STEP Maths I, II, III 1988 solutions Watch

insparato
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If you look at the two posts it took me 9 minutes therefore I win :proud:.

I suspect i just picked the two easiest questions out of the two papers. I feel guilty for doings maths .
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kabbers
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ok doing II/1if no one else is
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insparato
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I've had a look at it, on first glance it seems like not a bad question! You go for it though.
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kabbers
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cheers yep, a very straightforward one (all the more embarrasing any mistakes will be :p:)

ok firstly, work out the sum to infinity (of a geometric sequence)

\displaystyle\sum^n_{n=0} r^n = 1 + r + r^2 + .. + r^n

 = \frac{r-1}{r-1}(1 + r + r^2 + .. + r^n)

 = \frac{1}{r-1}(r - 1 + r^2 - r + r^3 - r^2 + .. + r^{n+1} - r^n)

 = \frac{r^{n+1} - 1}{r-1}


therefore, iff |r| < 1, as n \to \infty, \frac{r^{n+1} - 1}{r-1} \to \frac{1}{1-r} (*)

Also note that

\displaystyle\sum^n_{n=1} r^n = r + r^2 + .. + r^n

 = \frac{r-1}{r-1}(r + r^2 + r^3 + .. + r^n)

 = \frac{1}{r-1}(r^2 - r + r^3 - r^2 + r^4 + .. + r^{n+1} - r^n)

(**) = \frac{r^{n+1}-r}{r-1}

therefore, iff |r| < 1, as n \to \infty, \frac{r^{n+1} - r}{r-1} \to \frac{r}{1-r} (**)


part 1,

\displaystyle\sum^{\infty}_{n=0} x^n - \frac{1}{2}\sum^{\infty}_{n=0} (\frac{x}{2})^n

|x| < 1, so by (*),

= \frac{1}{1-x} - \frac{1}{2} \frac{1}{1-\frac{x}{2}}

= \frac{1}{1-x} - \frac{1}{2} \frac{2}{2-x}

= \frac{1}{1-x} - \frac{1}{2-x}

= \frac{1}{x-2} - \frac{1}{x-1}

= \frac{x-1}{(x-2)(x-1)} - \frac{x-2}{(x-1)(x-2)}

= \frac{1}{(x-2)(x-1)} = f(x) as required.


part 2,


\displaystyle\sum^{\infty}_{n=1} x^{-n} - \frac{1}{2}\sum^{\infty}_{n=0} (\frac{x}{2})^n

\displaystyle= \sum^{\infty}_{n=1} (\frac{1}{x})^{n} - \frac{1}{2}\sum^{\infty}_{n=0} (\frac{x}{2})^n

since 1<|x|<2, both 1/x and x/2 fall within the applicable range (|r|<1), so by (**) & (*)

- \frac{\frac{1}{x}}{1-\frac{1}{x}} - \frac{1}{2-x} (I just copied the second part from above)

=- \frac{1}{x-1} - \frac{1}{2-x}

=- \frac{1}{x-1} + \frac{1}{x-2}

= \frac{x-1}{(x-2)(x-1)} - \frac{x-2}{(x-1)(x-2)}

= \frac{1}{(x-2)(x-1)} = f(x) as required.



part 3,


for |x| > 2, \displaystyle\sum^{\infty}_{n=1} x^{-n} is ok, but \displaystyle\frac{1}{2}\sum^{\i  nfty}_{n=0} (\frac{x}{2})^n is not.

so, we need something that will give us 1/(x-2)

I guessed \displaystyle\sum^{\infty}_{n=1} (\frac{2}{x})^n, it seemed to fit the way the question progresssed

\displaystyle\sum^{\infty}_{n=1} (\frac{2}{x})^n

by (**),
= \frac{\frac{2}{x}}{1-\frac{2}{x}}

= \frac{2}{x-2} hooray it works


Therefore,


\displaystyle\frac{1}{2}\sum^{\i  nfty}_{n=1} (\frac{2}{x})^n - \sum^{\infty}_{n=1} (\frac{1}{x})^n

= \frac{1}{x-2} - \frac{1}{x-1}

= f(x) (as above)


So
\displaystyle\frac{1}{2}\sum^{\i  nfty}_{n=1} (\frac{2}{x})^n - \sum^{\infty}_{n=1} (\frac{1}{x})^n

is the desired expansion
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Square
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Right so I've been working on II/7 but I'm a little bit confused.

It looks to me as if there have been a few substitutions going on here but the questions has been using the same variables across.

For example the first part is fairly easy, transforming the integral I into the first equation is easily done by the substitution x=1+y, which gives the same answer but in terms of y, however the question gives it in terms of x, the original variable which makes me think I've perhaps missed something and I'm not supposed to use a substitution after all?
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squeezebox
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STEP III - Question 7

When n=0: \frac{d^{2}y_{0}}{dx^{2}} - \omega^{2}x^{2}y_{0} + \omegay_{0} = 0 (*).

We have: y_{0}(x) = e^{-\lambda x^{2}} (1)

\Rightarrow y_{0}'(x) = -2\lambdaxe^{-\lambda x^{2}} (2)

\Rightarrow y_{0}''(x) = -2\lambdae^{-\lambda x^{2}} + 4\lambda^{2}x^{2}e^{-\lambda x^{2}} (3).


And so, subsituting (1) and (3) into (*):

 -2\lambdae^{-\lambda x^{2}} + 4\lambda^{2}x^{2}e^{-\lambda x^{2}} -\omega^{2}x^{2}e^{-\lambda x^{2}} + \omegae^{-\lambda x^{2}} = 0 (**).


So for (**) to be zero for all x;

(4)  4\lambda^{2} -\omega^{2} = 0 and (5)  -2\lambda + \omega = 0.

(5) \Rightarrow \lamda = \frac{\omega}{2} .

Hence \lambda is a postive constant, \frac{\omega}{2}. (since  \omega &gt; 0 ).


So we now have: y_{0}(x) = e^{-\frac{\omega}{2} x^{2}}. And since  \omega &gt; 0 , y_{0} \longrightarrow 0 as x \longrightarrow \pm \infty.

Also, y_{0}'(x) = -\omegaxe^{-\frac{\omega}{2} x^{2}}.

Now, as x \longrightarrow \pm \infty,  x^{k}e^{-x^{2}} \longrightarrow 0 \forall k. Hence, since \omega &gt; 0,  y_{0}'(x) \longrightarrow 0 as x \longrightarrow \pm \infty.


For n=1, y_{1}(x) = xe^{-\frac{\omega}{2} x^{2}}.

xe^{-\frac{\omega}{2} x^{2}} \longrightarrow 0 as x \longrightarrow \pm \infty, so y_{1}(x) \longrightarrow 0, as  x \longrightarrow \pm \infty.

Also, y_{1}'(x) = e^{-\frac{\omega}{2} x^{2}} - 2x^{2}e^{-\frac{\omega}{2} x^{2}} .
Both terms tend to zero and x tends to \pm \infty, and so y_{1}'(x) \longrightarrow 0 as x \longrightarrow \pm \infty.

So the conditions hold for n=1 and n=0, and  \lambda = \frac{\omega}{2} , where \omega is a positive constant.

\frac{d}{dx} [ y_{m}\frac{d y_{n}}{dx} - y_{n} \frac{dy_{m}}{dx} ] = \frac{dy_{m}}{dx} \times \frac{dy_{n}}{dx} + y_{m}\frac{d^{2}y_{n}}{dx} - \frac{dy_{n}}{dx} \times \frac{dy_{m}}{dx} - y_{n}\frac{d^{2}y_{m}}{dx} = y_{m}\frac{d^{2}y_{n}}{dx} - y_{n}\frac{d^{2}y_{m}}{dx} .


Now,  \frac{d^{2}y_{n}}{dx} = \omega^{2} x^{2}y_{n} - (2n+1) \omega y_{n}. Which is the same for y_{m}.

Hence: y_{m}\frac{d^{2}y_{n}}{dx} - y_{n}\frac{d^{2}y_{m}}{dx} = y_{m}[\omega^{2} x^{2}y_{n} - (2n+1) \omega y_{n}] - y_{n}[\omega^{2} x^{2}y_{m} - (2m+1) \omega y_{m}] = -(2n+1) \omega y_{n}y_{m} +  (2m+1) \omega y_{n}y_{m} = 2\omega y_{n}y_{m}[m-n]. .


If  m \neq n, then \frac{d}{dx} [ y_{m}\frac{d y_{n}}{dx} - y_{n} \frac{dy_{m}}{dx} ] \neq 0 .

And: 2\displaystyle\int^{\infty}_{-\infty} (m-n) \omega y_{m}y_{n} \, dx = \displaystyle\int^{\infty}_{-\infty} \frac{d}{dx} [ y_{m}\frac{d y_{n}}{dx} - y_{n} \frac{dy_{m}}{dx} ] \, dx

= \lim_{a\to \infty} [ y_{m}(a)\frac{d y_{n}(a)}{dx} - y_{n}(a) \frac{dy_{m}(a)}{dx} ] - \lim_{c\to -\infty} [ y_{m}(c)\frac{d y_{n}(c)}{dx} - y_{n}(c) \frac{dy_{m}(c)}{dx} ].

Now, we are given that as  x \longrightarrow \pm \infty, \frac{dy_{n}}{dx} and  y_{n}(x) \longrightarrow 0.

Hence: y_{m}(a)\frac{d y_{n}(a)}{dx} - y_{n}(a) \frac{dy_{m}(a)}{dx} \longrightarrow 0 as a \longrightarrow \infty

and

y_{m}(c)\frac{d y_{n}(c)}{dx} - y_{n}(c) \frac{dy_{m}(c)}{dx} \longrightarrow 0 as  c \longrightarrow -\infty .

and so: 2\displaystyle\int^{\infty}_{-\infty} (m-n) \omega y_{m}y_{n} \, dx  = 0

\Rightarrow \displaystyle\int^{\infty}_{-\infty}y_{m}y_{n} \, dx  = 0 , since omega, m and n are just constants.
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squeezebox
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(Original post by insparato)
If you look at the two posts it took me 9 minutes therefore I win :proud:.

I suspect i just picked the two easiest questions out of the two papers. I feel guilty for doings maths .
#zomg#
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squeezebox
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(Original post by Square)
Right so I've been working on II/7 but I'm a little bit confused.

It looks to me as if there have been a few substitutions going on here but the questions has been using the same variables across.

For example the first part is fairly easy, transforming the integral I into the first equation is easily done by the substitution x=1+y, which gives the same answer but in terms of y, however the question gives it in terms of x, the original variable which makes me think I've perhaps missed something and I'm not supposed to use a substitution after all?
You can just rename y as x, because it doesnt really matter what the dummy variable is called since the integral is going to eventually be evaluated at 0 and 1.
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insparato
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(Original post by squeezebox)
#zomg#
I don't do a maths related degree far from it, medicine actually. So when I do a sneaky STEP question or two i feel bad because i really should be doing something more humanie.
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Square
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(Original post by squeezebox)
You can just rename y as x, because it doesnt really matter what the dummy variable is called since the integral is going to eventually be evaluated at 0 and 1.
Good, just confused me slightly.
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insparato
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III/3

 |z - i| = 2 is the circle on the complex plane with radius 2 and centre (0,i)

so x^2 + (y - 1)^2 = 2

and therefore in paramatrised form

 x = \sqrt2 cos\theta

 y  = 1 + \sqrt2 sin \theta

I've omitted sketches however I have described them in sufficient detail to make an easy sketch if one chooses.

(i)

 w = \frac{z + i}{z - i}

 wz - wi = z + i

 wz - z = i + wi

 z(w - 1) = i(w + 1)

 z = \frac{i(w+1)}{w-1}

 z - i = \frac{i(w +1)}{w - 1} - i = \frac{i(w + 1) - i(w-1)}{w - 1}

 = \frac{1 + i}{w - 1}

 |z - i| = \left|\frac{1+i}{w-1}\right|

 2 = \frac{|1 + i|}{|w - 1|}

 |w - 1| = \frac{\sqrt2}{2}

Okay so its a circle of radius  \frac{\sqrt2}{2} and the centre is (1,0) [/latex]

(ii)

If z = x + iy and is real then y must = 0

so

 w = u + iv = \frac{x + i(y + 1)}{x + i(y - 1)} = \frac{x + i}{x - i}

 (u + iv)(x - i) = x + i

 ux - ui + vxi + v = x + i

 ux + v + i(-u + vx - 1) = x

-u + vx - 1 must = 0 (this is important bit on)

so

 ux + v = x

 x(1 - u) = v

 x = \frac{v}{1 - u}

Now from before

 -u + vx - 1 = 0

 vx = u + 1

 x = \frac{u+1}{v}

so

 \frac{v}{1 - u} = \frac{u+1}{v}

 v^2 = (u + 1)(1 - u)

 u^2 + v^2 = 1

So in the W plane, when z is real, you get a circle centre(0,0) with radius 1

(iii)

if z = x + iy and z is imaginary then x = 0

 u + vi = \frac{x + i(y + 1)}{x + i(y - 1)}

 u + vi = \frac{i(y + 1)}{i(y - 1)}

 (u + vi)(iy - i) = yi + i

 uyi - ui - vy + v = yi + i

 yi - uyi = -ui - vy + v - i

 y(i-ui) = i(-1 -u) - vy + v

 -vy + v = 0

 y = \frac{i(-1-u)}{i(1 - u)}

 y = \frac{-1 -u}{1 - u}

 1 = \frac{-1 - u}{1 - u}

 -1 - u = 1 -u

 0 = 2

Wahey! wtf ? I'm sure i got something different when jotting it down, something that made sense! hmm have i gone wrong somewhere? I think I'm doing the right method.
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Square
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II/7 - under construction! - sorry my latex skills aren't very good.

I=\displaystyle \int_1^2 \frac{(2-2x+x^2)^k}{x^{k+1}}dx

First part:

substitution:

\displaystyle x=1+y \\ \implies dx=dy \\ x^2=1+2y+y^2 \\ \\ \\ I=\int_0^1\frac{(2-2(1+y)+1+2y+y^2)^k}{(1+y)^{k+1}}  dy \\ \\

=\int_0^1\frac{(1+y^2)^k}{(1+y)^  {k+1}}dy

For the parts afterwards I get slightly confused, I've noticed that a substitution of y=tanx gets me close to the answer. However I only noticed that by exanding out the denominator of the second integral. Some advice on this would help.

iii) Take the third integral from the second part of the question:

\displaystyle 2\int_0^{\pi/8}\frac{1}{\sqrt{2}\cos \theta \cos (\frac{\pi}{4}-\theta)^{k+1}}d \theta

\displaystyle x=tan \theta \\ \implies \cos \theta=\frac{1}{\sqrt{x^2+1}} \\ \\ \implies \sin \theta=\frac{x}{\sqrt{x^2+1}}

\displaystyle \\ \cos(\frac{\pi}{4}-\theta)=\frac{1}{\sqrt{2}}(\cos \theta + \sin \theta) = \frac{1}{\sqrt{2}}(\frac{1+x}{\s  qrt{x^2+1}})

\displaystyle \theta=tan^{-1}x \\ \frac{d \theta}{dx}=\frac{1}{1+x^2} \\ d \theta=\frac{1}{1+x^2}dx

Substitute:

\displaystyle I=2\int_0^{\sqrt{2}-1} \frac{1}{1+x^2} \times \frac{1}{(\frac{1+x}{\sqrt{x^2+1  }}\frac{1+x}{\sqrt{x^2+1}})^{k+1  }}dx

\displaystyle \\ \\ I=2\int_0^{\sqrt{2}-1} \frac{1}{(1+x^2)(\frac{1+x}{1+x^  2})^{k+1}}

\\ \displaystyle I=2\int_0^{\sqrt{2}-1} \frac{(1+x^2)^k}{(1+x)^{k+1}}

Last part, first subtitute in y=x+1 to get one of the integrals, then into that substitute y=x^2 and then equate and cancel. I'll write up shortly.
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Swayum
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STEP I question 5.

I'm sure there must be a more elegant way of doing this, but here's my 2AM attempt:

Let \frac{1}{(1-ax)(1-bx)} = \frac{A}{1-ax} + \frac{B}{1-bx}

1 = A(1-bx) + B(1-ax)

A + B = 1 (comparing constant coefficients)

A = 1 - B


-Ab - Ba = 0 (comparing x coefficients)

Ab + Ba = 0

(1-B)b + Ba = 0

b - Bb + Ba = 0

B = \frac{-b}{a-b}

A = 1 - \frac{-b}{a-b} = \frac{a}{a-b}

\frac{1}{(1-ax)(1-bx)} = \frac{a}{(a - b)(1 - ax)} + \frac{-b}{(a - b)(1 - bx)}

\frac{a}{a-b}(1 - ax)^{-1} = \frac{a}{a-b}(1 + ax + a^2x^2 + a^3x^3 + ... + a^mx^m + ...)

\frac{-b}{a-b}(1 - bx)^{-1} = \frac{-b}{a-b}(1 + bx + b^2x^2 + b^3x^3 + ... + b^3x^3 + ...)

\frac{1}{(1-ax)(1-bx)} = 1 + \frac{a^2 - b^2}{a-b}x + \frac{a^3 - b^3}{a-b}x^2 + ... + \frac{a^{m+1} - b^{m+1}}{a-b} + ...

c_m = \frac{a^{m+1} - b^{m+1}}{a-b}



c^{2}_m = \frac{(a^{m+1})^2 - 2a^{m+1}b^{m+1} - (b^{m+1})^2}{(a - b)^2}

c^{2}_m = \frac{a^{2m + 2} - 2(ab)^{m+1} - b^{2m+2}}{(a - b)^2}



Let S = c^2_0 + c^2_1x + c^2_2x^2 + ... + c^2_mx^m + ...

S = \frac{a^2 - 2ab + b^2}{(a - b)^2} + \frac{a^4 - 2(ab)^2 + b^4}{(a - b)^2}x + \frac{a^6 - 2(ab)^3 + b^6}{(a - b)^2}x^2 + ...

S(a - b)^2 = (a^2 - 2ab + b^2) + (a^4 - 2(ab)^2 + b^4)x + (a^6 - 2(ab)^3 + b^6)x^2 + ...

S(a-b)^2 = (a^2 + a^4x + a^6x^2 + ... ) - 2(ab + (ab)^2x + (ab)^3x^2 + ...) + (b^2 + b^4x + b^6x^2 + ...)

a^2 + a^4x + a^6x^2 + ... = \frac{a^2}{1 - a^2x}

ab + (ab)^2x + (ab)^3x^2 + ... = \frac{ab}{1 - abx}

b^2 + b^4x + b^6x^2 + ... = \frac{b^2}{1 - b^2x}

S(a-b)^2 = \frac{a^2}{1 - a^2x} - \frac{2ab}{1 - abx} + \frac{b^2}{1 - b^2x}

\frac{a^2}{1 - a^2x} - \frac{2ab}{1 - abx} = \frac{a^2 - a^3bx - 2ab + 2a^3bx}{(1 - a^2x)(1 - abx)} = \frac{a^2 - 2ab + a^3bx}{(1 - a^2x)(1 - abx)}

S(a-b)^2 = \frac{a^2 - 2ab + a^3bx}{(1 - a^2x)(1 - abx)} + \frac{b^2}{1 - b^2x} = \frac{a^2 - 2ab + a^3bx - a^2b^2x + 2ab^3x - a^3b^3x^2 + b^2(1 - a^2x)(1 - abx)}{(1 - a^2x)(1 - b^2x)(1 - abx)}

S(a-b)^2 = \frac{a^2 - 2ab + a^3bx - a^2b^2x + 2ab^3x - a^3b^3x^2 + b^2 - ab^3x - a^2b^2x + a^3b^3x^2}{(1 - a^2x)(1 - b^2x)(1 - abx)}

S(a-b)^2 = \frac{a^2 - 2ab + b^2 + a^3bx - 2a^2b^2x + ab^3x}{(1 - a^2x)(1 - b^2x)(1 - abx)}

S(a-b)^2 = \frac{(a - b)^2 + abx(a^2 - 2ab + b^2)}{(1 - a^2x)(1 - b^2x)(1 - abx)} = \frac{(a - b)^2 + (a - b)^2abx}{(1 - a^2x)(1 - b^2x)(1 - abx)}

S(a-b)^2 = \frac{(a - b)^2(1 + abx)}{(1 - a^2x)(1 - b^2x)(1 - abx)}

S = \frac{1 + abx}{(1 - a^2x)(1 - b^2x)(1 - abx)}

WWWWW

Valid for b^2|x| &lt; 1
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Glutamic Acid
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I/7.


f'(x) = 2ax + b \\

f(1) = a + b + c \\

f(-1) = a - b + c \\

f(0) = c
Therefore
f(1)(x+1/2) + f(-1)(x - 1/2) - 2f(0)x = (a+b+c)(x + 1/2) + (a-b+c)(x - 1/2) - 2cx
 = ax + bx + cx + 1/2a + 1/2b + 1/2c + ax - bx + cx - 1/2a + 1/2b - 1/2c - 2cx = 2ax + b

If |f(x)| <= 1, then -1 <= f(x) <= 1.
If |f'(x)| <= 4, then -4 <= f(x) <= 4.

f'(x) = f(1)(x + 1/2) + f(-1)(x - 1/2) - 2f(0)x.

The largest values of f'(x) will be when x = -1 or 1 to maximize the values of (x + 1/2) and -(x-1/2). The largest value of f(1)(x + 1/2) can be 3/2 when f(1) = 1. The largest value of f(-1)(x - 1/2) can be 1/2 when f(-1) = 1. The largest value of -2f(0)x can be 2 when f(0) = -1. 3/2 + 1/2 + 2 = 4, so |f'(x)| <= is satisfied. When x = -1, the largest value can be (-1)(-1/2) + (-1)(-3/2) - (2)(1)(-1) = 4.

The smallest values of f'(x) will be when x = -1 or 1 likewise. When x = 1, be (-1)(3/2) + (-1)(1/2) - 2(1) with f(0) = 1, f(1) = -1 and f(-1) = -1. This gives -4. When x = -1, it will be (1)(-1/2) + (1)(-3/2) - 2(1) = -4, with f(1) = 1, f(-1) = 1 and f(0) = -1, giving -4. Therefore |f'(x) <= 4 is still satisfied.

Letting f(0) = -1, this gives c = -1. Letting f(1) = 1, this gives a + b - 1 = 1 hence a + b = 2. Letting f(-1) = 1, this gives a - b = 2. Hence 2a = 4, a = 2 and b = 0.
f(x) = 2x^2 - 1.

Sorry for any errors.
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kabbers
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relatively straightforward q, pointing out any mistakes will be appreciated as always

Ok first note that at the end of day n, the firm will owe everything it owed the previous day, the interest on this plus everything it has borrowed that day.

This may be expressed as \displaystyle S_{n+1} = (1+k)S_n + c

By evaluating the first few terms of this sequence, it becomes easy to 'guess' and then prove a total value for Sn:

\displaystyle S_0 = 0

\displaystyle S_1 = c

\displaystyle S_2 = (1+k)c + c = c((1+k) + 1)

\displaystyle S_3 = (1+k)((1+k)c + c) + c = c((1+k)^2 + (1+k) + 1)


ok my guess is that \displaystyle S_n = c \sum^{n-1}_{r=0} (1+k)^r = c \frac{(1+k)^n - 1}{(1+k) - 1} = c \frac{(1+k)^n - 1}{k} (by geometric series)

this fits the questions, but we have to prove it (induction!)


Basis case: \displaystyle S_0 =  c \frac{(1+k)^0 - 1}{k} = 0, therefore the basis case works.

Inductive step: \displaystyle S_{n+1} = (1+k)S_n + c = (1+k)c \frac{(1+k)^n - 1}{k} + c = c((1+k)\frac{(1+k)^n - 1}{k} + 1)

\displaystyle  = c(\frac{(1+k)^{n+1} - (k+1)}{k} + 1) = c(\frac{(1+k)^{n+1} - (k+1) + k}{k}) = c\frac{(1+k)^{n+1} - 1}{k})

This is in the same form as the above, so it works.


for part 2, firstly we can assert that \displaystyle S_T = c\frac{(1+k)^{T} - 1}{k}) (this is the initial amount owed)

the firm must pay interest on what it owes then pay back what it has earned.

\displaystyle S_{T+m+1} = (1+k)S_{T+m} - d

Again, let's try a few values:

\displaystyle S_{T+1} = (1+k)S_T - d

\displaystyle S_{T+2} = (1+k)((1+k)S_T - d) - d = (1+k)^2 S_T - d(1+k) - d

\displaystyle S_{T+3} = (1+k)((1+k)((1+k)S_T - d) - d) - d = (1+k)^3 S_T - d(1+k)^2 - d(1+k) - d

hypothesis: \displaystyle S_{T+m} = (1+k)^m S_T - d\sum^{m-1}_{r=0}(1+k)^r

\displaystyle  = (1+k)^m S_T - d\frac{(1+k)^m - 1}{k} = (1+k)^m(S_T - \frac{d}{k}) + \frac{d}{k} (again by geometric)


we need to prove this, again by induction.

basis case: when m = 0, \displaystyle S_T = (1+k)^0 (S_T - \frac{d}{k}) + \frac{d}{k} = S_T basis works

inductive step: \displaystyle S_{T+m+1} = (1+k)S_{T+m} - d = (1+k)((1+k)^m(S_T - \frac{d}{k}) + \frac{d}{k}) - d

\displaystyle  = (1+k)^{m+1}(S_T - \frac{d}{k}) + (1+k)\frac{d}{k} - d

\displaystyle  = (1+k)^{m+1}(S_T - \frac{d}{k}) + \frac{d}{k} + d - d

\displaystyle  = (1+k)^{m+1}(S_T - \frac{d}{k}) + \frac{d}{k}

Hence the sum works. However the question did not ask for us to express S_T, so we must substitute in (see our above def. for S_T)


\displaystyle S_{T+m} = (1+k)^m(c\frac{(1+k)^{T} - 1}{k} - \frac{d}{k}) + \frac{d}{k}

\displaystyle S_{T+m} = (1+k)^m(c\frac{(1+k)^{T} - 1 - d}{k}) + \frac{d}{k}

\displaystyle S_{T+m} = \frac{c}{k} (k+1)^{T+m} - \frac{c+d}{k}(k+1)^m + \frac{d}{k}



for part 3, note that if the sequence is decreasing, the firm will pay off its debt. so the conditions are:

\displaystyle S_{T+m+1} &lt; S_{T+m}

\displaystyle \frac{c}{k} (k+1)^{T+m+1} - \frac{c+d}{k}(k+1)^{m+1} + \frac{d}{k} &lt; \frac{c}{k} (k+1)^{T+m} - \frac{c+d}{k}(k+1)^m + \frac{d}{k}

Letting x = k+1


\displaystyle \frac{c}{k} x^{T+m+1} - \frac{c+d}{k}x^{m+1} + \frac{d}{k} &lt; \frac{c}{k} x^{T+m} - \frac{c+d}{k}x^m + \frac{d}{k}

\displaystyle \frac{c}{k} x^{T+m+1} - \frac{c+d}{k}x^{m+1} &lt; \frac{c}{k} x^{T+m} - \frac{c+d}{k}x^m

taking everything on to one side


\displaystyle x^{T+m}(\frac{c}{k}x - \frac{c}{k}) - x^m(\frac{c+d}{k}x-\frac{c+d}{k}) &lt; 0

\displaystyle (x-1)(\frac{c}{k}x^{T+m} - \frac{c+d}{k}x^m) &lt; 0

\displaystyle x^m(x-1)(\frac{c}{k}x^{T} - \frac{c+d}{k}) &lt; 0

k+1 > 0, so both x^m and (x-1) are greater than zero. Therefore, for the inequality to be satisfied,

\displaystyle \frac{c}{k}x^{T} - \frac{c+d}{k} &lt; 0

\displaystyle cx^{T} - (c+d) &lt; 0

\displaystyle x^{T} - 1 &lt; \frac{d}{c}

\displaystyle (1+k)^{T} - 1 &lt; \frac{d}{c} as required.
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squeezebox
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Doing II/5 and 4 - but if anyone else beats me to it, feel free .
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kabbers
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ok here's my attempt at III/6, i'm not completely sure if it's adequate for STEP.. any tips appreciated.

part 1:

rewrite f(x) as an addition of sines, using the identity \sin(a+b) + \sin(a-b) = 2\sin(a)\cos(b)

f(x) = \sin 2x \cos x = \frac{1}{2}(\sin(2x+x) + sin(2x-x)) = \frac{1}{2}(\sin 3x + \sin x)

note that (\sin 3x)' = 3 \cos 3x,
(3 \cos 3x)' = (-3^2 \sin 3x),
(-3^2 \sin 3x)' = (-3^3 \cos 3x),
(-3^3 \cos 3x)' = (3^4 \sin 3x),

and hence
(\sin 3x)^{(4)} = (3^4 \sin 3x) <=> (\sin 3x)^{(4n)} = (3^{4n} \sin 3x),

also note that (\sin x)' = \cos x,
(\cos x)' = -\sin x,
(-\sin x)' = -\cos x,
(-\cos x)' = \sin x,

and hence
(\sin x)^{(4)} = (\sin x) <=> (\sin x)^{(4n)} = \sin x,


Therefore, when m < 4,

f^{(4n+m)} (x) = \frac{1}{2} (\sin 3x + \sin x)^{(4n+m)} = \frac{1}{2} (3^{4n} \sin 3x + \sin x)^{(m)}


1988/4 = 497; therefore n = 497, m = 0

so f^{(1988)}(x) = \frac{1}{2} (3^{1988} \sin 3x + \sin x)




part 2: firstly note that if y is small, \sin y \approx y (this we can tell from the maclaurin expansion)
also note that if y is small, \cos y \approx 1

so, subbing in :

f^{(1988)}(x) = \frac{1}{2}(3^{1988} \sin 3x + \sin x)

= \frac{1}{2}(3^{1988} \sin 3(\frac{\pi}{3} + \epsilon) + \sin (\frac{\pi}{3} + \epsilon))

= \frac{1}{2}(3^{1988} (\sin 3\frac{\pi}{3}\cos 3\epsilon + \sin 3\epsilon \cos 3\frac{\pi}{3}) + \sin \frac{\pi}{3}\cos \epsilon + \sin \epsilon \cos \frac{\pi}{3})

= \frac{1}{2}(3^{1988} (\sin \pi\cos 3\epsilon + \sin 3\epsilon \cos \pi) + \sin \frac{\pi}{3}\cos \epsilon + \sin \epsilon \cos \frac{\pi}{3})

= \frac{1}{2}(3^{1988} (\sin \pi\cos 3\epsilon + \sin 3\epsilon \cos \pi) + \sin \frac{\pi}{3}\cos \epsilon + \sin \epsilon \cos \frac{\pi}{3})

= \frac{1}{2}(-3^{1988} \sin 3\epsilon + \frac{\sqrt 3}{2}\cos \epsilon + \frac{1}{2}\sin \epsilon)

using the above approximations,

= \frac{1}{2}(-3^{1988} 3\epsilon + \frac{\sqrt 3}{2} + \frac{1}{2} \epsilon)


\frac{1}{2} \epsilon is small enough to ignore. subbing in the other values,


= -3^{1988} 3\frac{3^{-1989}\sqrt 3}{2} + \frac{\sqrt 3}{2}

= -3^{1988} 3^{-1988}\frac{\sqrt 3}{2} + \frac{\sqrt 3}{2}

= -\frac{\sqrt 3}{2} + \frac{\sqrt 3}{2}

= 0

To prove that \frac{\pi}{3} + \epsilon is the smallest value > 0: there are no roots of 3^1988 sin(3x) between 0 and pi/3, this is because the lowest x ( > 0) at which sin(x) = 0 is pi. Thus our root is going to be very near pi/3 (we can neglect the sin x over here as it is tiny compared to the sin 3x), in fact slightly greater to compensate for sin(x) being equal to sqrt(3)/2.

latex edits.
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squeezebox
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Has anyone else done III/2? I think I have a solution, but since I've never solved a difference equation or worked with them like this before today, I dunno If its correct or not..

EDIT: I'll post my solution when I have time.
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Glutamic Acid
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II/3.

- Both k and 1/k will satisfy the equation.
x^2 + bk + c = 0 \\

\dfrac{1}{k^2} + \dfrac{b}{k} + c = 0 \implies 1 + bk + ck^2 = 0 \implies k^2 + \dfrac{b}{c}k + \dfrac{1}{c} = 0.
So 1/c = c hence c^2 = 1 and c = +-1.
And b/c = b so b = cb. When c = 1, b can be any real value. When c = -1, b = 0.

- Since we know c = 1, the quadratic will be of the form k^2 + bk + 1 = 0, as k^2 - 1 = 0 does not satisfy the conditions (because -1 is a root, but 1--1 = 2 isn't a root).
Also (1-k)^2 + b(1-k) + 1 = 0 will satisfy it. This gives  1-2k+k^2+b-bk+1=0 \implies k^2 + (-2-b)k + (2+b) = 0

Comparing coefficients, -2-b = b ==> b = -1 This also satisfies 2 - 1 =1. So the quadratic is x^2 - x + 1 = 0.

- Call the roots of the equation \alpha, 1 - \alpha, \dfrac{1}{\alpha}.

Therefore
\alpha^3 + p\alpha^2 + q\alpha + r = 0 \\

(1-\alpha)^3 + p(1 - \alpha)^2 + q(1-\alpha) + r = 0 \\

\dfrac{1}{\alpha^3} + \dfrac{p}{\alpha^2} + \dfrac{q}{\alpha} + r = 0.

Multiplying through last equation by alpha^3 and dividing by r, you get:
\alpha^3 + \dfrac{q}{r}\alpha^2 + \dfrac{p}{r}\alpha + \dfrac{1}{r} = 0.

Comparing coefficients, 1/r = 1 so r = 1.
q/r = p p/r = q so p = q.

Replacing q by p and expanding the second equation we get.
1 - 3\alpha + 3\alpha^2 - \alpha^3 + p - 2\alpha p + p\alpha^2 + p - p\alpha + 1 = 0 \\

\implies \alpha^3 - 1 + 3\alpha - 3\alpha^2 - p + 2\alpha p - p \alpha^2 - p + p\alpha - 1 = 0 \\

\implies \alpha^3 + (-3-p)\alpha^2 + (3+3p)\alpha - (2p + 2) = 0

Comparing coefficients, -3-p = p ==> 2p = -3 and p = -3/2. Since q = p, q = -3/2.
So the cubic equation is x^3 - \dfrac{3}{2}x^2 - \dfrac{3}{2}x + 1 = 0.
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squeezebox
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UNDER CONSTRUCTION

STEP II - Question 5

Let z = c + is (Where for convenience, c = \cos(\theta) and  s = \sin(\theta).

Then: z^{7} = (c + is)^{7} = c^{7} + 7isc^{6} - 21s^{2}c^{5} - 35is^{3}c^{4} + 35c^{3}s^{4} + 21is^{3}c^{2} - 7s^{6}c - is^{7} .

Also, z^{7} = \cos(7\theta) + i\sin(7\theta). So equating the real and imaginary parts:

\cos(7\theta) = c^{7} - 21s^{2}c^{5} + 35c^{3}s^{4} -7s^{6}c. and  \sin(7\theta) = 7sc^{6} - 35s^{3}c^{4} + 21s^{5}c^{2} - s^{7} .

 \tan(7\theta) = \frac{\sin(7\theta)}{\cos(7\thet  a)} = \frac{ 7sc^{6} - 35s^{3}c^{4} + 21s^{5}c^{2} - s^{7}}{c^{7} - 21s^{2}c^{5} + 35c^{3}s^{4} -7s^{6}c} = \frac{7t - 35t^{3} + 21t^{5} - t^{7}}{1 - 21t^{2} + 35t^{4} - 7t^{6}} = \frac{t(7 - 35t^{2} + 21t^{4} - t^{6}}{1 - 21t^{2} + 35t^{4} - 7t^{6}} . Where t = \tan(\theta).


So if \tan(7\theta) = 0, then either \tan(\theta) = 0, or 7 - 35t^{2} + 21t^{4} - t^{6} = 0 (*).

If \tan(\theta) \neq 0 , then the values of theta for which (*) = 0 are given by \tan(7\theta) = 0 , provided  \theta \neq n \pi, where n is an integer.

\Rightarrow \theta = \frac{\pi}{7}, \frac{2\pi}{7}, \frac{3\pi}{7}, \frac{4\pi}{7}, \frac{5\pi}{7}, \frac{6\pi}{7} these willl give distinct values of \tan(\theta).

Hence the roots of (*) are : \tan(\frac{\pi}{7}), \tan(\frac{2\pi}{7}), \tan(\frac{3\pi}{7}), \tan(\frac{4\pi}{7}), \tan(\frac{5\pi}{7}), \tan(6\frac{\pi}{7})..

Product of the roots = -7. Hence: \tan(\frac{\pi}{7})\tan(\frac{2\  pi}{7})\tan(3\frac{\pi}{7})\tan(  4\frac{\pi}{7})\tan(\frac{5\pi}{  7})\tan(6\frac{\pi}{7}) = -7.

Now,  tan(\frac{6\pi}{7}) = -tan(\frac{\pi}{7}), \tan(\frac{5\pi}{7}) = -\tan(2\frac{\pi}{7}) and \tan(3\frac{\pi}{7}) = \tan(4\frac{\pi}{7})

So: \tan^{2}(\frac{2\pi}{7})\tan^{2}  (\frac{4\pi}{7})\tan^{2}(\frac{6  \pi}{7}) = 7

\Rightarrow \tan(\frac{2\pi}{7})\tan(\frac{4  \pi}{7})\tan(\frac{6\pi}{7}) = \sqrt{7}. (Chosen positive root since there are an even number of negative tans.)

Repeat the whole process for n=9, and we end up with:

\tan^{2}(\frac{2\pi}{9})\tan^{2}  (\frac{4\pi}{9})\tan^{2}(\frac{6  \pi}{9})\tan^{2}(\frac{8\pi}{9}) = 9

\Rightarrow \tan(\frac{2\pi}{9})\tan(\frac{4  \pi}{9})\tan(\frac{6\pi}{9})\tan  (\frac{8\pi}{9}) = 3.

And for n=11, we end up with:

\tan^{2}(\frac{2\pi}{11})\tan^{2  }(\frac{4\pi}{11})\tan^{2}(\frac  {6\pi}{11})\tan^{2}(\frac{8\pi}{  11})\tan^{2}(\frac{10\pi}{11}) = 11

\Rightarrow \tan(\frac{2\pi}{11})\tan(\frac{  4\pi}{11})\tan(\frac{6\pi}{11})\  tan(\frac{8\pi}{11})\tan(\frac{1  0\pi}{11}) = -\sqrt{11} (negative root since odd number of negative tans).

Phew, my hands hurt.
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