STEP Maths I,II,III 1987 Solutions

There isn't one. Some 1987 solutions were posted in the 1991 thread.

Reply 2

generalebriety

16

Square

Is there a thread for 1987 STEP solutions? I cant find it if there is.

If not we should make one :biggrin:

Go for it... :wink:

That is, if you're ready to handle updating the first post...

Reply 3

Well I'd need some contributions from other regulars, especially with II/III.

Got some of the STEP I questions written up though.

Reply 4

insparato

15

I can help keep it updated if you want. Go for it...

Reply 5

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.

Reply 6

Dystopia

Anyone have a link to the papers?

Edit: Got them.

Reply 7

nota bene

15

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 :wink:

edit: I've PMd the link

Reply 8

Bugger, think there might actually be a mistake in my solution, taken it down for now.

Reply 9

Dystopia

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.

Reply 10

Differentiating

\displaystyle f(x)=e^{\text{a}x}\text{cosb}x[br][br]f'(x)=\text{ae}^{\text{a}x}\text{cosb}x-\text{be}^{ax}\text{sinb}x[br][br]= \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 [br][br]\text{no solutions}

\displaystyle \text{acosb}x-\text{bsinb}x=0[br][br]\text{a}=\text{b}\tan\text{b}x[br][br]\text{b}x=\text{n}\pi+\arctan(\frac{a}{b})

Consider any turning point:

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

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

.

Maybe there is a tidier way to do it with GP formulae I dont know.

Reply 11

Swayum

18

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}...

Unparseable latex formula:

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

Unparseable latex formula:

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

Unparseable latex formula:

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

Unparseable latex formula:

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

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

Unparseable latex formula:

\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.

Reply 12

DFranklin

18

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.

Reply 13

coffeym