2} \, \mathrm{d}x

.

Using Integration by Parts with

u = x^{n-1}

and

v = -\frac{1}{3}\sqrt{1-x^2}(1-x^2) \, \mathrm{d}x

we have:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}I_n = \frac{n-1}{3} \int_0^1 x^{n-2} \sqrt{1-x^2} (1-x ) \, \mathrm{d}x \iff I_n = \frac{n-1}{3} I_{n-2} - \frac{n-2}{3}I_n\end{equation*}

So, some re-arranging yields:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}3I_n + (n-1)I_n = (n-1)I_{n-2} \iff (n+2)I_n = (n-1)I_{n-2}\end{equation*}

Aww yiss, this was lovely. Many thanks. :smile:

(edited 8 years ago)

Reply 148

Gregorius

Original post by Zacken

\displaystyle I_n = \int_0^1 x^n \sqrt{1-x^2} \, \mathrm{d}x = \int_0^1 x^{n-1} \cdot x \sqrt{1-x^2} \, \mathrm{d}x

.

Using Integration by Parts with

u = x^{n-1}

and

v = -\frac{1}{3}\sqrt{1-x^2}(1-x^2) \, \mathrm{d}x

we have:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}I_n = \frac{n-1}{3} \int_0^1 x^{n-2} \sqrt{1-x^2} (1-x ) \, \mathrm{d}x \iff I_n = \frac{n-1}{3} I_{n-2} - \frac{n-2}{3}I_n\end{equation*}

So, some re-arranging yields:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}3I_n + (n-1)I_n = (n-1)I_{n-2} \iff (n+2)I_n = (n-1)I_{n-2}\end{equation*}

Aww yiss, this was lovely. Many thanks. :smile:

Nice. Do you know this one? If

\displaystyle I_n = \int_{-1}^{+1} (1-x^2)^n \cos(\alpha x) \mathrm{d} x

show that

\displaystyle \alpha^2 I_n = 2n(2n-1) I_{n-1} - 4n(n-1)I_{n-2}

Reply 149

Zacken

Original post by Gregorius

Nice. Do you know this one? If

\displaystyle I_n = \int_{-1}^{+1} (1-x^2)^n \cos(\alpha x) \mathrm{d} x

show that

\displaystyle \alpha^2 I_n = 2n(2n-1) I_{n-1} - 4n(n-1)I_{n-2}

I do not, but it is a welcome distraction. Thank you! Biting into it now. :smile:

Reply 150

Ayman!

Original post by Zacken

...

haha don't thank me. you just successfully finished a subchapter!!! :banana:

Reply 151

Gregorius

Original post by Zacken

I do not, but it is a welcome distraction. Thank you! Biting into it now. :smile:

There's a rider to the question when you've finished... :colone:

Reply 152

luminarychild

Looks like we have something in common, me and you both started off in similar positions, I admire your hard work.
That is a fxck load of exams but it is very achievable I have witnessed it happened on more than one occasion.

Good Luck

Reply 153

Insight314

Original post by Zacken

Update:

I will be useful for once. I'm going to start looking over FP3 reduction formulae and do some of the exercises.

Ayyy we on the same exercise m8

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Reply 154

Zacken

Original post by Gregorius

Nice. Do you know this one? If

\displaystyle I_n = \int_{-1}^{+1} (1-x^2)^n \cos(\alpha x) \mathrm{d} x

show that

\displaystyle \alpha^2 I_n = 2n(2n-1) I_{n-1} - 4n(n-1)I_{n-2}

Took me far too long to spot the

-x^2 = 1 - x^2 - 1

trick. Anyways, this requires two bouts of integration, the first one yields, with

u = (1-x^2)^n

and

\displaystyle v = \frac{1}{\alpha}\sin \alpha x

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \frac{\alpha I_n}{2n} = \int_{-1}^{1} x(1-x^2)^{n-1} \sin \alpha x \, \mathrm{d}x[br]\end{equation}

Now, we bash it with IBP again, this time with

u = x(1-x^2)^{n-1}

and

Unparseable latex formula:

\displaystlye v = -\frac{1}{\alpha} \cos \alpha x

which gives us:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\frac{\alpha^2}{2n}I_n = I_{n-1} + 2(n-1)\int_{-1}^{1} \cos \alpha x (1-x^2)^{n-2} (-x^2) \, \mathrm{d}x \end{equation*}

Clever bit:

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \frac{\alpha^2}{2n}I_n = I_{n-1} + 2(n-1)\int_{-1}^{1} \cos \alpha x (1-x^2)^{n-2} (1-x^2 - 1) \, \mathrm{d}x \end{equation*}

So:

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \frac{\alpha^2}{2n}I_n = (2n-1)I_n - 2(n-1)I_{n-2} \iff \alpha^2 I_n = 2n(2n-1)I_{n-1} - 4n(n-1)I_{n-2} \end{equation*}

as required. Lovely.

Reply 155

Zacken