SimonM
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(Updated as far as #214) SimonM - 07.05.2009

STEP I:
1: Solution by Square & DFranklin
2: Solution by DFranklin
3: Solution by coffeym
4: Solution by DeathAwaitsU
5: Solution by Coffeym
6: Solution by brianeverit
7: Solution by ukgea & Solution by nota bene
8: Solution by kabbers
9: Solution by brianeverit
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by nota bene
15: Solution by brianeverit
16: Solution by brianeverit


STEP II:
1: Solution by Mazzacre
2: Solution by brianeverit
3: Solution by brianeverit
4: Solution by brianeverit
5: Solution by Dystopia
6: Solution by Dystopia
7: Solution by Dystopia
8: Solution by brianeverit
9: Solution by brianeverit
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
15: Solution by brianeverit
16: Solution by brianeverit


STEP III:
1: Solution by DeanK22 Solution by SimonM
2: Solution by SimonM
3: Solution by Dystopia
4: Solution by Dystopia
5: Solution by Dystopia
6: Solution by Dystopia
7: Solution by ukgea
8: Solution by SimonM
9: Solution by SimonM
10: Solution by Dystopia
11: Solution by ben-smith
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
15: Solution by brianeverit
16: Solution by SimonM


Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
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Dystopia
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There isn't one. Some 1987 solutions were posted in the 1991 thread.
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generalebriety
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(Original post by Square)
Is there a thread for 1987 STEP solutions? I cant find it if there is.

If not we should make one
Go for it... That is, if you're ready to handle updating the first post...
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Square
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Well I'd need some contributions from other regulars, especially with II/III.

Got some of the STEP I questions written up though.
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insparato
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I can help keep it updated if you want. Go for it...
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Square
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I'll just make it this thread then, post your solutions gogo.

Gonna copy the format from the other threads and gonna put it into my opening post, will change the thread name also.
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Dystopia
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Anyone have a link to the papers?

Edit: Got them.
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nota bene
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Go for it, I'll do some questions this night - gonna go out to the cinema and then go to the pub for dinner, but will be back before midnight

edit: I've PMd the link
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Square
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Bugger, think there might actually be a mistake in my solution, taken it down for now.
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Dystopia
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STEP II, Q7.

The trapezium rule approximates the area y=f(x) under the curve by trapezia, all having a base of width h. The side lengths are f(n), f(n+h) for n from the lower limit to h less than the upper limit. The area of each trapezium is therefore h/2(f(n) + f(n+1)), which, when added together, results in the expression given.

\displaystyle \int^{n}_{1} \ln x \; \mathrm{d}x \approx \frac{1}{2}\left(\ln 1 + 2\ln 2 + \cdots + 2\ln(n-1) + \ln n\right) = \ln(n!) - \frac{1}{2}\ln n

\displaystyle \int^{n}_{1} \ln x \; \mathrm{d}x = \left[x\ln x - x\right]^{n}_{1} = n\ln n + 1 - n

\Rightarrow \ln(n!) \approx (n + \frac{1}{2})\ln n + 1 - n

\Rightarrow n! \approx n^{n + \frac{1}{2}}  e^{1-n} = g(n)

f(x) = \ln x, \; f''(x) = -\frac{1}{x^{2}}

Therefore ln x is a convex function, so the area under the curve is underestimated by trapezia.

\Rightarrow n! < n^{n + \frac{1}{2}}  e^{1-n}

\displaystyle \int^{n}_{1} \ln x \; \mathrm{d}x \approx \frac{1}{2k}\left(\ln 1 + 2\ln \frac{k+1}{k} + 2\ln\frac{k+2}{k} + \cdots + 2\ln n\right) - \frac{1}{2k}\ln n

= \frac{1}{k}(\ln(k+1) + \ln(k+2) + \cdots + \ln\frac{kn}{k}) - \frac{k(n-1)}{k} \ln k - \frac{1}{2k}\ln n \approx n\ln n + 1 - n

\Rightarrow \ln\frac{(kn)!}{k!} \approx k(n-1)\ln k + (kn+\frac{1}{2})\ln n + k(1-n)

\Rightarrow (kn)! \approx k! \times k^{k(n-1)} \times n^{kn+\frac{1}{2}} \times e^{k(1-n)}

\Rightarrow (kn)! \approx k! \; n^{kn+\frac{1}{2}} \; \left(\frac{e}{k}\right)^{k(1-n)}

As required. This is closer to (kn)! than g(kn) because the intervals are closer together for the second approximation and so it is more accurate.
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Square
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Differentiating
\displaystyle f(x)=e^{\text{a}x}\text{cosb}x



f'(x)=\text{ae}^{\text{a}x}\text  {cosb}x-\text{be}^{ax}\text{sinb}x



= \text{e}^{\text{a}x}(\text{acosb  }x-\text{bsinb}x)

Finding x-ordinates of turning points:

\displaystyle f'(x)=0

\displaystyle \text{e}^{\text{a}x}=0 



\text{no solutions}

\displaystyle \text{acosb}x-\text{bsinb}x=0



\text{a}=\text{b}\tan\text{b}x



\text{b}x=\text{n}\pi+\arctan(\f  rac{a}{b})

Consider any turning point: x=\frac{1}{b}(\text{n}\pi+\arcta  n(\frac{a}{b}))

\displaystyle f(\frac{1}{b}(\text{n}\pi+\arcta  n(\frac{a}{b}))=e^{a(\frac{n\pi}  {b}+\frac{1}{b}\arctan(\frac{a}{  b}))}\times\cos(n\pi+\arctan(\fr  ac{a}{b})

Expand out using addition formula sin terms go to zero:

\displaystyle e^{\frac{an\pi}{b}}\times e^{\frac{1}{b}\arctan(\frac{a}{b  })}\times\cos n\pi \arctan\frac{a}{b}

Now if n is even the expression is going to be:

\displaystyle e^{\frac{an\pi}{b}}\times e^{\frac{1}{b}\arctan(\frac{a}{b  })}\times \arctan\frac{a}{b}

And if n is odd the expression will be:
\displaystyle -e^{\frac{an\pi}{b}}\times e^{\frac{1}{b}\arctan(\frac{a}{b  })}\times \arctan\frac{a}{b}

Now consider the next turning point directly after the one shown above ie. (n+1)pi

x=\frac{1}{b}(\text{(n+1)}\pi+\a  rctan(\frac{a}{b}))

This expression will be of the form:

e^{\frac{a(n+1)\pi}{b}} \times e^{\frac{1}{b}\arctan(\frac{a}{b  })}\times \cos((n+1)\pi)\times \arctan\frac{a}{b}

Which is going to equal:

\displaystyle e^{\frac{a\pi}{b}} \times e^{\frac{an\pi}{b}}\times e^{\frac{1}{b}\arctan(\frac{a}{b  })}\times \cos((n+1)\pi) \times \arctan\frac{a}{b}

So, depending on from the last part if n is even or odd, it is now the opposite here, which means that this next turning point will have the opposite sign from the previous one.

So this is going to equal:

\displaystyle -e^{\frac{a\pi}{b}}f(\frac{1}{b}(  \text{n}\pi+\arctan(\frac{a}{b})  )

Since this is true for n and n+1 then all the turning points must be part of a geometric progression with common ratio: -e^{\frac{a\pi}{b}} as required.

If anyone can think of a better way of doing the explaining bit at the end they are more than welcome. I've missed out quite abit of the individual working stages but have labelled what I have done so hopefully you can follow it .

Maybe there is a tidier way to do it with GP formulae I dont know.
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Swayum
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STEP I, question 4.

Let S = \mathrm{log}_2 e - \mathrm{log}_4 e + \mathrm{log}_{16} e...

Take each term to logarithm base 2 (change the base rule)

\displaystyle S = \frac{\mathrm{log}_2 e}{1} - \frac{\mathrm{log}_2 e}{2} +  \frac{\mathrm{log}_2 e}{4}...

S = (\mathrm{log}_2 e})(1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32...)

Split the second brackets into two parts.

P = 1 + 1/4 + 1/16...
N = -1/2 - 1/8 - 1/32...

So

S = (\mathrm{log}_2 e})(P + N)

P is a geometric series with first term 1 and common ratio 1/4. Its sum to infinity is:

P = 1/(1-1/4) = 4/3

N is a geometric series with first term -1/2 and common ratio 1/4. Its sum to infinity is:

N = (-1/2)/(1-1/4) = -2/3

S = (\mathrm{log}_2 e})(P + N) = (\mathrm{log}_2 e})(4/3 - 2/3)

S = 2/3(\mathrm{log}_2 e})

Change the base to e, so we have the natural logarithm.

\displaystyle S = 2/3(\mathrm{log}_2 e}) = 2/3(\frac{\mathrm{log}_e e}{\mathrm{log}_e 2}) = 2/3(\frac{1}{\mathrm{log}_e 2})
\displaystyle S = \frac{2}{3\mathrm{ln} 2}

Using the power rule

\displaystyle S = \frac{2}{3\mathrm{ln} 2} = \frac{2}{\mathrm{ln} 8}

\displaystyle S = \frac{2*\frac{1}{2}}{\frac{1}{2}  \mathrm{ln} 8} (multiplying top and bottom by 1/2)

\displaystyle S = \frac{1}{\frac{1}{2}\mathrm{ln} 8} = \frac{1}{\mathrm{ln} \sqrt 8}

\displaystyle S = \frac{1}{\mathrm{ln} 2\sqrt 2}

WWWWW.
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DFranklin
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(Original post by Square)
If anyone can think of a better way of doing the explaining bit at the end they are more than welcome.
Write c = \arctan(\frac{a}{b}), x_n = \frac{1}{b}(n\pi + c) (so x_n is the nth turning point).

Then f(x_n) = e^{a(n\pi+c)/b}\cos(n\pi + c), so

\displaystyle \frac{f(x_{n+1}}{f(x_n} = \frac{ e^{a((n+1)\pi+c)/b}\cos((n+1)\pi + c)}{ e^{a(n\pi+c)/b}\cos(n\pi + c)} = e^{a\pi/b} \frac{\cos((n+1)\pi + c)}{\cos(n\pi+c)}

But \cos(y+\pi) = \cos(y)\cos(\pi)-\sin(\pi)\sin(y) = -\cos(y), so \frac{\cos((n+1)\pi + c)}{\cos(n\pi+c)} = -1.

So \frac{f(x_{n+1})}{f(x_n)} = -e^{a\pi/b} and so the values of f at the turning points form a GP with ratio  -e^{a\pi/b} as required.
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coffeym
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STEP I Q3

Our differential equation, referred to from here on as (*) is:

\displaystyle x^3\frac{dv}{dx}+x^2v=\frac{1+x^  2v^2}{(1+x^2)v}

Let \displaystyle y=xv \implies v=\frac{y}{x}

\displaystyle \implies \frac{dv}{dx}=\frac{x\frac{dy}{d  x}-y}{x^2}

Hence in (*):

\displaystyle x^2\frac{dy}{dx}=\frac{x(1+y^2)}  {y(1+x^2)}

After cancelling an x and rearranging, (*) reduces to:

\displaystyle \frac{y}{1+y^2}\frac{dy}{dx}=\fr  ac{1}{x(1+x^2)}

Hence, after dealing with partial fractions (which should be routine)

\displaystyle \int{\frac{y}{1+y^2}}\,dy=\int{\  frac{1}{x}-\frac{x}{1+x^2}}\,dx

Thus \displaystyle \frac12 \ln{(1+y^2)}=\ln{|x|}-\frac12 \ln{(1+x^2)}+c

Now x=1,v=1 \implies y=1 \implies c=\ln{2}

\displaystyle \therefore \ln{(1+y^2)}=\ln{\left(\frac{4x^  2}{1+x^2}\right)}

\displaystyle \implies 1+x^2v^2=\frac{4x^2}{1+x^2}

\displaystyle \implies x^2v^2=\frac{3x^2-1}{1+x^2}

\displaystyle \implies v^2=\frac{3}{1+x^2}-\frac{1}{x^2(1+x^2)}

Hence as x \to \infty, v \to 0

This completes the question, although I would like someone to have a quick scan for errors as I did this quite roughly.
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ukgea
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III/7

\displaystyle \arctan t = \int_0^t \frac 1{1 + x^2}dx

\displaystyle = \int_0^t \frac{1 + (-1)^nx^{2n+2} + (-1)^{n+1}x^{2n+2}}{1+x^2}dx

\displaystyle = \int_0^t \frac{1 + (-1)^nx^{2n+2}}{1+x^2} dx + (-1)^{n+1}\int _0^t\frac{x^{2n+2}}{1+x^2}dx

(using formula for sum of geometric progression with common ratio -x^2)

\displaystyle = \int_0^t \left(1 - x^2 + x^4 - \cdots + (-1)^nx^{2n}\right) dx + (-1)^{n+1}\int_0^t \frac{x^{2n+2}}{1+x^2}dx

\displaystyle = t - \frac{t^3}{3} + \frac{t^5}{5} - \cdots + \frac{(-1)^n t^{2n+1}}{2n+1} + (-1)^{n+1}\int_0^t \frac{x^{2n+2}}{1+x^2}dx

as required.

Rearranging this, we get

\displaystyle \left| \arctan t - \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1}\right| = \int_0^t \frac{x^{2n+2}}{1+x^2}dx

But for 0 \leq t \leq 1

\displaystyle \int_0^t \frac{x^{2n+2}}{1+x^2}dx \geq \int_0^t \frac{x^{2n+2}}{2}dx = \frac{t^{2n+3}}{2(2n+3)}

and

\displaystyle \int_0^t \frac{x^{2n+2}}{1+x^2}dx \leq \int_0^t \frac{x^{2n+2}}{1}dx = \frac{t^{2n+3}}{2n+3}.

The last three statements together prove

\displaystyle \frac{t^{2n+3}}{2(2n+3)} \leq \left| \arctan t - \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1}\right| \leq \frac{t^{2n+3}}{2n+3} (*)

as required.

Let us set t=1 in (*). Using the right half, we get

\displaystyle \left|\frac{\pi}{4} - \sum_{r=0}^n \frac{(-1)^r}{2n+1} \right| \leq \frac{1}{2n+3}

Since the RHS of this inequality clearly approches zero as n grows towards positive infinity, we have by squeezing that

\displaystyle \sum_{r=0}^n \frac{(-1)^r}{2n+1}

approaches \pi / 4 as n \rightarrow \infty, and the next required result immediately follows.

Let the error in the approximation of \pi be E. The error is given by

\displaystyle E = \left|\pi - 4\sum_{r=0}^n \frac{(-1)^r}{2r+1}\right|,

which is exactly four times the middle term in (*) for t=1. Thus from the left part of (*) it follows

\displaystyle E \geq \frac{4}{2(2n+3)}

For n \leq 98 we have

\displaystyle \frac{4}{2(2n+3)} \geq \frac{4}{398} > 10^{-2}

Thus we have

\displaystyle E > 10^-2

as required.
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Swayum
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On a side note, is it bad if I can only do 1 question per STEP 1 at this stage (the stage being week before start of upper sixth)?
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generalebriety
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(Original post by DeathAwaitsU)
On a side note, is it bad if I can only do 1 question per STEP 1 at this stage (the stage being week before start of upper sixth)?
Depends. Are you talking about 1987? No. If you're talking about 2004-present, then maybe. But you'll improve.
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eponymous
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(Original post by generalebriety)
Depends. Are you talking about 1987? No. If you're talking about 2004-present, then maybe. But you'll improve.
what is good if you have only done c1 and c2?
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generalebriety
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(Original post by eponymous)
what is good if you have only done c1 and c2?
If you've only done C1 and C2, I'd be surprised if you could do more than the odd STEP I question at all. The 1992 STEP II paper had quite a few nice questions, I seem to remember.
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eponymous
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(Original post by generalebriety)
The 1992 STEP II paper had quite a few nice questions, I seem to remember.
thanks, i'll go look at them now
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