STEP I Question 4

Im[cos5t + isin5t] = Im[(cost + isint)^5]
sin5t = Im[cos^5t + 5isintcos^4t - 10sin^2tcos^3t - 10isin^3tcos^2t + 5sin^4tcost + isin^5t]
sin5t = 5sintcos^4t - 10sin^3tcos^2t + sin^5t
sin5t = sint[5cos^4t - 10cos^2t + 10cos^4t + 1 + cos^4t - 2cos^2t]
sin5t = sint[16cos^4t - 12cos^2t + 1]

Let t = pi/5
0 = sint[16cos^4t - 12cos^2t + 1]
sint =/= 0
16cos^4t - 12cos^2t + 1 = 0
cos^2t = [12 +- sqrt[144 - 64]]/32
= [12 +- sqrt80]/32
= [3 +- sqrt5]/8
A glance at the graph of cos^2 shows that the larger root is wanted (the other is very small).
cos^2t = [3 + sqrt5]/8

Consider (1/4)(1 + rt5)
= sqrt[(1/16)(6 + 2rt5)] = [(3 + sqrt5)/8]

Therefore cost = [1 + sqrt5]/4

Reply 3

insparato

15

Having a go at STEP I Question 3 now.

3 i)

Sum of Differences method

f(r) - f(r-1) = .....

f(1) - f(0)
f(2) - f(1)
f(3) - f(2)
.
.
.
f(n-1) - f(n-2)
f(n) - f(n-1)

All terms cancel except two f(n) and f(0)

Therefore

\displaystyle \sum_{r=1}^n f(r) - f(r-1) = f(n) - f(0)

ii)

f(r) = r^2(r+1)^2

f(r-1) = r^2(r-1)^2

f(r) - f(r-1) = r^2(r+1)^2 - r^2(r-1)^2

= r^2(r^2+2r+1 -r^2+2r-1) = 4r^3

Therefore

\displaystyle \sum_{r=1}^n f(r) - f(r-1) = f(n) - f(0) = n^2(n+1)^2

\displaystyle \sum_{r=1}^n f(r)-f(r-1) = \sum_{r=1}^n 4r^3 = n^2(n+1)^2

\displaystyle \sum_{r=1}^n r^3 = \frac{1}{4}n^2(n+1)^2

iii)

\displaystyle \sum_{r=1}^n (2n-1)^3 - \sum_{r=1}^n (2n)^3

\frac{1}{4}(2n-1)^2(2n)^2 - \frac{1}{4}(2n)^2(2n+1)^2

n^2(4n^2-4n+1 -(4n^2+4n+1))

n^2(-8n)

-8n^3

Hmm ive gone wrong somewhere on the last part.

Oh looking at the paper 1^3 - 2^3 + 3^3 .......(2n+1)^3 hmmm where does (2n+1)^3

surely sum (2n-1)^3 - sum(2n)^3 should work ?

Reply 4

Speleo

7

STEP I Question 5

f'(x) = n - nC2x + nC3x^2 - nC4x^3 + ... + nCn(-1)^(n-1).x^n-1

But (1-x)^n = 1 - nx + nC2x^2 - nC3x^3 + ... + (-1)^(n)x^n
1 - (1-x)^n = nx - nC2x^2 + nC3x^3 + ... + (-1)^(n-1)x^n
[1 - (1-x)^n]/x = n - nC2x + nC3x^2 - nC4x^3 + ... + nCn(-1)^(n-1).x^n-1
as required.

f(x) = INT f'(x) dx limits 0 and x
f(x) = INT [1 - (1-x)^n]/x dx limits 0 and x
Let y = 1 - x; dy/dx = -1
-> -INT [1 - y^n]/[1 - y] dy limits 1 and 1 - x
= INT [1 - y^n]/[1 - y] dy limits 1-x and 1

f(x) = INT [1-y^n]/[1-y] dy = INT 1 + y + y^2 + ... + y^n-1 dy
= [y + y^2/2 +y^3/3 + ... + y^n/n]
x = 1 so limits are 0->1
f(1) => 1 + 1/2 + 1/3 + 1/4 + ... + 1/n

Reply 5

Speleo

7

STEP I Question 6

i) y^-2.dy/dx + y^-1 = e^2x
Let u = 1/y
du/dy = -y^-2
dy/dx = du/dx.dy/du = -y^2.du/dx = -u^-2.du/dx
-du/dx + u = e^2x
du/dx - u = -e^2x
(e^-x)du/dx - (e^-x)u = -e^(x)
d/dx(ue^-x) = -e^x
ue^-x = A - e^x
u = Ae^x - e^2x
y = 1/[Ae^x - e^2x]

ii) y^-3.dy/dx + y^-2 = e^2x
Let u = y^-2
du/dy = -2y^-3
dy/dx = du/dx.dy/du = du/dx.-(1/2)y^3
= -(1/2)u^(-3/2).du/dx
-(1/2)u^(3/2)u^(-3/2).du/dx + u = e^2x
du/dx - 2u = -2e^2x
(e^-2x)du/dx - (e^-2x)u = -2
d/dx[ue^-2x] = -2
ue^-2x = B - 2x
u = Be^2x - 2xe^2x
y = 1/sqrt[Be^2x - 2xe^2x]

Reply 6

Speleo

7

STEP I Question 8

y'' + fy' + gy = h

y=x
f + gx = h

y=1
g = h
f = g(1-x)

y=1/x
2/x^3 - f/x^2 + g/x = h
2/x^3 - g(1-x)/x^2 + g/x = g
2/x^3 = g(1 - 1/x + (1-x)/x^2)
2 = g(x^3 - x^2 - x^2 + x)
g = h = 2/[x^3 - 2x^2 + x] = 2/x(x-1)^2
f = -2/x(x-1)

y = ax + b + c/x
y' = a - c/x^2
y'' = 2c/x^3

2c/x^3 - [2/x(x-1)][a - c/x^2] + [2/x(x-1)^2][ax + b + c/x] = [2/x(x-1)^2]

2c - [2/x-1][ax^2 - c] + x[2/(x-1)^2][ax^2 + bx + c] = 2x^2/(x-1)^2
2c(x-1)^2 - (x-1)(2ax^2 - 2c) + 2x(ax^2 + bx + c) = 2x^2
2cx^2 - 4cx + 2c - 2ax^3 + 2ax^2 + 2cx - 2c + 2ax^3 + 2bx^2 + 2xc = 2x^2
2cx^2 + 2ax^2 + 2bx^2 = 2x^2
a + b + c = 1

Reply 7

khaixiang

5

Question 1 STEP III, 1995 is found in Siklos' Advanced Problems in Mathematics, question 14. (Not the booklet that can be downloaded for free)

EDIT: I've attached it here for those who didn't order the booklet.

STEP II Question 4

By parts:
u_n = [-sin^(n-1)tcost] + (n-1) INT sin^(n-2)tcos^2t dt
Putting cos^2t = 1 - sin^2t and using limits on the [...] gives:
u_n = 0 + (n-1)(u_n-2 - u_n)
u_n = (n-1)(u_n-2) - (n-1)u_n
nu_n = (n-1)(u_n-2)

nu_n.u_n-1 = (n-1)u_n-1.u_n-2

^ This proves that mu_m.u_m-1 is the same for all m, so it is only necessary to prove it for one case, n = 2, and the rest follow by induction.
2 INT sin^2t dt . INT sint dt
= INT 1 - cos2t dt . [-cost]
= [t - 1/2 sin2t].[1]
= [pi/2 - 0]
= pi/2

*graph*. sin^(n-1) is a flattened sin curve and sin^n is one flattened slightly more, so just below it but having the same end points.
Since sin^nt is between 0 and 1 in the range of the integrals, higher powers mean lower areas, but they will all still be greater than 0, so 0 < u_n < u_n-1.

n.u_n.u_n-1 = pi/2
But u_n < u_n-1, so:
n.(u_n)^2 < pi/2
and
n.{u_n-1)^2 > pi/2

Second half of the next inequality has already been shown.
Changing n to n+1 we have pi/2 < (n+1)(u_n)^2 by the second half of the previous inequality.
npi/2 < n(n+1)(u_n)^2
(n/n+1)pi/2 < n(u_n)^2
So the entire inequality has been shown.

As n-> infinity, n/n+1 = 1 - 1/n goes to 1.
Therefore, pi/2 <= n(u_n)^2 <= pi/2
n(u_n)^2 = pi/2

Reply 9

Speleo

7

STEP II Question 7

i) AIB = (1/2)rc; BIC = (1/2)ra; CIA = (1/2)rb
Area of triangle = Z = r(1/2)(a+b+c) = rs

ii) Z = (1/2)bcsinA
Z^2 = (1/4)b^2c^2sin^2A
Z^2 = (1/4)(b^2c^2 - (bccosA)^2)
Z^2 = (1/16)(4b^2c^2 - (2bccosA)^2)

But 2bccosA = b^2 + c^2 - a^2

Z^2 = (1/16)(4b^2c^2 - (b^2 + c^2 - a^2)^2)
Difference of two squares:
Z^2 = (1/16)(2bc + b^2 + c^2 - a^2)(2bc - b^2 - c^2 + a^2)
Z^2 = (1/16)((b+c)^2 -a^2)(a^2 - (b-c)^2)
Difference of two squares again:
Z^2 = (1/16)(b+c-a)(b+c+a)(a-b+c)(a+b-c)
Z^2 = (1/16)(a+b+c-2a)(a+b+c)(a+b+c-2b)(a+b+c-2c)
Z^2 = (s-a)s(s-b)(s-c)
Z = sqrt[s(s-a)(s-b)(s-c)]

iii) Draw a diagram, x^2 = R^2 - r^2
But r = Z/s = sqrt[(1/s)(s-a)(s-b)(s-c)]
x^2 = R^2 - (1/s)(s-a)(s-b)(s-c)
x = sqrt[R^2 - (1/s)(s-a)(s-b)(s-c)]

Reply 10

Speleo

7

STEP II Question 8

dx/dt = (2K-x)x
INT 1/(2K-x)x dx = INT 1 dt

Partial fractions, 1/(2K-x)x = A/(2K-x) + B/x
1 = Ax + B(2K-x)
2Kb = 1
b = 1/(2K)
0 = A - B
A = 1/(2K)

INT 1/2K(2K-x) + 1/2Kx dx = INT 1 dt
-(1/2K)ln[2K(2K-x)] + (1/2K)ln[2Kx] = t + C
ln[Ax] = lnx as far as we are concerned (an arbitrary constant is involved).
(1/2K).[lnx - ln(2K-x)] = t + C'
When t = 0, x = K
C' = (1/2K).[ln(K/K)]
C' = 0

ln[x/(2K-x)] = 2Kt
x/(2K-x) = e^2Kt
2K-x = xe^-2Kt
2K = x(1 + e^-2Kt)
x = 2K/(1 + e^-2Kt)
x = (K + Ke^-2Kt + K - Ke^-2Kt)/(1 + e^-2Kt)
x = K + K(1-e^-2Kt)/(1+e^-2Kt)

*graph* starts at x=K, goes up and becomes asymptotic to x=2K.
t-> infinity, x-> 2K

dx/dt = (2K-x)x - L
dy/dt = dx/dt
dy/dt = (2K - y - K + a)(y + K - a) - L
dy/dt = (K - y + a)(K + y - a) - L
dy/dt = K^2 + Ky - Ka - Ky - y^2 + ay + Ka + ay - a^2 - L
dy/dt = K^2 - y^2 + 2ay - a^2 - L
dy/dt = a^2 - y^2 + 2ay - a^2
dy/dt = (2a-y)y

When t = 0, y = K - K + a = a
Equations are symmetrical in (x,K) and (y,a)

y = a + a(1-e^-2at)/(1+e^-2at)
x - K + a = a + a(1-e^-2at)/(1+e^-2at)
x = K + a(1-e^-2at)/(1+e^-2at)
t-> infinity
x -> K + a

Reply 11

Speleo

7

7 questions is enough for now :tongue:

, but I'll come back and do STEP II 1 and 6 in a bit if no-one else has (please do, I don't particularly want to type them out).

Reply 12

insparato

15

You've done alot tonight Speleo go watch some tv or something before STEP rots your brains :p:

.

Ill have alook at STEP II 1 and 6 however i dont expect to do any of them :smile:

.

Reply 13

DFranklin

18

Step I, Q3, last bit:

Unparseable latex formula:

\displaystyle 1^3-2^3+3^3-4^3+...-(2n+1)^3 \\[br]= 1^3+2^3+...+(2n+1)^3 - 2\big\{(2^3)+(4^3)+(6^3)+...+(2n)^3\big\} \\ [br]= \sum_1^{2n+1} r^3 - 16 \sum_1^n r^3 \\[br]= \frac{1}{4}(2n+1)^2(2n+2)^2 - 4 n^2(n+1)^2 \\[br]= (2n+1)^2(n+1)^2-4n^2(n+1)^2[br]= (4n+1)(n+1)^2

Alternatively:

Unparseable latex formula:

\displaystyle 1^3-2^3+3^3-4^3+...-(2n+1)^3 \\[br]= 1 + (3^3-2^3)+(5^3-4^3)+...+((2n+1)^3-(2n)^3) \\[br]= 1 + \sum_1^n (2r+1)^3-(2r)^3 \\[br]= 1 + \sum_1^n 12r^2+6r + 1 \\[br]= 1 + 2n(n+1)(2n+1) + 3n(n+1)+n \\[br]= (n+1) (2n(2n+1)+3n+1) = (n+1)(4n^2+5n+1)=(n+1)^2(4n+1)

.

Reply 14

insparato

15

Grrr i should revise finite series :smile:

. That question was definitely not beyond me.

Reply 15

Speleo

7

STEP III Question 3

(sorry, it was calling to me :tongue:

)

No marks for clarity unfortunately.

Aux. quadratic:
m^2 + 2km + 1 = 0
m = [-2k +- sqrt(4k^2 - 4)]/2
m = -k +- sqrt(k^2 - 1)

i) k > 1
x = e^(-kt).[Ae^(sqrt(k^2 - 1)t) + Be^(-sqrt(k^2 - 1)t)]

ii) k = 1
x = (At + B)e^(-kt)

iii) 0 < k < 1
x = e^(-kt).[Acos(sqrt(k^2 - 1)t) + Bsin(sqrt(k^2 - 1)t)]

x(0) = 0
0 = A
x = Be^(-kt)sin(sqrt(k^2-1)t)

dx/dt = -kBe^(-kt)sin(sqrt(k^2-1)t) + sqrt(k^2-1)Be^(-kt)cos(sqrt(k^2-1)t) = 0 at max/min.
ksin(sqrt(k^2-1)t) = sqrt(k^2-1)cos(sqrt(k^2-1)t)
tan(sqrt(k^2-1)t) = sqrt(k^2-1)/k
sqrt(k^2-1)t = arctan[sqrt(k^2-1)/k] + mpi
t = [1/sqrt(k^2-1)]arctan[sqrt(k^2-1)/k] + mpi/sqrt(k^2-1)

When dividing successive x terms, the difference in the e term will just be pi/sqrt(k^2-1).
x_n+1/x_n = e^(-kpi/sqrt(k^2-1)) sin[arctan[sqrt(k^2-1)/k] + (m+1)pi]/sin[arctan[sqrt(k^2-1)/k] + mpi]
sin(t + pi) = -sinpi
x_n+1/x_n = -e^(-kpi/sqrt(k^2-1))
e^(-kpi/sqrt(k^2-1)) = a.
-kpi/sqrt(k^2-1) = lna
k^2/(k^2-1) = (lna)^2/(pi)^2
k^2 = k^2(lna)^2/(pi)^2 - (lna)^2/(pi)^2
k^2((lna)^2/(pi)^2 - 1) = (lna)^2/(pi)^2
k^2[((lna)^2 - (pi)^2)/(pi)^2] = (lna)^2/(pi)^2
k^2 = (lna)^2/[(pi)^2 + (lna)^2

Reply 16

DFranklin

18

Paper I, Q1 (no sketch):

(i) f(x) = x^3-4x^2-x-4. It's easy to spot f(1),f(-1),f(4) = 0. So f(x) = (x+1)(x-1)(x-4). So f(x)>0 for -1<=x<=1, x>=4.

(ii) g(x,y) = x^3 - 4x^2y-xy^2+4y^3 = (x+y)(x-y)(x-4y) (noting the similarities with (i)). So g(x,y) = 0 on the lines y=x,y=-x,y=x/4.

(iii) Draw the 3 lines we found in (ii). Note that g(1,0) > 0, and g is going to change sign every time you cross a line. (Or if not feeling brave in trusting that, find a test point in each region).

Reply 17

Speleo

7

STEP II Question 1

I just can't stop myself D:

i) (1 - x)(1 + x + x^2 + x^3 + ... + x^n)
= (1 + x + x^2 + x^3 + ... + x^n) - (x + x^2 + x^3 + x^4 + ... + x^(n+1))
= 1 - x^(n+1)
1 + x + x^2 + x^3 + ... + x^n = [1 - x^(n+1)]/(1-x)

ii) 1 + 2x + 3x^2 + ... + nx^(n-1) = [-(1-x)(n+1)(x^n) - (x^(n+1) - 1)]/(1-x)^2
Let x = -1
1 - 2 + 3 - ... + (-1)^(n-1)n = [-(2)(n+1)(-1)^n - (-1)^(n+1) + 1]/4
If n is odd,
S = 2(n+1)/4 = (n+1)/2
If n is even,
S = [-2(n+1) - 2]/4
= [-2n - 2 + 2]/4 = n/2

iii) Group pairs.
If n is even:
= SUM 1 - 4r from 1 to n/2
= n/2 - n(n/2 + 1)
= n/2 - n^2/2 - n
= -n/2 - n^2/2

If n is odd:
= SUM 1 - 4r from 1 + (n-1)/2 + n^2
= (n-1)/2 - (n-1)((n-1)/2 + 1) + n^2
= n/2 - 1/2 - (1/2)(n-1)(n+1) + n^2
= n/2 - 1/2 - (1/2)(n^2 - 1) + n^2
= n/2 + n^2/2

i.e. A = B = 1/2

Reply 18

Speleo

7

STEP II Question 6

What the hell is wrong with me?

z^2 + az + b = 0
and (z-u)(z-v) = 0
-u-v = a
uv = b

From this point on a means alpha.

a^7 - 1 = cis(2pi/7)^7 - 1 = cis(2pi) - 1 = 1 - 1 = 0

Other roots: a^0, a^2, a^3, a^4, a^5, a^6

(a + a^2 + a^4)^2 + A(a + a^2 + a^4) + B = 0
a^2 + a^4 + a + 2a^3 + 2a^5 + 2a^6 + Aa + Aa^2 + Aa^4 + B = 0
B + (A+1)a + (A+1)a^2 + 2a^3 + (A+1)a^4 + 2a^5 + 2a^6 = 0

Guess other root as a^3 + a^5 + a^6
(a^3 + a^5 + a^6)^2 + A(a^3 + a^5 + a^6) + B = 0
a^6 + a^3 + a^5 + 2a + 2a^2 + 2a^4 + Aa^3 + Aa^5 + Aa^6 + B = 0
B + 2a + 2a^2 + (A+1)a^3 + 2a^4 + (A+1)a^5 + (A+1)a^6 = 0

Clearly B=2 and A=1, as the sum of the powers of a is 0.
Take one from the other if you want confirmation (I think).

z^2 + z + 2 = 0 has root a + a^2 + a^4
z = [-1 +- sqrt(-7)]/2
Looking at the diagram of the 7th roots of unity, a + a^2 + a^4 has Im(z) > 0, so a = [-1 + isqrt7]/2

Re(a) = sum of cosines = -(1/2)
Im(a) = sum of sines = (sqrt7)/2