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Help with a whole bunch of questions
I've received sheets and sheets of maths questions from Oxford, and although I've managed to do a lot of them, many I can't do, since I hate induction and never did P5 (but I did P6).
If you can help me with any of these I'll be very appreciative.
INDUCTION EXERCISES
2) Prove that SUM 1/r^2 (from r=1 to r=n) <= 2 - 1/n
I tried changing the boundaries of the sum thingy to r=n+1 and hence adding 1/(n+1)^2 to the right hand side of the equation, but I can't seem to get it to work.
3) Prove root(n) <= SUM 1/root(k) [from] <= 2root(n) - 1
ALGEBRA EXERCISES
3i) Find a 2x2 matrix A such that A^5 = I but A^[1 to 4] doesn't = I.
I think this has something to do with a rotation matrix, any ideas?
3iv) Find 2x2 matrixes A and B such that AB is invertible and BA is singular.
I'm pretty sure no such matrixes exist, but I need to prove it.
5c) A is a 2x2 matrix again. Suppose A^n = 0 for some n>=2. Prove detA = 0.
Deduce from this that A^2 = 0.
CALCULUS EXERCISES 3 - TECHNIQUES OF INTEGRATION
(I'll use '|' as the integral sign here. Boundaries indicated in square brackets)
This is where I really struggled, knowing very little about hyperbolic substitutions.
1b) | x sec^2(x) dx
d) |[1 0] tan^-1(x) dx
e) |[1 0] dx/(e^x + 1)
2b) |[2 1] dx/root(x^2 - 1)
I get arcsin2 - arcsin1 for this. Can this be simplified to 2i.arcsinh2 - arcsin1 or does this not help?
d) |[infinity 2] dx/(x^2 - 1)^(3/2)
4) Let t=tan(theta/2). Show that
sin(theta) = 2t/(1 + t^2)
cos(theta) = (1-t^2) / (1 + t^2)
tan(theta) = 2t/(1 + t^2)
and that dtheta = 2dt / (1+t^2)
I can see just from the double angle formula that tan(theta) is true, but surely sin(theta) / cos(theta) = tan(theta), so how can the two others be true?
I used the identity sec^2(theta) = 1 + tan^2(theta) to find cos(theta), and ended up with 1+t^2 / 1-t^2, i.e. the reciprocal of what it says I should have.
COMPLEX NUMBERS
2) By substituting z=x+iy or x=re^itheta into the following equation, sketch the following regions on the complex plane on separate Argand diagrams.
d) e^z = 1
Does z just equal 0? Hence I'm guessing that y=-x?
Just thought I'd check with this one.
4a) Given that e^itheta = costheta + isintheta, prove that
cos(alpha + beta) = cos(alpha)cos(beta) - sin(alpha)sin(beta)
If you can help me with any of these I'll be very appreciative.
INDUCTION EXERCISES
2) Prove that SUM 1/r^2 (from r=1 to r=n) <= 2 - 1/n
I tried changing the boundaries of the sum thingy to r=n+1 and hence adding 1/(n+1)^2 to the right hand side of the equation, but I can't seem to get it to work.
3) Prove root(n) <= SUM 1/root(k) [from] <= 2root(n) - 1
ALGEBRA EXERCISES
3i) Find a 2x2 matrix A such that A^5 = I but A^[1 to 4] doesn't = I.
I think this has something to do with a rotation matrix, any ideas?
3iv) Find 2x2 matrixes A and B such that AB is invertible and BA is singular.
I'm pretty sure no such matrixes exist, but I need to prove it.
5c) A is a 2x2 matrix again. Suppose A^n = 0 for some n>=2. Prove detA = 0.
Deduce from this that A^2 = 0.
CALCULUS EXERCISES 3 - TECHNIQUES OF INTEGRATION
(I'll use '|' as the integral sign here. Boundaries indicated in square brackets)
This is where I really struggled, knowing very little about hyperbolic substitutions.
1b) | x sec^2(x) dx
d) |[1 0] tan^-1(x) dx
e) |[1 0] dx/(e^x + 1)
2b) |[2 1] dx/root(x^2 - 1)
I get arcsin2 - arcsin1 for this. Can this be simplified to 2i.arcsinh2 - arcsin1 or does this not help?
d) |[infinity 2] dx/(x^2 - 1)^(3/2)
4) Let t=tan(theta/2). Show that
sin(theta) = 2t/(1 + t^2)
cos(theta) = (1-t^2) / (1 + t^2)
tan(theta) = 2t/(1 + t^2)
and that dtheta = 2dt / (1+t^2)
I can see just from the double angle formula that tan(theta) is true, but surely sin(theta) / cos(theta) = tan(theta), so how can the two others be true?
I used the identity sec^2(theta) = 1 + tan^2(theta) to find cos(theta), and ended up with 1+t^2 / 1-t^2, i.e. the reciprocal of what it says I should have.
COMPLEX NUMBERS
2) By substituting z=x+iy or x=re^itheta into the following equation, sketch the following regions on the complex plane on separate Argand diagrams.
d) e^z = 1
Does z just equal 0? Hence I'm guessing that y=-x?
Just thought I'd check with this one.
4a) Given that e^itheta = costheta + isintheta, prove that
cos(alpha + beta) = cos(alpha)cos(beta) - sin(alpha)sin(beta)
Scroll to see replies
(2) Without induction ...
(sum over r from 1 to n) 1/r^2
= 1 + (sum over r from 2 to n) 1/r^2
<= 1 + (sum over r from 2 to n) 1/(r(r - 1))
= 1 + (sum over r from 2 to n) [1/(r - 1) - 1/r]
= 1 + 1 - 1/n
= 2 - 1/n.
With induction ...
OK for n = 1. Suppose k >= 1 and the result holds when n = k. Then
(sum over r from 1 to k + 1) 1/r^2
= [(sum over r from 1 to k) 1/r^2] + 1/(k + 1)^2
<= 2 - 1/k + 1/(k + 1)^2 . . . by the inductive hypothesis
= 2 - [1/(k + 1)] * [(k + 1)/k - 1/(k + 1)]
= 2 - [1/(k + 1)] * [1 + 1/k - 1/(k + 1)]
<= 2 - 1/(k + 1) . . . since [1 + 1/k - 1/(k + 1)] >= 1.
So the result holds when n = k + 1. QED.
(sum over r from 1 to n) 1/r^2
= 1 + (sum over r from 2 to n) 1/r^2
<= 1 + (sum over r from 2 to n) 1/(r(r - 1))
= 1 + (sum over r from 2 to n) [1/(r - 1) - 1/r]
= 1 + 1 - 1/n
= 2 - 1/n.
With induction ...
OK for n = 1. Suppose k >= 1 and the result holds when n = k. Then
(sum over r from 1 to k + 1) 1/r^2
= [(sum over r from 1 to k) 1/r^2] + 1/(k + 1)^2
<= 2 - 1/k + 1/(k + 1)^2 . . . by the inductive hypothesis
= 2 - [1/(k + 1)] * [(k + 1)/k - 1/(k + 1)]
= 2 - [1/(k + 1)] * [1 + 1/k - 1/(k + 1)]
<= 2 - 1/(k + 1) . . . since [1 + 1/k - 1/(k + 1)] >= 1.
So the result holds when n = k + 1. QED.
Ok for the induction quesyions, 2) add 1/(n+1)² to get
S(n+1) < 2 - 1/n + 1/(n+1)²
= 2 - (1/(n+1))[(n+1)² - n)/(n(n+1)] by factorising
= 2 - (1/(n+1))[(n² + n + 1)/(n²+n)]
Now as n is postive clearly n² + n + 1 > n²+n so [(n² + n + 1)/(n²+n)] > 1
Therefore the RHS < 2 - 1/(n+1) and thus so is S(n+1)
For 3) note that 2sqrt(n) > sqrt(n+1)
S(n+1) < 2 - 1/n + 1/(n+1)²
= 2 - (1/(n+1))[(n+1)² - n)/(n(n+1)] by factorising
= 2 - (1/(n+1))[(n² + n + 1)/(n²+n)]
Now as n is postive clearly n² + n + 1 > n²+n so [(n² + n + 1)/(n²+n)] > 1
Therefore the RHS < 2 - 1/(n+1) and thus so is S(n+1)
For 3) note that 2sqrt(n) > sqrt(n+1)
(3) The lower bound is easy. For the upper bound, we have to prove that
2 sqrt(n) + 1/sqrt(n + 1) <= 2 sqrt(n + 1)
or equivalently, multiplying through by sqrt(n + 1),
2 sqrt(n(n + 1)) + 1 <= 2(n + 1)
ie, sqrt(n(n + 1)) <= n + 1/2
ie, n(n + 1) <= (n + 1/2)^2
ie, n^2 + n <= n^2 + n + 1/4.
The last statement is true. So the first statement is also true.
It might be better to write the proof down backwards!
2 sqrt(n) + 1/sqrt(n + 1) <= 2 sqrt(n + 1)
or equivalently, multiplying through by sqrt(n + 1),
2 sqrt(n(n + 1)) + 1 <= 2(n + 1)
ie, sqrt(n(n + 1)) <= n + 1/2
ie, n(n + 1) <= (n + 1/2)^2
ie, n^2 + n <= n^2 + n + 1/4.
The last statement is true. So the first statement is also true.
It might be better to write the proof down backwards!
Jamie Frost
ALGEBRA EXERCISES
3i) Find a 2x2 matrix A such that A^5 = I but A^[1 to 4] doesn't = I.
I think this has something to do with a rotation matrix, any ideas?
3iv) Find 2x2 matrixes A and B such that AB is invertible and BA is singular.
I'm pretty sure no such matrixes exist, but I need to prove it.
ALGEBRA EXERCISES
3i) Find a 2x2 matrix A such that A^5 = I but A^[1 to 4] doesn't = I.
I think this has something to do with a rotation matrix, any ideas?
3iv) Find 2x2 matrixes A and B such that AB is invertible and BA is singular.
I'm pretty sure no such matrixes exist, but I need to prove it.
(3i) Rotate clockwise by 2pi/5:
A =
cos(2pi/5) -sin(2pi/5)
sin(2pi/5) cos(2pi/5)
(3iv)
det(AB) = det(A)det(B)
det(BA) = det(B)det(A)
So det(AB) = det(BA). So the determinants are either both zero (in which case AB and BA are singular) or both nonzero (in which case AB and BA are invertible).
Jamie Frost
Does z just equal 0? Hence I'm guessing that y=-x?
Just thought I'd check with this one.
4a) Given that e^itheta = costheta + isintheta, prove that
cos(alpha + beta) = cos(alpha)cos(beta) - sin(alpha)sin(beta)
Does z just equal 0? Hence I'm guessing that y=-x?
Just thought I'd check with this one.
4a) Given that e^itheta = costheta + isintheta, prove that
cos(alpha + beta) = cos(alpha)cos(beta) - sin(alpha)sin(beta)
e^iθ = cosθ + i*sinθ
e^i(a+b) = (e^ia)(e^ib)
Using the given equation
e^i(a+b) = cos(a+b) + i*sin(a+b)
= (cos a + i*sin a)(cos b + i*sin b)
= (cos a * cos b - sin a * sin b) + i(sin a * cos b) + cos a * sin b)
equate the real parts,
cos(a+b) = (cos a * cos b - sin a * sin b)
Jamie Frost
COMPLEX NUMBERS
2) By substituting z=x+iy or x=re^itheta into the following equation, sketch the following regions on the complex plane on separate Argand diagrams.
d) e^z = 1
Does z just equal 0? Hence I'm guessing that y=-x?
Just thought I'd check with this one.
2) By substituting z=x+iy or x=re^itheta into the following equation, sketch the following regions on the complex plane on separate Argand diagrams.
d) e^z = 1
Does z just equal 0? Hence I'm guessing that y=-x?
Just thought I'd check with this one.
I don't get this one... how can you have the complex number itself as a power of e?
mik1a, I don't know what's happened to you. Once upon a time, you were a nice innocent child who had only done P1 and had just completed Decision 1, now you're doing this Further Maths stuff! What's happened to you! 
mik1a
I don't get this one... how can you have the complex number itself as a power of e?
The Euler formula tells you how to evaluate the exponential of a complex number.
e^ix = cos(x) + i*sin(x)
So if you want 1 then you want sin(x) to be zero and cos(x) to be 1, so x is (2n)*pi where n is an integer. So the answer is 0, or 2pi or 4pi, or -2pi etc etc.
mik1a
god knows....
edit
maybe a new sort of disease.. I call it mathitis
edit
maybe a new sort of disease.. I call it mathitis
You've done P2 and P3 already, and you're doing P4 now? P4 is complex numbers...
Invisible
You've done P2 and P3 already, and you're doing P4 now? P4 is complex numbers...
Yeah, although I think complex numbers in exponential form isn't in my p4 book (still useful though
If z is real, θ is either pi or 0, and r = a (z = a + ib)
re^iθ = r(cos θ + i.sinθ ) = r(-1 + 0) = -r
re^i.pi = -r
so
e^(i*pi) = -1
see its so elegant
I can't stop learning lol
AntiMagicMan
The Euler formula tells you how to evaluate the exponential of a complex number.
e^ix = cos(x) + i*sin(x)
So if you want 1 then you want sin(x) to be zero and cos(x) to be 1, so x is (2n)*pi where n is an integer. So the answer is 0, or 2pi or 4pi, or -2pi etc etc.
e^ix = cos(x) + i*sin(x)
So if you want 1 then you want sin(x) to be zero and cos(x) to be 1, so x is (2n)*pi where n is an integer. So the answer is 0, or 2pi or 4pi, or -2pi etc etc.
I thought z meant the complex number, so e^z ? I thought z = re^iθ, so why would this go as the power of e again? or is z meant to represent the angle?
4) use trig to get tan(A) in terms of t then draw a right triangle of height 2t and base 1 - t^2 and angle A then use pythagoros to find the hypotenuse. use simple trig to find cos(A) and sin(A)
LOL, I actually wrote in my notes before that cos(theta)=1 and sin(theta)=0, but it didn't occur to me at the time that theta didnt just have to be 0.
Hmmm, I swear I tried that rotation matrix before you suggested but it didn't work - I'll try it again.
Thanks for your help so far!
Hmmm, I swear I tried that rotation matrix before you suggested but it didn't work - I'll try it again.
Thanks for your help so far!
Jamie Frost
CALCULUS EXERCISES 3 - TECHNIQUES OF INTEGRATION
(I'll use '|' as the integral sign here. Boundaries indicated in square brackets)
This is where I really struggled, knowing very little about hyperbolic substitutions.
1b) | x sec^2(x) dx
d) |[1 0] tan^-1(x) dx
e) |[1 0] dx/(e^x + 1)
2b) |[2 1] dx/root(x^2 - 1)
I get arcsin2 - arcsin1 for this. Can this be simplified to 2i.arcsinh2 - arcsin1 or does this not help?
d) |[infinity 2] dx/(x^2 - 1)^(3/2)
CALCULUS EXERCISES 3 - TECHNIQUES OF INTEGRATION
(I'll use '|' as the integral sign here. Boundaries indicated in square brackets)
This is where I really struggled, knowing very little about hyperbolic substitutions.
1b) | x sec^2(x) dx
d) |[1 0] tan^-1(x) dx
e) |[1 0] dx/(e^x + 1)
2b) |[2 1] dx/root(x^2 - 1)
I get arcsin2 - arcsin1 for this. Can this be simplified to 2i.arcsinh2 - arcsin1 or does this not help?
d) |[infinity 2] dx/(x^2 - 1)^(3/2)
1b)
∫ xsec²x dx
xtanx - ∫tanx dx
xtanx + ln(cosx)
============
1d)
∫[1 0] tan(-1)x dx
let u = tan(-1)x
x = tanu
dx = sec²u du
at x = 1, u = π/4
at x = 0, u = 0
∫[π/4 0] u.sec²u du
using the result from 1b) this becomes,
[utanu + ln(cosu)][π/4 0]
{(π/4*1 + ln(1/√2)) - (0 + ln(1))}
π/4 - ½ln(2)
=========
1e)
∫[1 0] 1/(e^x + 1) dx
∫[1 0] (e^x + 1)/(e^x + 1) - (e^x)/(e^x + 1) dx
∫[1 0] 1 - (e^x)/(e^x + 1) dx
[x - ln(e^x + 1)] [1 0]
{(1 - ln(e + 1)) - (0 - ln(1+1))}
1 + ln(2/(e+1))
===========
2b)
∫[2 1] dx/root(x^2 - 1)
let x = cosht
dx = sinht dt
root(x^2 - 1)=√(cosh²t - 1) = √(sinh²t) = sinht.
at x = 2, t = cosh(-1)(2)
at x = 1, t = cosh(-1)(1)
∫sinht/sinht dt = ∫ 1 dt = [t][cosh(-1)(2) cosh(-1)(1)]
cosh(-1)(2) - cosh(-1)(1)
===================
d)
∫[∞ 2] dx/(x^2 - 1)^(3/2)
let x = cosht
dx = sinht dt
root(x^2 - 1)=√(cosh²t - 1) = √(sinh²t) = sinht.
at x = ∞, t = cosh(-1)(∞
= ∞at x = 2, t = cosh(-1)(2)
∫sinht/sinh³t dt = ∫ 1/sinh²t dt = [cotht] [∞ 2]
1 - coth(2)
=========
5(c) A^n = AAAAA ... n times. Since det (AB) = det(A)det(B), if A^n = 0, then det(A^n) = 0, so det(A)det(A)...det(A) n times = 0, so det(A) = 0.
It follows that if det(A) = 0, then if the matrix is written a,b,c,d, that ad = bc.
I think it follows from here, can't be bothered writing out the latex for matrix algebra.
It follows that if det(A) = 0, then if the matrix is written a,b,c,d, that ad = bc.
I think it follows from here, can't be bothered writing out the latex for matrix algebra.
Fermat
1b)
∫ xsec²x dx
xtanx - ∫tanx dx
xtanx + ln(cosx)
============
1d)
∫[1 0] tan(-1)x dx
let u = tan(-1)x
x = tanu
dx = sec²u du
at x = 1, u = π/4
at x = 0, u = 0
∫[π/4 0] u.sec²u du
using the result from 1b) this becomes,
[utanu + ln(cosu)][π/4 0]
{(π/4*1 + ln(1/√2)) - (0 + ln(1))}
π/4 - ½ln(2)
=========
1e)
∫[1 0] 1/(e^x + 1) dx
∫[1 0] (e^x + 1)/(e^x + 1) - (e^x)/(e^x + 1) dx
∫[1 0] 1 - (e^x)/(e^x + 1) dx
[x - ln(e^x + 1)] [1 0]
{(1 - ln(e + 1)) - (0 - ln(1+1))}
1 + ln(2/(e+1))
===========
2b)
∫[2 1] dx/root(x^2 - 1)
let x = cosht
dx = sinht dt
root(x^2 - 1)=√(cosh²t - 1) = √(sinh²t) = sinht.
at x = 2, t = cosh(-1)(2)
at x = 1, t = cosh(-1)(1)
∫sinht/sinht dt = ∫ 1 dt = [t][cosh(-1)(2) cosh(-1)(1)]
cosh(-1)(2) - cosh(-1)(1)
===================
d)
∫[∞ 2] dx/(x^2 - 1)^(3/2)
let x = cosht
dx = sinht dt
root(x^2 - 1)=√(cosh²t - 1) = √(sinh²t) = sinht.
at x = ∞, t = cosh(-1)(∞ = ∞
at x = 2, t = cosh(-1)(2)
∫sinht/sinh³t dt = ∫ 1/sinh²t dt = [cotht] [∞ 2]
1 - coth(2)
=========
∫ xsec²x dx
xtanx - ∫tanx dx
xtanx + ln(cosx)
============
1d)
∫[1 0] tan(-1)x dx
let u = tan(-1)x
x = tanu
dx = sec²u du
at x = 1, u = π/4
at x = 0, u = 0
∫[π/4 0] u.sec²u du
using the result from 1b) this becomes,
[utanu + ln(cosu)][π/4 0]
{(π/4*1 + ln(1/√2)) - (0 + ln(1))}
π/4 - ½ln(2)
=========
1e)
∫[1 0] 1/(e^x + 1) dx
∫[1 0] (e^x + 1)/(e^x + 1) - (e^x)/(e^x + 1) dx
∫[1 0] 1 - (e^x)/(e^x + 1) dx
[x - ln(e^x + 1)] [1 0]
{(1 - ln(e + 1)) - (0 - ln(1+1))}
1 + ln(2/(e+1))
===========
2b)
∫[2 1] dx/root(x^2 - 1)
let x = cosht
dx = sinht dt
root(x^2 - 1)=√(cosh²t - 1) = √(sinh²t) = sinht.
at x = 2, t = cosh(-1)(2)
at x = 1, t = cosh(-1)(1)
∫sinht/sinht dt = ∫ 1 dt = [t][cosh(-1)(2) cosh(-1)(1)]
cosh(-1)(2) - cosh(-1)(1)
===================
d)
∫[∞ 2] dx/(x^2 - 1)^(3/2)
let x = cosht
dx = sinht dt
root(x^2 - 1)=√(cosh²t - 1) = √(sinh²t) = sinht.
at x = ∞, t = cosh(-1)(∞
= ∞at x = 2, t = cosh(-1)(2)
∫sinht/sinh³t dt = ∫ 1/sinh²t dt = [cotht] [∞ 2]
1 - coth(2)
=========
Just few errors and simplifications of ur answers:
1b)
∫ xsec²x dx
xtanx - ∫tanx dx
xtanx - ln(secx)
==========================================================
1d) is fine
==========================================================
1e) I couldn't do but seems fine LOL
==========================================================
2b) Has a better simplified answer as ln(2+√3). This is obtained using the fact that the integral integrates to arcoshx (standard result) and using
arcoshx = ln [x+√(x²-1)]
which gives:
ln [2+√(2²-1)] - ln [1+√(1²-1)]
ln [2+√3] - ln[1]
ln(2+√3)
==========================================================
2d) Has a better simplified answer as: (2 - √3)/√3. Unfortunately your answer isn't correct, because you used x=2 as t=2, whereas it should have been t=arcosh(2). The error occurs here:
at x = ∞, t = arcosh ∞ = ∞
at x = 2, t = arcosh(2)
∫sinht/sinh³t dt = ∫ 1/sinh²t dt = [-cotht] [∞ arcosh2]
[-coth ∞] - [-coth(arcosh(2))]
-1 + (2/√3)
(2 - √3)/√3
==========================================================
By the way did you notice we got the exact score for the maths competition, arch rivals eh?! LOL
integral_neo
Just few errors and simplifications of ur answers:
1b)
∫ xsec²x dx
xtanx - ∫tanx dx
xtanx + ln(secx)
================================================== ========
1b)
∫ xsec²x dx
xtanx - ∫tanx dx
xtanx + ln(secx)
================================================== ========
I beg to differ
y = ln(cosx) - my solution
let u = cosx
du/dx = -sinx
y = lnu
dy/dx = dy/du.du/dx
dy/dx = (1/u).(-sinx)
dy/dx = -sinx/cosx = -tanx
===================
y = ln(secx) - your solution
let u = secx
du/dx = secxtanx
y = lnu
dy/dx = dy/du.du/dx
dy/dx = (1/u).(secxtanx)
dy/dx = (secxtanx)/secx = +tanx (wrong sign - simple typo
===================
Hmmm, apart from the sign bit thingy, I'm not sure why it it would be simpler ?? Or is that just for the others?
integral_neo
2b) Has a better simplified answer as ln(2+√3).
Yes. Much neater.
integral_neo
This is obtained using the fact that the integral integrates to arcoshx (standard result)
I havent got that standard. I only use this one sometimes. And it's a bit sparse.
Is there another/better site on-line I could use?
integral_neo
2d) Has a better simplified answer as: (2 - √3)/√3. Unfortunately ...
Ah yes, silly error
integral_neo
By the way did you notice we got the exact score for the maths competition, arch rivals eh?! LOL
Well, yes. But i got all the hard ones right!
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