MA 212 Further Mathematical Method (Calculus) - London School of Economics. Solutions Section A and

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MA212 (2017) - Solutions Section A Question 1 (a) The Laplace transform is de…ned by Lff (t)g = get

We have Lff 0 (t)g = sLff (t)g 0

Lff (t)g =

Z 1

e

st 0 st

f

f (t)]1 0

Z 1

e st f (t)dt.

t=0

f (0). Indeed, integrating by parts, we Z 1

( s)e st f (t)dt = sLff (t)g

f (0):

t=0

t=0

We used that f is of exponential growth at most to deduce that e st f (t) ! 0 as t ! 1.

Z t

h(u) du, we have g(0) = 0 and g 0 (t) = h(t) (by the

For g(t) =

u=0

Fundamental Theorem of Calculus). Hence ~ h(s) = Lfg 0 (t)g = sLfg(t)g

g(0) = sLfg(t)g

~ and g~(s) = 1s h(s) follows. (b) Set H = H(s) = Lfh(t)g. By applying the Laplace Transform to both sides of the equation and using (a), we obtain that sH

1 h(0) + 4 H = Lf1g: s

We use the fact that Lf1g = 1s and h(0) = 1 to obtain: sH Hence, 4 1 H s+ = +1 s s and, consequently, H=

s s+1 1 2 = + 2 : 2 2 2 s +4 2s +2 s + 22

We deduce that h(t) = 12 sin(2t) + cos(2t).

1

1 1 1+4 H = : s s

(c) We have problems at x = 1 and at 1. Hence we split the integral into two and dealZwith each separately: 2 1 x 1 We know that dx converges for p < 1. Also, lim 3 = = p p x!1 (x 1) x) (x 1)p 1 (x x = 1=2p . Hence, lim 2 x!1 (x + x)p Z 2 Z 2 1 x x dx = dx 3 p p 2 x) 1) (x + x)p 1 (x 1 (x converges i¤ p < 1 by the limit comparison test.

Similarly, Z 1 Z 1 Z 1 1 1 x 1 x dx = dx = dx: 3 p 3p 2 p 3p 1 (x x) x (1 + x ) x (1 + x 2 )p 2 2 2 1 1 = 3p 1 = lim = 1. By the limit comparison x!1 x (1 + x 2 )p test, this integral converges i¤ 3p 1 > 1.

Consequently, the original integral converges i¤ 2=3 < p < 1.

Also, lim

x!1 (x3

x

x)p

2

Question 2 (a)(i) x sin x 2 + cos 2x2 + cos x sin x 2+0+0 1 x2 x2 = lim = = : 2 2 2 x!1 x!1 (2x + 1) (2 + 1=x) (2 + 0) 2

lim

Here we used the fact that sin x and cos x are bounded. (ii) The next limit is 0=0-type and we use L’Hospital Rule: sin(1=x)

1

ln(cos(1=x)) sin(1=x) cos(1=x) x2 lim x ln(cos(1=x)) = lim = lim = lim = 0: x!1 x!1 x!1 x!1 cos(1=x) 1=x 1=x2 (iii) The next limit is 0=0-type and we use L’Hospital Rule: 2

2

2

ex 2x ex 2 ex 1 = lim = lim sin x lim = 1: x!0 sin x + x cos x x!0 x!0 x sin x + cos x x u

(b)(i) we use the substitution t = e …rst:

Z 2 1

1 dt = ln t

Z ln 2 0

1 u e du. Now, u

1 u e lim u 1 = lim eu = 1;

u!0

u

u!0

R ln 2 the integral 0 u1 du diverges, hence the original integral also diverges by LCT. (ii) we have a problem at 0 and at 1, so we divide into two integrals: Z 1 R 1 dt sin3 t sin3 t 7=2 dt converges absolutely because 1=t and 7=2 1 t7=2 t t7=2 1 converges.

Z 1 sin3 t dt also converges. We use that limt!0 sin t=t = 1 and obtain t7=2 0 that sin3 t sin3 t 7=2 lim t 1 = lim 3 = 1: t!0 p t!0 t t Z 1 Z 1 1 sin3 t Since dt converges, by LCT, dt converges as well.

1=2 t7=2 0 t 0 3

Question 3 (a)(i) We have Z 1r 0

=

Z

x 1

x

dx =

=2

0

2

2 sin t dt =

Z

Z

=2

0

s

sin2 t 2 sin t cos t dt = 1 sin2 t

=2

(1

cos(2t)) dt = t

0

Z

=2

0

sin(2t) 2 0

sin t 2 sin t cos t dt cos t

=2

=

2

:

The region of integration D is bounded by lines y = 0, y = 1 and x = y, and the curve x = y 2 . (Picture to follow.)

p From this we deduce that x ranges between 0 and 1 and y between x and x.

By changing the order of integration, we can write Z 1 Z px y y p p dx dy = dy dx x) x y 2 x) x y 2 x=0 y=x (1 y=0 x=y 2 (1 #px p p Z 1 Z 1 " p 2 x y x x x x2 = dx = + dx = (1 x) 1 x 1 x x=0 x=0 x Z 1p Z 1r x(1 x) x p dx = dx = : (1 x) 2 (1 x)2 0 0

Z 1 Z y

(b) The region D of integration is the sector of the unit circle x2 + y 2 = 1 bounded by lines y = 0, y = x passing through the origin. (Picture to follow.)

4

We have u = x2 + y 2 and v = x2 y 2 from which we deduce 0 u 1.

Since y x, we have v 0. Finally, we have x2 = (u + v)=2 and y 2 = (u v)=2 0, so we must have u v.

Hence, this transformation changes D into triangle = f(u; v) j 0 u 1; 0 v ug: p p We also get that x = (u + v)=2 and y = (u p v)=2. Notice that p the transp 0 0 formation is injective on : if (u + v)=2 = (u + v )=2 and (u v)=2 = p 0 0 (u v )=2, we get u + v = u0 + v 0 and u v = u0 v 0 from which u = u0 and v = v 0 easily follow.

To get the Jacobian for this transformation, we calculate @u = 2y; @y @v = 2y; @y

@u = 2x; @x @v = 2x; @x and the Jacobian @(u; v) = 2x ( 2y) @(x; y)

2x 2y =

8xy:

Hence, 1 @(x; y) 1 = @(u;v) = : @(u; v) 8xy @(x;y)

Notice that Jacobian does not change its sign on , so we obtain Z Z Z Z Z Z x2 y 2 v xye dxdy = xye dxdyj@(x; y)=@(u; v)j du dv = xyev D Z Z Z Z u Z 1 v 1 1 1 1 u v = e du dv = e dv du = (e 8 8 u=0 v=0 8 u=0 e 2 1 = [eu u]10 = : 8 8

5

1 du dv 8xy 1) du

Question 4 (a)(i) Here (x) = ( 1)[x] has jumps as follows: the jump in at x = 1 is 1 1 = 2, at x = 2 is 1 ( 1) = +2, at x = 3 is 1 1 = 2, and at x = 4 is 1 ( 1) = +2. Hence, Z 4 x2 d(( 1)[x] ) = ( 2) 12 + 2 22 + ( 2) 32 + 2 42 = 2 + 8 18 + 32 = 20: 0

p (ii) We use substitution y = x and integration by parts (twice): Z 4 Z 2 Z 2 Z 2 p 2 2 2 4 2 2 xd([ x]x) = y d([y]y ) = y [y] 0 y [y]d(y ) = 32 2 y 3 [y]dy = 0 0 0 0 Z 2 Z 2 1 1 2 y 4 d[y] = 16+ (14 1+24 1) = 49=2: [y]d(y 4 =2) = 32 [y]y 4 =2 0 + 32 2 2 0 0 This is also approachable by other means. (b)We apply Leibnitz Rule: !

Z x2 Z x2 cos(xt) @ cos(xt) cos(x3 ) 2 cos(x3 ) dt = (2x)+ dt = + t x2 t x t=0 t=0 @x t=0 2 Z 2 x x 2 cos(x3 ) cos(xt) 2 cos(x3 ) 3 cos(x3 ) 1 sin(xt) dt = : = x x x x t=0 t=0 d dx

Z x2

@ cos(xt) Our manipulation is justi…ed because = sin(xt) is jointly con@x t tinuous everywhere. (c)(i) A function K(t; x) is dominatedly convergent for t a and c x d if there exists a function (t), de…ned for every t a, such that jK(t; x)j (t) for every t Z 1 (t)dt converges.

a and c

x

d;

t=a

(ii) For t

Z 1

1 and x 2 [0; 1], we have jK(t; x)j =

j sin(tx)j t2 + x2

1

t2 + x2

1 =: (t): t2

12 dt converges, (t) satis…es two conditions above and K(t; x) is t 1 dominatedly convergent.

Since

6

sin(xt) t dt = t

Section B Question 5 (a) The square matrices A and B are similar if they are of the same size and B = P 1 AP for some non-singular matrix P: Suppose this holds.

Since det P 1 det P = det P 1 P = det I = 1 pB (x) = det(xI B) = det P 1 (xI A)P = det P 1 det(xI A) det P = det(xI

A) = pA (x):

Since the characteristic polynomial is uniquely characterized (up to a multiplicative non-zero constant) by its roots, i.e. the respective eigenvalues, A and B have identical sets of eigenvalues. (b) Expanding by the bottom row 2 A) = det 4

det(xI

= xf(x

From

we obtain

2

1 4 1 0

x

1 1 0

1 x+1 0

3 0 1 5 x

1)(x + 1) + 1g = x3 :

32 3 2 3 1 0 x 0 5 4 5 4 1 1 y = 0 5 0 0 z 0

x

x y = 0; y + z = 0:

So z = 0; and E(A; 0) = N (A); the null space of A; is generated by 2 3 1 x4 1 5: 0 7

On the other hand 2

1 A2 = 4 1 0

32 1 0 1 1 1 54 1 0 0 0

3 2 1 0 0 0 1 1 5=4 0 0 0 0 0 0

3 1 1 5: 0

We take v3 = e3 which witnesses A2 v3 6= 0 . Then we form the cascade: 2 3 0 v2 = Av3 = Ae3 = 4 1 5 = e2 0 2 3 1 v1 = Av2 = A2 v3 = A2 v3 = 4 1 5 = (e1 + e2 ) 0 Av1 = A3 v3 = 0:

So for

2

3 1 0 0 P = [v1 ; v2 ; v3 ] = [ (e1 + e2 ); e2 ; e3 ] = 4 1 1 0 5 0 0 1 2 3 0 1 0 1 4 P AP = J = 0 0 1 5 0 0 0

Note for later use that (expanding by the bottom row) det P = so

2 3 1 1 1 0 0 4 1 1 0 5 = 1 4 1 1 0 0 0 1 2

(d) Solving

2

3 2 3 0 1 0 0 0 5 = 4 1 1 0 5: 1 0 0 1

3 0 1 0 y_ = 4 0 0 1 5 y 0 0 0

yields y30 (t) y20 (t) y10 (t) where

0 1 0

1; and

= = = =

0 so that y3 (t) = y3 (0) = ; say y3 (t) = so that y2 (t) = t + ; say, where = y2 (0): y2 (t) = t + so that y1 (t) = t2 =2 + t + ; y1 (0): 8

and so the solution to y_ =Jy is 2 32 1 t t2 =2 t 54 y =etJ y(0)= 4 0 1 0 0 1 So

3

2

32 3 1 t t2 =2 y1 (0) 5=4 0 1 t 5 4 y2 (0) 5 : 0 0 1 y3 (0)

2

3 1 t t2 =2 t 5: etJ = 4 0 1 0 0 1

In particular, if y(0) = ( ; ; )T = ( 1; 0; 1)T 2 32 3 2 1 t t2 =2 1 t 54 0 5 = 4 y(t) =etJ y(0)= 4 0 1 0 0 1 1

3 1 + t2 =2 5: t 1

(e) The equation x_ =Ax may be transformed by taking x = P y to yield, P y_ = x_ = Ax =AP y;

i.e. y_ =P 1 AP y =Jy: Here

2

3 2 32 3 2 3 y1 (0) 1 0 0 1 1 4 y2 (0) 5 = P 1 x(0) = 4 1 1 0 5 4 1 5 = 4 0 5 y3 (0) 0 0 1 1 1

The equation x_ =Ax may be transformed by taking x = P y to yield y_ =P 1 AP y =Jy . So 2 32 32 3 1 0 0 1 t t2 =2 y1 (0) t 5 4 y2 (0) 5 x = Py =4 1 1 0 54 0 1 0 0 1 0 0 1 y3 (0) 2 32 3 2 3 2 1 0 0 1 + t =2 1 t2 =2 5 = 4 1 + t t2 =2 5 : t = 4 1 1 0 54 0 0 1 1 1 Remark. For the record, as P here is an elementary matrix corresponding to ‘multiply …rst row by 10 and subtract row 1 from row 2, to …nd P 1 9

regard P 1 as reversing these row operations. So take 2 3 1 0 0 P 1=4 1 1 0 5 0 0 1 Check:

2

32 3 2 3 1 0 0 1 0 0 1 0 0 P 1P = 4 1 1 0 5 4 1 1 0 5 = 4 0 1 0 5 0 0 1 0 0 1 0 0 1

10

Question 6 (a)

(i) An n

n matrix M is orthogonal if M 1 = M T .

(ii) Let M and N be orthogonal. Then (M N )T M N = N T M T M N = N T N = In ; since M 1 = M T and N 1 = N T . This veri…es that (M N ) 1 = (M N )T , so M N is orthogonal. (iii) We calculate that jjM xjj2 = (M x) (M x) = (M x)T M x = xT M T M x = xT x = jjxjj2 : Thus jjM xjj = jjxjj.

11

(iv) First observe that a real orthogonal matrix M is also unitary, i.e.

M 1 = M . This means that we may use the standard inner product in Cn : Let be an eigenvalue with corresponding eigenvector x, i.e. M x = x, for x 6= 0. Then with reference to part (iii), which may be repeated using complex inner products, jjxjj2 = (M x)T M x = ( x)T x =

jjxjj2 ;

from which it follows that j j = 1.

Remark. Complex eigenvalues arise whenever the orthogonal matrix represents a rotation other than by : The matrix M of part (b) is just such a case: 2 3 x 1 0 pM (x) = det(xI M ) = det 4 0 x 1 5 = x(x2 0) + ( 1) 1 0 x = x3

1 = (x

1)(x2 + x + 1)

This characteristic polynomial pM (x) has just one real eigenvalue = 1: The other two p eigenvalues, being of unit modulus, are of here is the angle of rotation; the form e i = ( 1 i 3)=2: The but we cannot say which sign to select for the rotation until we have picked the vector generating the axis of rotation. See later. (b)

(i) We calculate that 2 0 1 T 4 0 0 M M= 1 0

32 0 0 5 4 1 1 0 0

0 0 1

3 2 3 1 1 0 0 0 5 = 4 0 1 0 5; 0 0 0 1

which shows that M 1 = M T , i.e. that M is orthogonal. The transformation is orientation-preserving since the determinant of M is 1.

– Notice that det(xI

2

x 4 M ) = det 0 1

3 1 0 x 1 5 = x(x2 0 x 12

0) + ( 1) = x3

1:

(ii) We know that the axis of rotation is an eigenvector of M with eigenvalue 1. Thus we want to …nd x 6= 0 such that (M I)x = 0, i.e. to solve 2 32 3 2 3 1 0 1 x1 0 4 1 5 4 5 4 1 0 x2 = 0 5 : 0 1 1 x3 0 This gives x1 = x2 and x3 = corresponding to = 1 is

x1 . Thus a normalised eigenvector 2

3 1 1 v = p 4 1 5: 3 1

The axis of rotation is then the line de…ned as the linear span of v. In order to …nd the angle of rotation, we choose a normalised vector u which is orthogonal to v (i.e. u lies in the plane of rotation). We can for instance choose 2 3 1 1 4 5 0 ; u= p 2 1 Then u has norm 1 and u v = 0. We calculate that 2 2 32 3 3 0 0 1 1 1 1 1 0 54 0 5 = p 4 1 5 Mu = p 4 1 0 2 0 2 1 0 0 1

Since jjujj = jjM ujj = 1, we have that

cos = jjujj jjM ujj cos = u (M u) = Alternatively, notice that det M = +1; so 1 + 2 cos cos

13

= trace(M ) = 0 1 = : 2

1 : 2

Since we are not asked to say whether the rotation is clockwise or anticlockwise (see the cosine graph below), we answer that the angle of rotation is 23 (120 degrees).

Just for the record: 2 3 2 3 e 1 e2 e 3 1 14 1 1 1 0 1 5 = ( e1 e2 + e3 ) = 4 1 5 u Mu = 2 2 2 1 1 0 1 2 3 2 3 p p 1 1 3 1 4 3 14 5 p 1 1 5= v; = = 2 2 3 2 1 1

so relative to the direction v this is a clockwise rotation; here we get p 3 : sin = 2

14

(iii) The axis and angle of rotation of the transformation represented

by M 1 is the same as that of M , but the direction of the rotation represented by M 1 is the opposite of the direction of the rotation represented by M so 23 (since the transformation represented by M 1 reverses the action of the transformation represented by M ). (iv) The transformation represented by M 6 is the transformation with the same axis of rotation as that of M , and which has an angle of rotation 6 23 = 4 (or 6 120 = 2 360 degrees). Thus this transformation is simply the identity transformation, which corresponds to the transformation represented by the identity matrix.

Remark. The sense of the angle of rotation ‘about v’is clockwise (negative). But if we replace v by v as the generator of the axis the sense of the rotation beacome anti-clockwise (positive).

This explains why we cannot tell the sense from the roots of the characteristic polynomial pM (x):

15

Question 7 (a)

(i) We calculate AA ui = AA

1 1 = p AA Avi = p A i vi =

1 p Avi i

i

i

1

ip

Avi = i

which veri…es that ui is an eigenvector of AA with corresponding eigenvalue i . (ii) We calculate ui uj =

1 p Avi

1 p Avj

i

=p

j

1

=p

Avi

Avj

Avj

Avi

i j

1 i j

=p

1

vj A Avi

i j

=p

i

vj vi

i j

=

s

i

vi vj :

j

Therefore if i 6= j, then ui uj = 0, since vi and vj are orthogonal.

Moreover, if i = j, then ui ui = vi vi = 1, since v1 ; ; vn are orthonormal vectors. This veri…es that the vectors u1 ; : : : ; uk form an orthonormal set. (iii) We have that In = v1 v1 +

+ vk vk + vk+1 vk+1 +

+ vn vn :

We then multiply this identity by A to obtain A = Av1 v1 +

+ Avk vk + Avk+1 vk+1 +

+ Avn vn :

Since i = 0, for i = k + 1; : : : ; n, the identity above therefore reduces to A = Av1 v1 + + Avk vk : 16

i ui ;

But then, as Avi =

p

i ui for 1

i

p

1 u1 v1 +

A =

2 1

A=

+

k, p

k uk vk ;

by the expression for ui above. This is the singular values decomposition of A. (b) First observe that 1 0 0 1

:

Thus A A=

2 1

2

3 1 0 5= 1

2 4 1 0

1 0 0 1

5 2

2 2

:

Our next task is to …nd the eigenvalues and eigenvectors of A A. First we calculate that pA A (x) = det(xI

A) = (x

5)(x

4 = x2

2)

7x + 6:

Thus pA A (x) = 0 i¤ x = 1 or x = 6. The eigenvalues of A A are therefore 1 = 1 and 2 = 6.

Now, AA

4 2

I2 =

2 1

:

So in order to …nd the …rst eigenvector, we want to solve 4 2

2 1

x1 x2

0 0

=

;

which is equivalent to x2 = 2x1 . This means that the …rst normalised eigenvector of A A corresponding to 1 = 1 is 1 v1 = p 5

1 2

6I2 =

1 2

:

Next, we observe that AA

17

2 4

:

In order to …nd the second eigenvector, we want to solve 1 2

2 4

x1 x2

0 0

=

;

which is equivalent to x1 = 2x2 . Therefore the normalised eigenvector of A A corresponding to 2 = 6 is given by 1 v2 = p 5

2 1

:

Our next task is to calculate u1 and u2 . From the formula in (a) (iii), we calculate that 2 2 3 3 2 1 0 1 1 1 1 u1 = p Av1 = p 4 1 0 5 = p 4 1 5: 2 1 5 5 0 1 2

Similarly, we calculate that 2 2 1 1 u2 = p Av2 = p 4 1 6 30 0

3 1 0 5 1

2 1

2

3 5 1 = p 4 2 5: 30 1

We therefore conclude that the singular values decomposition of A is given by 2 3 2 3 0 5 p 1 1 1 1 1 2 + 6p 4 2 5 p 2 1 : A= p 4 1 5p 5 5 30 5 2 1

18

Question 8 (a) (i) For the given matrix B 0 1 2 3 1 1 1 1

BT B =

1 1 2 (14 9)

2

3 1 1 7 7= 1 5 1

0 6 1 6 4 2 3

2 3 3 7 1 2 L = (B T B) 1 B T = 3 10 1 3 1 1 3 = 7 4 1 2 10

(B T B) 1 =

(a)(ii) (x jjx

u) as (x

P x)?(P x ujj2 = jjx

3 7

14 6 6 4

=2

0 1 2 3 1 1 1 1

P x) 2 R(B)? and (P x

P xjj2 + jjP x

ujj2

7 3 3 2

jjx

and the lower bound is realized when u = P x: Thus jjx by taking u = P x: (b) Write 2 3 2 6 3 7 7 z=6 4 5 5: 6

P xjj2 ;

u) 2 R(B)

ujj is minimized

(i) Setting b = 0; the (inconsistent) system of equations zi = ati + c in matrix form reads:

2

3 2 3 2 3 2 0 1 6 3 7 6 1 7 6 1 7 7 6 7 6 7 z=6 4 5 5 =a 4 2 5 + c 4 1 5 : 6 3 1

For given a; c the error e between the left and right hand sides is 3 2 3 2 3 2 3 2 2 3 e1 2 0 1 2 6 e2 7 6 3 7 6 1 7 6 1 7 6 3 7 a 7 6 7 6 7 6 7 6 7 e =6 4 e3 5 := 4 5 5 a 4 2 5 c 4 1 5 = 4 5 5 B c e4 6 3 1 6 19

;

where, as in part (a),

2

3 1 1 7 7: 1 5 1

0 6 1 B=6 4 2 3

Thus B includes the column of t readings. So e measures the distance of z from possible points in R(B): Since BL is the orthogonal projection, this distance is minimized when B

a c

= BLz:

That is, the minimized squares are: e21 + e22 + e23 + e24 : So, since LB = I; a c

= LB

a c

= LBLz = Lz:

So here a c

= Lz =

1 10

3 7

1 1 4 1

3 2

Check:

2

3 2 6 3 7 6 7= 1 4 5 5 10 6

14 19

0 1 2 3 1:9 3:3 4:7 6:1

(ii) This time b need not be 0; and we consider the equations zi = ati + byi + c: The column of t readings and y readings are included in A: Write 2 3 2 3 0 1 1 0 1 6 1 2 1 7 6 1 1 7 1 1 0 7 6 7 A=6 4 2 3 1 5=4 2 1 5 0 1 1 : 3 4 1 3 1 20

1 1 0 0 1 1

CC T = (CC T ) 1 =

2

1 3

2 1

1 2

3 1 0 4 1 1 5= 0 1 2

3 1 0 1 R = C T (CC T ) 1 = 4 1 1 5 3 0 1 2 3 2 1 14 1 1 5: = 3 1 2

Here the error for given a; b; c 3 2 3 2 2 3 2 3 2 e1 0 1 6 3 7 6 e2 7 6 1 7 6 2 7 7 6 7 6 6 7 6 7 4 e3 5 := 4 5 5 a 4 2 5 b 4 3 5 e4 6 3 4

2

2 1 1 2

2 1

1 2

3 2 3 1 2 6 1 7 6 3 7 7 6 7 c6 4 1 5=4 5 5 1 6

2

3 a A4 b 5: c

again measures the distance of z from R(B): As AG is an orthogonal projection, this is minimized by taking 2 3 a 4 A b 5 = AGz: c So if

2

then we have

2

3 a 4 b 5 = Gz; c

3 a A 4 b 5 = AGz: c

Remark. The solution Gz to the last equation is in fact the one with smallest norm in R(G); indeed if (a; b; c)T is any solution to the preceeding 21

equation, then

2

3 a GA 4 b 5 = GAGz = Gz; c

and GA projects parallel to N (A) and onto R(G):

22