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STEP Maths I, II, III 1988 solutions

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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 :frown:.
Reply 21
Avatar for kabbers
kabbers
4
ok doing II/1if no one else is
I've had a look at it, on first glance it seems like not a bad question! You go for it though.
Reply 23
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kabbers
4
cheers :smile: 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)

∑n=0nrn=1+r+r2+..+rn\displaystyle\sum^n_{n=0} r^n = 1 + r + r^2 + .. + r^n

=r−1r−1(1+r+r2+..+rn) = \frac{r-1}{r-1}(1 + r + r^2 + .. + r^n)

=1r−1(r−1+r2−r+r3−r2+..+rn+1−rn) = \frac{1}{r-1}(r - 1 + r^2 - r + r^3 - r^2 + .. + r^{n+1} - r^n)

=rn+1−1r−1 = \frac{r^{n+1} - 1}{r-1}


therefore, iff |r| < 1, as n→∞,rn+1−1r−1→11−r(∗)n \to \infty, \frac{r^{n+1} - 1}{r-1} \to \frac{1}{1-r} (*)

Also note that

∑n=1nrn=r+r2+..+rn\displaystyle\sum^n_{n=1} r^n = r + r^2 + .. + r^n

=r−1r−1(r+r2+r3+..+rn) = \frac{r-1}{r-1}(r + r^2 + r^3 + .. + r^n)

=1r−1(r2−r+r3−r2+r4+..+rn+1−rn) = \frac{1}{r-1}(r^2 - r + r^3 - r^2 + r^4 + .. + r^{n+1} - r^n)

(∗∗)=rn+1−rr−1(**) = \frac{r^{n+1}-r}{r-1}

therefore, iff |r| < 1, as n→∞,rn+1−rr−1→r1−r(∗∗)n \to \infty, \frac{r^{n+1} - r}{r-1} \to \frac{r}{1-r} (**)


part 1,

∑n=0∞xn−12∑n=0∞(x2)n\displaystyle\sum^{\infty}_{n=0} x^n - \frac{1}{2}\sum^{\infty}_{n=0} (\frac{x}{2})^n

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

=11−x−1211−x2= \frac{1}{1-x} - \frac{1}{2} \frac{1}{1-\frac{x}{2}}

=11−x−1222−x= \frac{1}{1-x} - \frac{1}{2} \frac{2}{2-x}

=11−x−12−x= \frac{1}{1-x} - \frac{1}{2-x}

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

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

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


part 2,


∑n=1∞x−n−12∑n=0∞(x2)n\displaystyle\sum^{\infty}_{n=1} x^{-n} - \frac{1}{2}\sum^{\infty}_{n=0} (\frac{x}{2})^n

=∑n=1∞(1x)n−12∑n=0∞(x2)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 (**) & (*)

−1x1−1x−12−x- \frac{\frac{1}{x}}{1-\frac{1}{x}} - \frac{1}{2-x} (I just copied the second part from above)

=−1x−1−12−x=- \frac{1}{x-1} - \frac{1}{2-x}

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

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

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



part 3,


for |x| > 2, ∑n=1∞x−n\displaystyle\sum^{\infty}_{n=1} x^{-n} is ok, but 12∑n=0∞(x2)n\displaystyle\frac{1}{2}\sum^{\infty}_{n=0} (\frac{x}{2})^n is not.

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

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

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

by (**),
=2x1−2x= \frac{\frac{2}{x}}{1-\frac{2}{x}}

=2x−2= \frac{2}{x-2} hooray it works


Therefore,


12∑n=1∞(2x)n−∑n=1∞(1x)n\displaystyle\frac{1}{2}\sum^{\infty}_{n=1} (\frac{2}{x})^n - \sum^{\infty}_{n=1} (\frac{1}{x})^n

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

=f(x)= f(x) (as above)


So
12∑n=1∞(2x)n−∑n=1∞(1x)n\displaystyle\frac{1}{2}\sum^{\infty}_{n=1} (\frac{2}{x})^n - \sum^{\infty}_{n=1} (\frac{1}{x})^n

is the desired expansion
Reply 24
Avatar for Square
Square
13
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?
STEP III - Question 7

When n=0:
Unparseable latex formula:

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

.

We have: y0(x)=e−λx2(1)y_{0}(x) = e^{-\lambda x^{2}} (1)

Unparseable latex formula:

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



Unparseable latex formula:

\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 (*):

Unparseable latex formula:

-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) 4λ2−ω2=0 4\lambda^{2} -\omega^{2} = 0 and (5) −2λ+ω=0 -2\lambda + \omega = 0.

(5)
Unparseable latex formula:

\Rightarrow \lamda = \frac{\omega}{2}

.

Hence λ\lambda is a postive constant, ω2\frac{\omega}{2}. (since ω>0 \omega > 0 ).


So we now have: y0(x)=e−ω2x2y_{0}(x) = e^{-\frac{\omega}{2} x^{2}}. And since ω>0 \omega > 0 , y0⟶0y_{0} \longrightarrow 0 as x⟶±∞x \longrightarrow \pm \infty.

Also,
Unparseable latex formula:

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

.

Now, as x ⟶±∞\longrightarrow \pm \infty, xke−x2⟶0∀k x^{k}e^{-x^{2}} \longrightarrow 0 \forall k. Hence, since ω>0\omega > 0, y0′(x)⟶0 y_{0}'(x) \longrightarrow 0 as x⟶±∞x \longrightarrow \pm \infty.


For n=1, y1(x)=xe−ω2x2.y_{1}(x) = xe^{-\frac{\omega}{2} x^{2}}.

xe−ω2x2⟶0xe^{-\frac{\omega}{2} x^{2}} \longrightarrow 0 as x⟶±∞x \longrightarrow \pm \infty, so y1(x)⟶0y_{1}(x) \longrightarrow 0, as x⟶±∞ x \longrightarrow \pm \infty.

Also, y1′(x)=e−ω2x2−2x2e−ω2x2y_{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 y1′(x)⟶0y_{1}'(x) \longrightarrow 0 as x⟶±∞x \longrightarrow \pm \infty.

So the conditions hold for n=1 and n=0, and λ=ω2 \lambda = \frac{\omega}{2} , where ω\omega is a positive constant.

ddx[ymdyndx−yndymdx]=dymdx×dyndx+ymd2yndx−dyndx×dymdx−ynd2ymdx=ymd2yndx−ynd2ymdx.\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, d2yndx=ω2x2yn−(2n+1)ωyn. \frac{d^{2}y_{n}}{dx} = \omega^{2} x^{2}y_{n} - (2n+1) \omega y_{n}. Which is the same for ymy_{m}.

Hence: ymd2yndx−ynd2ymdx=ym[ω2x2yn−(2n+1)ωyn]−yn[ω2x2ym−(2m+1)ωym]=−(2n+1)ωynym+(2m+1)ωynym=2ωynym[m−n].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≠n m \neq n, then ddx[ymdyndx−yndymdx]≠0\frac{d}{dx} [ y_{m}\frac{d y_{n}}{dx} - y_{n} \frac{dy_{m}}{dx} ] \neq 0 .

And: 2∫−∞∞(m−n)ωymyn dx=∫−∞∞ddx[ymdyndx−yndymdx] dx2\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→∞[ym(a)dyn(a)dx−yn(a)dym(a)dx]−lim⁡c→−∞[ym(c)dyn(c)dx−yn(c)dym(c)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⟶±∞ x \longrightarrow \pm \infty, dyndx\frac{dy_{n}}{dx} and yn(x)⟶0 y_{n}(x) \longrightarrow 0.

Hence: ym(a)dyn(a)dx−yn(a)dym(a)dx⟶0y_{m}(a)\frac{d y_{n}(a)}{dx} - y_{n}(a) \frac{dy_{m}(a)}{dx} \longrightarrow 0 as a⟶∞a \longrightarrow \infty

and

ym(c)dyn(c)dx−yn(c)dym(c)dx⟶0y_{m}(c)\frac{d y_{n}(c)}{dx} - y_{n}(c) \frac{dy_{m}(c)}{dx} \longrightarrow 0 as c⟶−∞ c \longrightarrow -\infty .

and so: 2∫−∞∞(m−n)ωymyn dx=02\displaystyle\int^{\infty}_{-\infty} (m-n) \omega y_{m}y_{n} \, dx = 0

⇒∫−∞∞ymyn dx=0\Rightarrow \displaystyle\int^{\infty}_{-\infty}y_{m}y_{n} \, dx = 0 , since omega, m and n are just constants.
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 :frown:.


#zomg#
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.
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.
Reply 29
Avatar for Square
Square
13
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.
III/3

∣z−i∣=2 |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=2cosθ x = \sqrt2 cos\theta

y=1+2sinθ 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=z+iz−i w = \frac{z + i}{z - i}

wz−wi=z+i wz - wi = z + i

wz−z=i+wi wz - z = i + wi

z(w−1)=i(w+1) z(w - 1) = i(w + 1)

z=i(w+1)w−1 z = \frac{i(w+1)}{w-1}

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

=1+iw−1 = \frac{1 + i}{w - 1}

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

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

∣w−1∣=22 |w - 1| = \frac{\sqrt2}{2}

Okay so its a circle of radius 22 \frac{\sqrt2}{2} and the centre is (1,0)

(ii)

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

so

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

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

ux−ui+vxi+v=x+i ux - ui + vxi + v = x + i

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

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

so

ux+v=x ux + v = x

x(1−u)=v x(1 - u) = v

x=v1−u x = \frac{v}{1 - u}

Now from before

−u+vx−1=0 -u + vx - 1 = 0

vx=u+1 vx = u + 1

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

so

v1−u=u+1v \frac{v}{1 - u} = \frac{u+1}{v}

v2=(u+1)(1−u) v^2 = (u + 1)(1 - u)

u2+v2=1 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=x+i(y+1)x+i(y−1) u + vi = \frac{x + i(y + 1)}{x + i(y - 1)}

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

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

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

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

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

−vy+v=0 -vy + v = 0

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

y=−1−u1−u y = \frac{-1 -u}{1 - u}

1=−1−u1−u 1 = \frac{-1 - u}{1 - u}

−1−u=1−u -1 - u = 1 -u

0=2 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.
Reply 31
Avatar for Square
Square
13
II/7 - under construction! - sorry my latex skills aren't very good.

I=∫12(2−2x+x2)kxk+1dxI=\displaystyle \int_1^2 \frac{(2-2x+x^2)^k}{x^{k+1}}dx

First part:

substitution:

Unparseable latex formula:

\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 \\ \\[br]=\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:

2∫0π/812cos⁡θcos⁡(π4−θ)k+1dθ\displaystyle 2\int_0^{\pi/8}\frac{1}{\sqrt{2}\cos \theta \cos (\frac{\pi}{4}-\theta)^{k+1}}d \theta

x=tanθ  ⟹  cos⁡θ=1x2+1  ⟹  sin⁡θ=xx2+1\displaystyle x=tan \theta \\ \implies \cos \theta=\frac{1}{\sqrt{x^2+1}} \\ \\ \implies \sin \theta=\frac{x}{\sqrt{x^2+1}}

cos⁡(π4−θ)=12(cos⁡θ+sin⁡θ)=12(1+xx2+1)\displaystyle \\ \cos(\frac{\pi}{4}-\theta)=\frac{1}{\sqrt{2}}(\cos \theta + \sin \theta) = \frac{1}{\sqrt{2}}(\frac{1+x}{\sqrt{x^2+1}})

θ=tan−1xdθdx=11+x2dθ=11+x2dx\displaystyle \theta=tan^{-1}x \\ \frac{d \theta}{dx}=\frac{1}{1+x^2} \\ d \theta=\frac{1}{1+x^2}dx

Substitute:

I=2∫02−111+x2×1(1+xx2+11+xx2+1)k+1dx\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

I=2∫02−11(1+x2)(1+x1+x2)k+1\displaystyle \\ \\ I=2\int_0^{\sqrt{2}-1} \frac{1}{(1+x^2)(\frac{1+x}{1+x^2})^{k+1}}

I=2∫02−1(1+x2)k(1+x)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.
Reply 32
Avatar for Swayum
Swayum
18
STEP I question 5.

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

Let 1(1−ax)(1−bx)=A1−ax+B1−bx\frac{1}{(1-ax)(1-bx)} = \frac{A}{1-ax} + \frac{B}{1-bx}

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

A+B=1A + B = 1 (comparing constant coefficients)

A=1−BA = 1 - B


−Ab−Ba=0-Ab - Ba = 0 (comparing x coefficients)

Ab+Ba=0Ab + Ba = 0

(1−B)b+Ba=0(1-B)b + Ba = 0

b−Bb+Ba=0b - Bb + Ba = 0

B=−ba−bB = \frac{-b}{a-b}

A=1−−ba−b=aa−bA = 1 - \frac{-b}{a-b} = \frac{a}{a-b}

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

aa−b(1−ax)−1=aa−b(1+ax+a2x2+a3x3+...+amxm+...)\frac{a}{a-b}(1 - ax)^{-1} = \frac{a}{a-b}(1 + ax + a^2x^2 + a^3x^3 + ... + a^mx^m + ...)

−ba−b(1−bx)−1=−ba−b(1+bx+b2x2+b3x3+...+b3x3+...)\frac{-b}{a-b}(1 - bx)^{-1} = \frac{-b}{a-b}(1 + bx + b^2x^2 + b^3x^3 + ... + b^3x^3 + ...)

1(1−ax)(1−bx)=1+a2−b2a−bx+a3−b3a−bx2+...+am+1−bm+1a−b+...\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} + ...

cm=am+1−bm+1a−bc_m = \frac{a^{m+1} - b^{m+1}}{a-b}



cm2=(am+1)2−2am+1bm+1−(bm+1)2(a−b)2c^{2}_m = \frac{(a^{m+1})^2 - 2a^{m+1}b^{m+1} - (b^{m+1})^2}{(a - b)^2}

cm2=a2m+2−2(ab)m+1−b2m+2(a−b)2c^{2}_m = \frac{a^{2m + 2} - 2(ab)^{m+1} - b^{2m+2}}{(a - b)^2}



Let S=c02+c12x+c22x2+...+cm2xm+...S = c^2_0 + c^2_1x + c^2_2x^2 + ... + c^2_mx^m + ...

S=a2−2ab+b2(a−b)2+a4−2(ab)2+b4(a−b)2x+a6−2(ab)3+b6(a−b)2x2+...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=(a2−2ab+b2)+(a4−2(ab)2+b4)x+(a6−2(ab)3+b6)x2+...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=(a2+a4x+a6x2+...)−2(ab+(ab)2x+(ab)3x2+...)+(b2+b4x+b6x2+...)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 + ...)

a2+a4x+a6x2+...=a21−a2xa^2 + a^4x + a^6x^2 + ... = \frac{a^2}{1 - a^2x}

ab+(ab)2x+(ab)3x2+...=ab1−abxab + (ab)^2x + (ab)^3x^2 + ... = \frac{ab}{1 - abx}

b2+b4x+b6x2+...=b21−b2xb^2 + b^4x + b^6x^2 + ... = \frac{b^2}{1 - b^2x}

S(a−b)2=a21−a2x−2ab1−abx+b21−b2xS(a-b)^2 = \frac{a^2}{1 - a^2x} - \frac{2ab}{1 - abx} + \frac{b^2}{1 - b^2x}

a21−a2x−2ab1−abx=a2−a3bx−2ab+2a3bx(1−a2x)(1−abx)=a2−2ab+a3bx(1−a2x)(1−abx)\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=a2−2ab+a3bx(1−a2x)(1−abx)+b21−b2x=a2−2ab+a3bx−a2b2x+2ab3x−a3b3x2+b2(1−a2x)(1−abx)(1−a2x)(1−b2x)(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=a2−2ab+a3bx−a2b2x+2ab3x−a3b3x2+b2−ab3x−a2b2x+a3b3x2(1−a2x)(1−b2x)(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=a2−2ab+b2+a3bx−2a2b2x+ab3x(1−a2x)(1−b2x)(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=(a−b)2+abx(a2−2ab+b2)(1−a2x)(1−b2x)(1−abx)=(a−b)2+(a−b)2abx(1−a2x)(1−b2x)(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=(a−b)2(1+abx)(1−a2x)(1−b2x)(1−abx)S(a-b)^2 = \frac{(a - b)^2(1 + abx)}{(1 - a^2x)(1 - b^2x)(1 - abx)}

S=1+abx(1−a2x)(1−b2x)(1−abx)S = \frac{1 + abx}{(1 - a^2x)(1 - b^2x)(1 - abx)}

WWWWW

Valid for b2∣x∣<1b^2|x| < 1
I/7.
Unparseable latex formula:

[br]f'(x) = 2ax + b \\[br]f(1) = a + b + c \\[br]f(-1) = a - b + c \\[br]f(0) = c[br]


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)−2cxf(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 = 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.
Reply 34
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kabbers
4
relatively straightforward q, pointing out any mistakes will be appreciated as always :smile:

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 Sn+1=(1+k)Sn+c\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:

S0=0\displaystyle S_0 = 0

S1=c\displaystyle S_1 = c

S2=(1+k)c+c=c((1+k)+1)\displaystyle S_2 = (1+k)c + c = c((1+k) + 1)

S3=(1+k)((1+k)c+c)+c=c((1+k)2+(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 Sn=c∑r=0n−1(1+k)r=c(1+k)n−1(1+k)−1=c(1+k)n−1k\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: S0=c(1+k)0−1k=0\displaystyle S_0 = c \frac{(1+k)^0 - 1}{k} = 0, therefore the basis case works.

Inductive step: Sn+1=(1+k)Sn+c=(1+k)c(1+k)n−1k+c=c((1+k)(1+k)n−1k+1)\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)

=c((1+k)n+1−(k+1)k+1)=c((1+k)n+1−(k+1)+kk)=c(1+k)n+1−1k)\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 ST=c(1+k)T−1k)\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.

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

Again, let's try a few values:

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

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

ST+3=(1+k)((1+k)((1+k)ST−d)−d)−d=(1+k)3ST−d(1+k)2−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: ST+m=(1+k)mST−d∑r=0m−1(1+k)r\displaystyle S_{T+m} = (1+k)^m S_T - d\sum^{m-1}_{r=0}(1+k)^r

=(1+k)mST−d(1+k)m−1k=(1+k)m(ST−dk)+dk\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, ST=(1+k)0(ST−dk)+dk=ST\displaystyle S_T = (1+k)^0 (S_T - \frac{d}{k}) + \frac{d}{k} = S_T basis works

inductive step: ST+m+1=(1+k)ST+m−d=(1+k)((1+k)m(ST−dk)+dk)−d\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

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

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

=(1+k)m+1(ST−dk)+dk\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)


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

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

ST+m=ck(k+1)T+m−c+dk(k+1)m+dk\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:

ST+m+1<ST+m\displaystyle S_{T+m+1} < S_{T+m}

ck(k+1)T+m+1−c+dk(k+1)m+1+dk<ck(k+1)T+m−c+dk(k+1)m+dk\displaystyle \frac{c}{k} (k+1)^{T+m+1} - \frac{c+d}{k}(k+1)^{m+1} + \frac{d}{k} < \frac{c}{k} (k+1)^{T+m} - \frac{c+d}{k}(k+1)^m + \frac{d}{k}

Letting x = k+1


ckxT+m+1−c+dkxm+1+dk<ckxT+m−c+dkxm+dk\displaystyle \frac{c}{k} x^{T+m+1} - \frac{c+d}{k}x^{m+1} + \frac{d}{k} < \frac{c}{k} x^{T+m} - \frac{c+d}{k}x^m + \frac{d}{k}

ckxT+m+1−c+dkxm+1<ckxT+m−c+dkxm\displaystyle \frac{c}{k} x^{T+m+1} - \frac{c+d}{k}x^{m+1} < \frac{c}{k} x^{T+m} - \frac{c+d}{k}x^m

taking everything on to one side


xT+m(ckx−ck)−xm(c+dkx−c+dk)<0\displaystyle x^{T+m}(\frac{c}{k}x - \frac{c}{k}) - x^m(\frac{c+d}{k}x-\frac{c+d}{k}) < 0

(x−1)(ckxT+m−c+dkxm)<0\displaystyle (x-1)(\frac{c}{k}x^{T+m} - \frac{c+d}{k}x^m) < 0

xm(x−1)(ckxT−c+dk)<0\displaystyle x^m(x-1)(\frac{c}{k}x^{T} - \frac{c+d}{k}) < 0

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

ckxT−c+dk<0\displaystyle \frac{c}{k}x^{T} - \frac{c+d}{k} < 0

cxT−(c+d)<0\displaystyle cx^{T} - (c+d) < 0

xT−1<dc\displaystyle x^{T} - 1 < \frac{d}{c}

(1+k)T−1<dc\displaystyle (1+k)^{T} - 1 < \frac{d}{c} as required.
Doing II/5 :smile: and 4 - but if anyone else beats me to it, feel free :smile:.
Reply 36
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kabbers
4
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)=2sin⁡(a)cos⁡(b)\sin(a+b) + \sin(a-b) = 2\sin(a)\cos(b)

f(x)=sin⁡2xcos⁡x=12(sin⁡(2x+x)+sin(2x−x))=12(sin⁡3x+sin⁡x)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)′=3cos⁡3x(\sin 3x)' = 3 \cos 3x,
(3cos⁡3x)′=(−32sin⁡3x)(3 \cos 3x)' = (-3^2 \sin 3x),
(−32sin⁡3x)′=(−33cos⁡3x)(-3^2 \sin 3x)' = (-3^3 \cos 3x),
(−33cos⁡3x)′=(34sin⁡3x)(-3^3 \cos 3x)' = (3^4 \sin 3x),

and hence
(sin⁡3x)(4)=(34sin⁡3x)(\sin 3x)^{(4)} = (3^4 \sin 3x) <=> (sin⁡3x)(4n)=(34nsin⁡3x)(\sin 3x)^{(4n)} = (3^{4n} \sin 3x),

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

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


Therefore, when m < 4,

f(4n+m)(x)=12(sin⁡3x+sin⁡x)(4n+m)=12(34nsin⁡3x+sin⁡x)(m)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)=12(31988sin⁡3x+sin⁡x)f^{(1988)}(x) = \frac{1}{2} (3^{1988} \sin 3x + \sin x)




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

so, subbing in :

f(1988)(x)=12(31988sin⁡3x+sin⁡x)f^{(1988)}(x) = \frac{1}{2}(3^{1988} \sin 3x + \sin x)

=12(31988sin⁡3(π3+ϵ)+sin⁡(π3+ϵ))= \frac{1}{2}(3^{1988} \sin 3(\frac{\pi}{3} + \epsilon) + \sin (\frac{\pi}{3} + \epsilon))

=12(31988(sin⁡3π3cos⁡3ϵ+sin⁡3ϵcos⁡3π3)+sin⁡π3cos⁡ϵ+sin⁡ϵcos⁡π3)= \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})

=12(31988(sin⁡πcos⁡3ϵ+sin⁡3ϵcos⁡π)+sin⁡π3cos⁡ϵ+sin⁡ϵcos⁡π3)= \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})

=12(31988(sin⁡πcos⁡3ϵ+sin⁡3ϵcos⁡π)+sin⁡π3cos⁡ϵ+sin⁡ϵcos⁡π3)= \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})

=12(−31988sin⁡3ϵ+32cos⁡ϵ+12sin⁡ϵ)= \frac{1}{2}(-3^{1988} \sin 3\epsilon + \frac{\sqrt 3}{2}\cos \epsilon + \frac{1}{2}\sin \epsilon)

using the above approximations,

=12(−319883ϵ+32+12ϵ)= \frac{1}{2}(-3^{1988} 3\epsilon + \frac{\sqrt 3}{2} + \frac{1}{2} \epsilon)


12ϵ\frac{1}{2} \epsilon is small enough to ignore. subbing in the other values,


=−3198833−198932+32= -3^{1988} 3\frac{3^{-1989}\sqrt 3}{2} + \frac{\sqrt 3}{2}

=−319883−198832+32= -3^{1988} 3^{-1988}\frac{\sqrt 3}{2} + \frac{\sqrt 3}{2}

=−32+32= -\frac{\sqrt 3}{2} + \frac{\sqrt 3}{2}

=0= 0

To prove that π3+ϵ\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.
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.
II/3.

- Both k and 1/k will satisfy the equation.
Unparseable latex formula:

x^2 + bk + c = 0 \\[br]\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 k2+bk+1=0k^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(1-k)^2 + b(1-k) + 1 = 0 will satisfy it. This gives 1−2k+k2+b−bk+1=0  ⟹  k2+(−2−b)k+(2+b)=0 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 x2−x+1=0x^2 - x + 1 = 0.

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

Therefore
Unparseable latex formula:

\alpha^3 + p\alpha^2 + q\alpha + r = 0 \\[br](1-\alpha)^3 + p(1 - \alpha)^2 + q(1-\alpha) + r = 0 \\[br]\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:
α3+qrα2+prα+1r=0\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.
Unparseable latex formula:

1 - 3\alpha + 3\alpha^2 - \alpha^3 + p - 2\alpha p + p\alpha^2 + p - p\alpha + 1 = 0 \\[br]\implies \alpha^3 - 1 + 3\alpha - 3\alpha^2 - p + 2\alpha p - p \alpha^2 - p + p\alpha - 1 = 0 \\[br]\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 x3−32x2−32x+1=0x^3 - \dfrac{3}{2}x^2 - \dfrac{3}{2}x + 1 = 0.
UNDER CONSTRUCTION

STEP II - Question 5

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

Then: z7=(c+is)7=c7+7isc6−21s2c5−35is3c4+35c3s4+21is3c2−7s6c−is7.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, z7=cos⁡(7θ)+isin⁡(7θ).z^{7} = \cos(7\theta) + i\sin(7\theta). So equating the real and imaginary parts:

cos⁡(7θ)=c7−21s2c5+35c3s4−7s6c.\cos(7\theta) = c^{7} - 21s^{2}c^{5} + 35c^{3}s^{4} -7s^{6}c. and sin⁡(7θ)=7sc6−35s3c4+21s5c2−s7 \sin(7\theta) = 7sc^{6} - 35s^{3}c^{4} + 21s^{5}c^{2} - s^{7} .

tan⁡(7θ)=sin⁡(7θ)cos⁡(7θ)=7sc6−35s3c4+21s5c2−s7c7−21s2c5+35c3s4−7s6c=7t−35t3+21t5−t71−21t2+35t4−7t6=t(7−35t2+21t4−t61−21t2+35t4−7t6 \tan(7\theta) = \frac{\sin(7\theta)}{\cos(7\theta)} = \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⁡(θ)\tan(\theta).


So if tan⁡(7θ)=0\tan(7\theta) = 0, then either tan⁡(θ)=0\tan(\theta) = 0, or 7−35t2+21t4−t6=0(∗)7 - 35t^{2} + 21t^{4} - t^{6} = 0 (*).

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

⇒θ=π7,2π7,3π7,4π7,5π7,6π7\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⁡(θ)\tan(\theta).

Hence the roots of (*) are : tan⁡(π7),tan⁡(2π7),tan⁡(3π7),tan⁡(4π7),tan⁡(5π7),tan⁡(6π7).\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⁡(π7)tan⁡(2π7)tan⁡(3π7)tan⁡(4π7)tan⁡(5π7)tan⁡(6π7)=−7\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(6π7)=−tan(π7),tan⁡(5π7)=−tan⁡(2π7)andtan⁡(3π7)=tan⁡(4π7) 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(2π7)tan⁡2(4π7)tan⁡2(6π7)=7\tan^{2}(\frac{2\pi}{7})\tan^{2}(\frac{4\pi}{7})\tan^{2}(\frac{6\pi}{7}) = 7

⇒tan⁡(2π7)tan⁡(4π7)tan⁡(6π7)=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(2π9)tan⁡2(4π9)tan⁡2(6π9)tan⁡2(8π9)=9\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

⇒tan⁡(2π9)tan⁡(4π9)tan⁡(6π9)tan⁡(8π9)=3\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(2π11)tan⁡2(4π11)tan⁡2(6π11)tan⁡2(8π11)tan⁡2(10π11)=11\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

⇒tan⁡(2π11)tan⁡(4π11)tan⁡(6π11)tan⁡(8π11)tan⁡(10π11)=−11\Rightarrow \tan(\frac{2\pi}{11})\tan(\frac{4\pi}{11})\tan(\frac{6\pi}{11})\tan(\frac{8\pi}{11})\tan(\frac{10\pi}{11}) = -\sqrt{11} (negative root since odd number of negative tans).

Phew, my hands hurt.

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