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STEP III 2006, Question 2

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STEP III 2006, Question 3

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(edited 12 years ago)
STEP III 8


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We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows; [br]ddxf(x)=limdx0f(x+dx)f(x)dx [br]\displaystyle \frac{d}{dx} f(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}

We will now prove property 1

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We now prove property 2 is met by this operation

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We prove property three is met by the above operation.

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We now prove the final property;

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We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

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As required
I/8:

(i) Let the base be OAC, so area = 1/2ca. Height is b, so the area is b*1/3*(1/2ca) = abc/6.

(ii) cosθ\cos \theta is the angle between BC and AC. BC=(0b0)+(00c)=(0bc)BC = \left( \begin{array}{c} 0 \\ -b \\ 0 \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ c \end{array} \right) = \left( \begin{array}{c} 0 \\ -b \\ c \end{array} \right)

AC=(a00)+(00c)=(a0c)AC = \left( \begin{array}{c} -a \\ 0 \\ 0 \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ c \end{array} \right) = \left( \begin{array}{c} -a \\ 0 \\ c \end{array} \right)

(0bc).(a0c)=a2+c2b2+c2cosθcosθ=c2(a2+c2)(b2+c2)\left( \begin{array}{c} 0 \\ -b \\ c \end{array} \right) . \left( \begin{array}{c} -a \\ 0 \\ c \end{array} \right) = \sqrt{a^2 + c^2} \sqrt{b^2 + c^2} \cos \theta \Rightarrow \cos \theta = \dfrac{c^2}{\sqrt{(a^2 + c^2)(b^2 + c^2)}}

Area of triangle = "12absinθ\dfrac{1}{2}ab \sin \theta". Note that sin2θ=1cos2θ\sin^2 \theta = 1 - \cos^2 \theta, so sin2θ=1c4(a2+c2)(b2+c2)=a2b2+a2c2+b2c2(a2+c2)(b2+c2)\sin^2 \theta = 1 - \dfrac{c^4}{(a^2 + c^2)(b^2 + c^2)} = \dfrac{a^2b^2 + a^2c^2 + b^2c^2}{(a^2 + c^2)(b^2 + c^2)}

area=12a2+c2b2+c2a2b2+b2c2+a2c2(a2+c2)(b2+c2)=12a2b2+b2c2+a2c2\text{area} = \dfrac{1}{2} \sqrt{a^2 + c^2} \sqrt{b^2 + c^2} \dfrac{\sqrt{a^2b^2 + b^2c^2 +a^2c^2}}{\sqrt{(a^2 + c^2)(b^2 + c^2)}} = \dfrac{1}{2}\sqrt{a^2b^2 + b^2c^2 +a^2c^2}

The area will clearly be equal to abc/6, only that the height of the new triangle is d.

16abc=16(a2b2+b2c2+a2c2)1/2da2b2c2=d2(a2b2+b2c2+a2c2)\dfrac{1}{6}abc = \dfrac{1}{6}(a^2b^2 + b^2c^2 +a^2c^2)^{1/2}d \Rightarrow a^2b^2c^2 = d^2(a^2b^2 + b^2c^2 +a^2c^2)

1d2=a2b2+b2c2+a2c2a2b2c21d2=1a2+1b2+1c2\Rightarrow \dfrac{1}{d^2} = \dfrac{a^2b^2 + b^2c^2 +a^2c^2}{a^2b^2c^2} \Rightarrow \dfrac{1}{d^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}
DeanK22
STEP III 8


Spoiler



We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows; [br]ddxf(x)=limdx0f(x+dx)f(x)dx [br]\displaystyle \frac{d}{dx} f(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}

We will now prove property 1

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We now prove property 2 is met by this operation

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We prove property three is met by the above operation.

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We now prove the final property;

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We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

Spoiler



As required


I'm pretty sure you don't need to prove that each of the four rules works for differentiation. You just need to prove, by induction as you have, that Δ(xn)=nxn1 \Delta(x^n) = nx^{n-1} , using the four rules. Then let h(x)=k=0nakxk \displaystyle h(x) = \sum_{k=0}^n a_k x^k , and apply the operation to this function, making use of rules ii) and iii) to show that Δ(h(x))=k=0nkakxk1 \displaystyle \Delta(h(x)) = \sum_{k=0}^n k a_k x^{k-1} .
Reply 7
STEP III 2006, Question 4

initial part

(i)

(ii)

(iii)

Reply 8
2006 I Question 13
Part (i)
The number of diamonds in one kilogram has a Poisson distribution of DPoisson(1)D \sim \text{Poisson}(1). We find that per 100 grams, DPoisson(λ=0.1)D \sim \text{Poisson}(\lambda = 0.1). For 100T grams, DTPoisson(λ=0.1T)D_{T} \sim \text{Poisson}(\lambda = 0.1T). TT is the distribution of scores of the die.

Drawing 1, we have P(T=1D1=0)=16(e0.1)P(T = 1 \cap D_{1}=0) = \frac{1}{6}(e^{-0.1}).
Similarly, P((T=1D1)(T=2D2=0)(T=6D6=0))=t=1616e0.1t=\displaystyle P((T=1 \cap D_{1})\cup(T=2\cap D_{2}=0)\cup\ldots\cup(T=6\cap D_{6}=0)) = \sum_{t=1}^6 \frac{1}{6}e^{-0.1t}=
=16e0.1t=16e(0.1)(t1)=e0.161e0.61e0.1\displaystyle = \frac{1}{6}e^{-0.1}\sum_{t=1}^6 e^{(-0.1)(t-1)} = \frac{e^{-0.1}}{6}\frac{1-e^{-0.6}}{1-e^{-0.1}} (recognising a finite geometric sum).

Part (i) - expectation
E[DT]=t=1616E[Dt]=16t=160.1t=16(0.1+0.2+0.3+0.4+0.5+0.6)=2.16=0.72=0.35\displaystyle E[D_T] = \sum_{t=1}^6 \frac{1}{6}E[D_t] = \frac{1}{6}\sum_{t=1}^6 0.1t = \frac{1}{6}(0.1+0.2+0.3+0.4+0.5+0.6) = \frac{2.1}{6} = \frac{0.7}{2} = 0.35

Part (ii)
This time, TT is a geometric distribution with p=56p = \frac{5}{6}. Thus, P(T=t)=16(56)t1P(T=t)=\frac{1}{6}\left(\frac{5}{6}\right)^{t-1}.

This time, our probability is:
P((T=1D1=0)(T=2D2=0))=t=1P(T=tDt=0)=\displaystyle P((T=1 \cap D_{1}=0)\cup(T=2\cap D_{2}=0)\cup\ldots) = \sum_{t=1}^{\infty}P(T=t\cap D_{t}=0) =
=t=1(56)t116e0.1t=e0.16t=1(e0.156)t1=\displaystyle = \sum_{t=1}^{\infty} \left(\frac{5}{6}\right)^{t-1}\frac{1}{6}e^{-0.1t} = \frac{e^{-0.1}}{6}\sum_{t=1}^{\infty} \left(e^{-0.1}\frac{5}{6}\right)^{t-1} =
=e0.161156e0.1=e0.165e0.1\displaystyle = \frac{e^{-0.1}}{6}\frac{1}{1-\frac{5}{6}e^{-0.1}} = \frac{e^{-0.1}}{6-5e^{-0.1}}, as required.

Part (ii) - expectation
E[DT]=t=1E[Dt]P(T=t)=t=10.1t16(56)t1=160t=1t(56)t1\displaystyle E[D_T] = \sum_{t=1}^\infty E[D_t]P(T=t) = \sum_{t=1}^\infty 0.1t\frac{1}{6}\left(\frac{5}{6}\right)^{t-1} =\frac{1}{60}\sum_{t=1}^\infty t\left(\frac{5}{6}\right)^{t-1}.

Consider: ddx((1x)1)=(1x)2=ddxt=0xt=t=1txt1\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} ((1-x)^{-1}) = (1-x)^{-2} = \frac{\mathrm{d}}{\mathrm{d}x} \sum_{t=0}^\infty x^t = \sum_{t=1}^\infty tx^{t-1}.

Thus, E[DT]=160(156)2=3660=0.6\displaystyle E[D_T] = \frac{1}{60}\left(1-\frac{5}{6}\right)^{-2} = \frac{36}{60} = 0.6.
Reply 9
2006 I Question 14
Either I missed something crucial, or this question is incredibly easy. I'll write up the incredibly-easy interpretation of this question:

Part (i)
This is a geometric progression, where P(Red on rth attempt)=pqr1\displaystyle P(\text{Red on rth attempt}) = pq^{r-1}, where p=1n\displaystyle p=\frac{1}{n}, q=n1n\displaystyle q = \frac{n-1}{n}, so P(Red on rth attempt)=(n1n)r11n=(n1)r1nr\displaystyle P(\text{Red on rth attempt}) = \left(\frac{n-1}{n}\right)^{r-1}\frac{1}{n} = \frac{(n-1)^{r-1}}{n^r}.
Differentiating, we get: dPdn=nr(r1)(n1)r2rnr1(n1)r1n2r\displaystyle \frac{\mathrm{d}P}{\mathrm{d}n} = \frac{n^r(r-1)(n-1)^{r-2} - rn^{r-1}(n-1)^{r-1}}{n^{2r}}. Equating to zero, nr(r1)(n1)r2=rnr1(n1)r1    n(r1)=r(n1)    n(r1r)=r    n=r    n=rn^r(r-1)(n-1)^{r-2} = rn^{r-1}(n-1)^{r-1} \iff n(r-1) = r(n-1) \iff n(r-1-r) = -r \iff -n=-r \iff n = r. We know that this is the maximum as r tends to infinity, P tends to zero.

Part (ii)

P(Red on 1st)=1nP(\text{Red on 1st}) = \frac{1}{n}
P(Red on 2nd)=n1n1n1=1nP(\text{Red on 2nd}) = \frac{n-1}{n}\cdot\frac{1}{n-1} = \frac{1}{n}
P(Red on rth)=n1nn2n1n(r2)n(r3)n(r1)n(r2)1n(r1)=1nP(\text{Red on rth}) = \frac{n-1}{n}\cdot\frac{n-2}{n-1}\cdots\frac{n-(r-2)}{n-(r-3)}\cdot\frac{n-(r-1)}{n-(r-2)}\cdot\frac{1}{n-(r-1)}=\frac{1}{n} . Clearly maximum at n = 1.

Did I make a mistake here? It just looks too simple for a STEP question, I'm afraid.

Edit: see GHOSH-5's post here: http://www.thestudentroom.co.uk/showpost.php?p=19496483&postcount=41
Daniel Freedman
I'm pretty sure you don't need to prove that each of the four rules works for differentiation. You just need to prove, by induction as you have, that Δ(xn)=nxn1 \Delta(x^n) = nx^{n-1} , using the four rules. Then let h(x)=k=0nakxk \displaystyle h(x) = \sum_{k=0}^n a_k x^k , and apply the operation to this function, making use of rules ii) and iii) to show that Δ(h(x))=k=0nkakxk1 \displaystyle \Delta(h(x)) = \sum_{k=0}^n k a_k x^{k-1} .
Actually, it's worse than that.

They've said "given that the operator Δ\Delta has these properties; show that...".

Dean has said "A particular operator δ\delta has those properties, and for δ\delta, we have ...."

But he hasn't shown that δ\delta is the only possible operator with the properties.

This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a ton of marks for this (even if the Examiner sympathised).
DFranklin
Actually, it's worse than that.

They've said "given that the operator Δ\Delta has these properties; show that...".

Dean has said "A particular operator δ\delta has those properties, and for δ\delta, we have ...."

But he hasn't shown that δ\delta is the only possible operator with the properties.

This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a ton of marks for this (even if the Examiner sympathised).


Thanks for pointing that out - certainly a mistake that will not be made this Summer (hopefully). Unfortunately I do not really know how to prove the Uniquness of this operator - could you please shed some light on to how you do that? Thanks.
Well, that's effectively what the question is asking you to show. I would:

(1) Prove by induction that Δxn=nxn1\Delta x^n = n x^{n-1} (using (i) and (iv)).
(2) Then use (iii) to prove Δanxn=nanxn1\Delta a_n x^n = n a_n x^{n-1}
(3) Then use (ii) to prove Δ0Nanxn=0Nnanxn1\Delta \sum_0^N a_n x^n = \sum_0^N n a_n x^{n-1}.

[Which effectively proves uniqueness of Δ\Delta for polynomials].

Then simply observe this is the same as the derivative in the case of polynomials.
I have different answers to STEP I Q 14 Im not to sure who is wrong...

Part (i)

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Part (ii)

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STEP I: Q5

Part (i)



Part (ii)

Reply 15
STEP III 2006, Q5

We wish to show that α,β,γ\alpha, \beta, \gamma form an equilateral iff

α2+β2+γ2αββγγα=0\displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = 0

Multiplying by two and factorising, we wish to show that,

(αβ)2+(βγ)2+(γα)2=0\displaystyle (\alpha - \beta)^2 + (\beta - \gamma)^2 + (\gamma - \alpha)^2 = 0

holds for all equilateral triangles.

Note that αβ,βγ,γα\alpha - \beta, \beta - \gamma, \gamma - \alpha are the sides of the triangle. If they are shown on an Argand diagram, the angle between any two consecutive sides is π3\frac{\pi}{3}. Note also that the lengths of the sides are all of equal length, so the magnitudes of the complex numbers that represent them will also be equal.

Using this, we reconstruct the equation:

(reiθ)2+(rei(θ+π3))2+(rei(θ+2π3))2=0\displaystyle \left(re^{i \theta}\right)^2 + \left(re^{i (\theta +\frac{\pi}{3})}\right)^2 + \left(re^{i (\theta +\frac{2\pi}{3})}\right)^2 = 0

or, more simply,

r2(e2iθ+e2i(θ+π3)+e2i(θ+2π3))=0\displaystyle r^2 \left( e^{2 i \theta} + e^{2 i (\theta +\frac{\pi}{3})} + e^{2 i (\theta +\frac{2 \pi}{3})} \right) = 0

r2e2iθ(1+eiπ3+ei2π3)=0\displaystyle r^2 e^{2 i \theta} \left( 1 + e^{i \frac{\pi}{3}} + e^{i \frac{2 \pi}{3}} \right) = 0

The bracket evaluates to 0, so we are done.

For the next part, we take the roots of the cubic to be α,β,γ\alpha, \beta, \gamma, and expand, yielding Vieta's formula:

z3(α+β+γ)z2+(αβ+βγ+γα)zαβγ=0\displaystyle z^3 - (\alpha + \beta + \gamma) z^2 + (\alpha \beta + \beta \gamma + \gamma \alpha) z - \alpha \beta \gamma = 0

Algebra shows that:

α2+β2+γ2αββγγα=(α+β+γ)23(αβ+βγ+γα)=a23b=0\displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = (\alpha + \beta + \gamma)^2 - 3 (\alpha \beta + \beta \gamma + \gamma \alpha) = a^2 - 3b = 0

and the result follows immediately.

Now we write,

p=Aeiϕp = Ae^{i \phi}

q=B+Ciq = B + Ci

The transformation z=pw+qz = p w + q has the effect of rotating our equilateral triangle by the angle ϕ\phi, magnifying it by a factor of AA and translating it through the vector represented by qq (would I need to show this?). Under these transformations, the equilateral triangle formed by the roots of the initial cubic remains an equilateral triangle.
Reply 16
STEP I 2006 Question 2

Solution

RE; Noudai

opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker.
Reply 18
DeanK22
RE; Noudai

opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker.

I realised that a bit too late :p: I've chopped it out of the answer though.
Reply 19
STEP III 2006 Q7

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