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]\displaystyle \frac{d}{dx} f(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}

We will now prove property 1

Spoiler

We now prove property 2 is met by this operation

Spoiler

We prove property three is met by the above operation.

Spoiler

We now prove the final property;

Spoiler

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

Reply 5

Glutamic Acid

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)

\cos \theta

is the angle between BC and AC.

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 = \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)

\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 = "

\dfrac{1}{2}ab \sin \theta

". Note that

\sin^2 \theta = 1 - \cos^2 \theta

, so

\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)}

\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.

\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)

\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}

Reply 6

Daniel Freedman

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]\displaystyle \frac{d}{dx} f(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}

We will now prove property 1

Spoiler

We now prove property 2 is met by this operation

Spoiler

We prove property three is met by the above operation.

Spoiler

We now prove the final property;

Spoiler

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

\Delta(x^n) = nx^{n-1}

, using the four rules. Then let

\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

\displaystyle \Delta(h(x)) = \sum_{k=0}^n k a_k x^{k-1}

Reply 7

Dadeyemi

STEP III 2006, Question 4

initial part

(i)

(ii)

(iii)

Reply 8

Aurel-Aqua

2006 I Question 13
Part (i)
The number of diamonds in one kilogram has a Poisson distribution of

D \sim \text{Poisson}(1)

. We find that per 100 grams,

D \sim \text{Poisson}(\lambda = 0.1)

. For 100T grams,

D_{T} \sim \text{Poisson}(\lambda = 0.1T)

T

is the distribution of scores of the die.

Drawing 1, we have

P(T = 1 \cap D_{1}=0) = \frac{1}{6}(e^{-0.1})

.
Similarly,

\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}=

\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

\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,

T

is a geometric distribution with

p = \frac{5}{6}

. Thus,

P(T=t)=\frac{1}{6}\left(\frac{5}{6}\right)^{t-1}

.

This time, our probability is:

\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) =

\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} =

\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

\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:

\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,

\displaystyle E[D_T] = \frac{1}{60}\left(1-\frac{5}{6}\right)^{-2} = \frac{36}{60} = 0.6

Reply 9

Aurel-Aqua

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

\displaystyle P(\text{Red on rth attempt}) = pq^{r-1}

, where

\displaystyle p=\frac{1}{n}

\displaystyle q = \frac{n-1}{n}

, so

\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:

\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,

n^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(\text{Red on 1st}) = \frac{1}{n}

P(\text{Red on 2nd}) = \frac{n-1}{n}\cdot\frac{1}{n-1} = \frac{1}{n}

P(\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

Reply 10

DFranklin

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

\Delta(x^n) = nx^{n-1}

, using the four rules. Then let

\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

\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

\Delta

has these properties; show that...".

Dean has said "A particular operator

\delta

has those properties, and for

\delta

, we have ...."

But he hasn't shown that

\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).

Reply 11

Oh I Really Don't Care

DFranklin

Actually, it's worse than that.

They've said "given that the operator

\Delta

has these properties; show that...".

Dean has said "A particular operator

\delta

has those properties, and for

\delta

, we have ...."

But he hasn't shown that

\delta

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.

Reply 12

DFranklin

Well, that's effectively what the question is asking you to show. I would:

(1) Prove by induction that

\Delta x^n = n x^{n-1}

(using (i) and (iv)).
(2) Then use (iii) to prove

\Delta a_n x^n = n a_n x^{n-1}

(3) Then use (ii) to prove

\Delta \sum_0^N a_n x^n = \sum_0^N n a_n x^{n-1}

.

[Which effectively proves uniqueness of

\Delta

for polynomials].

Then simply observe this is the same as the derivative in the case of polynomials.

Reply 13

darkness9999

I have different answers to STEP I Q 14 Im not to sure who is wrong...

Part (i)

Spoiler

Part (ii)

Spoiler

Reply 14

darkness9999

STEP I: Q5

Part (i)

Part (ii)

Reply 15

Elongar

STEP III 2006, Q5

We wish to show that

\alpha, \beta, \gamma

form an equilateral iff

\displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = 0

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

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

holds for all equilateral triangles.

Note that

\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

\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:

\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,

\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

\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

\alpha, \beta, \gamma

, and expand, yielding Vieta's formula:

\displaystyle z^3 - (\alpha + \beta + \gamma) z^2 + (\alpha \beta + \beta \gamma + \gamma \alpha) z - \alpha \beta \gamma = 0

Algebra shows that:

\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 = Ae^{i \phi}

q = B + Ci

The transformation

z = p w + q

has the effect of rotating our equilateral triangle by the angle

\phi

, magnifying it by a factor of

A

and translating it through the vector represented by

q

(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

nuodai

STEP I 2006 Question 2

Solution

Reply 17

Oh I Really Don't Care

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

nuodai

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.