SimonM
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(Updated as far as #100.) SimonM - 23.03.2009

For explanation, scroll down:

STEP Mathematics I
1: Solution by DFranklin
2: Solution by Insparato
3: Solution by Insparato and DFranklin
4: Solution by Speleo
5: Solution by Speleo
6: Solution by Speleo
7: Solution by DFranklin
8: Solution by Speleo
9: Solution by SimonM
10: Solution by ad absurdum
11: Solution by waxwing
12: Solution by SimonM
13: Solution by SimonM
14: Solution by SimonM

STEP Mathematics II
1: Solution by Speleo
2: Solution by DFranklin
3: Solution by Rabite
4: Solution by Speleo
5: Solution by DFranklin
6: Solution by Speleo
7: Solution by Speleo
8: Solution by Speleo
9: Solution by waxwing
10: Solution by brianeverit
11: Solution by waxwing
12: Solution by nota bene
13: Solution by nota bene
14: Solution by brianeverit


STEP Mathematics III
1: Solved In Siklos Booklet, uploaded by khaixiang
2: Solution by Insparato
3: Solution by Speleo
4: Solution by Speleo
5: Solution by Speleo
6: Solved By Khaixiang, waxwing
7: Solution by DFranklin
8: Solution by DFranklin
9: Solution by DFranklin
10: Solution by DFranklin
11: Solution by DFranklin
12: Solution by brianeverit
13:
14: Solution by DFranklin



Explanation:
In this thread, we will try to collectively provide solutions for the questions from the STEP Maths I, II and III papers from 1996, seeing as no solutions are available on the net.

If you want to contribute, simply find a question which is marked with "unsolved" above (you might want to check for new solutions after the time the list was last updated as well), solve it, and post your solution in the thread. I will try to update the list as often as possible. If you have a solution to a question that uses a different idea from one already posted, submit it as well, as it's often useful to see different approaches to the same problem.

If you find an error in one of the solutions, simply say so, and hope that whoever wrote the solution in the first place corrects it.

Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
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insparato
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STEP I Question 2

i)
\displaystyle S = \int \frac{cosx}{cosx+sinx}

\displaystyle T = \int \frac{sinx}{cosx+sinx}

\displaystyle S+T = \int \frac{cosx+sinx}{cosx+sinx} = \int dx

 = x

 S-T = \frac{cosx-sinx}{cosx+sinx} = ln (cosx+sinx)

 S+T + S-T = 2S

 2S = x + ln(cosx+sinx)

 s = \frac{1}{2}x + \frac{1}{2}ln(cosx+sinx)

 S+T - (S-T) = 2T

 2T = x - ln(cosx+sinx)

 T = \frac{1}{2} - \frac{1}{2}ln(cosx+sinx)

ii)

 \displaystyle \int_{\frac{1}{4}}^{\frac{1}{2}} (1-4x)\sqrt{(\frac{1}{x}-1)}

 x = sin^2t

 \frac{dx}{dt} = 2sintcost

 dx = 2sintcost dt

 sin^2t = \frac{1}{2}

 t = \frac{\pi}{4}

 sin^2t = \frac{1}{4}

 t = \frac{\pi}{6}

 \displaystyle \int (1-4sin^2t)\sqrt{\frac{1}{sin^2t}-1}.2sintcost

 \displaystyle \int (1-4sin^2t)\sqrt{cot^2t}.2sintcost

 \displaystyle \int (1-4sin^2t)\frac{cost}{sint}.2sintc  ost

 \displaystyle \int (1-4sin^2t)2cos^2t

 \displaystyle \int (1-2(1-cos2x))(1+cos2x)

 \displaystyle \int (2cos2x-1)(1+cos2x)

 \displaystyle \int cos2x -1 + 2cos^22x

 \displaystyle \int cos2x + cos4x

 [\frac{1}{2}sin2x + \frac{1}{4}sin4x]

In the limits pi/4 and pi/6

 [\frac{1}{2}+0] - [\frac{\sqrt3}{4}+\frac{\sqrt3}{8  }]

 \frac{1}{2} - \frac{3\sqrt3}{8}
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Speleo
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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
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insparato
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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 ?
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Speleo
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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
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Speleo
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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]
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Speleo
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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
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khaixiang
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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.
Attached files
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Speleo
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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
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Speleo
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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)]
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Speleo
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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
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Speleo
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7 questions is enough for now , 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).
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insparato
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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 .
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DFranklin
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Step I, Q3, last bit:

\displaystyle 1^3-2^3+3^3-4^3+...-(2n+1)^3 \\

= 1^3+2^3+...+(2n+1)^3 - 2\big\{(2^3)+(4^3)+(6^3)+...+(2n  )^3\big\} \\ 

= \sum_1^{2n+1} r^3 - 16 \sum_1^n r^3 \\

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

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

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

Alternatively:

\displaystyle 1^3-2^3+3^3-4^3+...-(2n+1)^3 \\

= 1 + (3^3-2^3)+(5^3-4^3)+...+((2n+1)^3-(2n)^3) \\

= 1 + \sum_1^n (2r+1)^3-(2r)^3 \\

= 1 + \sum_1^n 12r^2+6r + 1 \\

= 1 + 2n(n+1)(2n+1) + 3n(n+1)+n \\

= (n+1) (2n(2n+1)+3n+1) = (n+1)(4n^2+5n+1)=(n+1)^2(4n+1).
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insparato
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Grrr i should revise finite series . That question was definitely not beyond me.
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Speleo
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STEP III Question 3

(sorry, it was calling to me )

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
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DFranklin
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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).
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Speleo
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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
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Speleo
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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
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#20
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#20
(Original post by Speleo)
STEP II Question 6

What the hell is wrong with me?
You have a condition called STEPitis. Its quite contagious no cure im afraid.
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