Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 8 docx

Appendix OFormulas from Linear Algebra Kramer’s Rule.. “Professors often present difficult problems that have simple, elegant solutions.. Note that mathematicians never refer to problems

I.5 Table of Fourier Cosine Transforms

π

Z ∞ 0

f (x) cos (ωx) dx

2

Z ∞ 0

I.6 Table of Fourier Sine Transforms

π

Z ∞ 0

f (x) sin (ωx) dx

2

Z ∞ 0



1 1

πω2

W [eaxcos(bx), eaxsin(bx)] b e2ax

W [eaxcosh(bx), eaxsinh(bx)] b e2ax

W [sin(c(x − a)), sin(c(x − b))] c sin(c(b − a))

W [cos(c(x − a)), cos(c(x − b))] c sin(c(b − a))

W [sin(c(x − a)), cos(c(x − b))] −c cos(c(b − a))

W [sinh(c(x − a)), sinh(c(x − b))] c sinh(c(b − a))

W [cosh(c(x − a)), cosh(c(x − b))] c cosh(c(b − a))

W [sinh(c(x − a)), cosh(c(x − b))] −c cosh(c(b − a))

Wedx

sin(c(x − a)), edxsin(c(x − b)) c e2dxsin(c(b − a))

Wedx

cos(c(x − a)), edxcos(c(x − b)) c e2dxsin(c(b − a))

Wedxsin(c(x − a)), edxcos(c(x − b))

−c e2dxcos(c(b − a))

Wedxsinh(c(x − a)), edxsinh(c(x − b))

c e2dxsinh(c(b − a))

Wedxcosh(c(x − a)), edxcosh(c(x − b)) −c e2dxsinh(c(b − a))

Wedxsinh(c(x − a)), edxcosh(c(x − b)) −c e2dxcosh(c(b − a))

W [(x − a) ecx, (x − b) ecx] (b − a) e2cx

, n ∈ N

, n ∈ N

hy0, y0i = b − a, hyn, yni = b − a

2 for n ∈ N

p(t) dt

H(x − ξ)

• y00+ c2y = 0, y(a) = y(b) = 0, c 6= b−anpi, n ∈ N

G(x|ξ) = sin(c(x<− a)) sin(c(x>− b))

c sin(c(b − a))

• y00+ c2y = 0, y(a) = y0(b) = 0, c 6= (2n−1)pi2(b−a) , n ∈ N

G(x|ξ) = −sin(c(x<− a)) cos(c(x>− b))

c cos(c(b − a))

• y00+ c2y = 0, y0(a) = y(b) = 0, c 6= (2n−1)pi2(b−a) , n ∈ N

G(x|ξ) = cos(c(x<− a)) sin(c(x>− b))

c cos(c(b − a))

• y00+ 2cy0+ dy = 0, y(a) = y(b) = 0, c2 > d

Function Sum and Difference Identities

sin x + sin y = 2 sin1

2(x + y) cos

1

2(x − y)sin x − sin y = 2 cos1

2(x + y) sin

1

2(x − y)cos x + cos y = 2 cos1

2(x + y) cos

1

2(x − y)cos x − cos y = −2 sin1

2(x + y) sin

1

2(x − y)Double Angle Identities

sin 2x = 2 sin x cos x, cos 2x = cos2x − sin2xHalf Angle Identities

sin x sin y = 1

2cos(x − y) −

1

2cos(x + y)cos x cos y = 1

2cos(x − y) +

1

2cos(x + y)sin x cos y = 1

2sin(x + y) +

1

2sin(x − y)cos x sin y = 1

2sin(x + y) −

1

2sin(x − y)Exponential Identities

eıx = cos x + ı sin x, sin x = e

ıx− e−ıx

ı2 , cos x =

eıx+ e−ıx2

1tanh xPythagorean Identities

cosh2x − sinh2x = 1, tanh2x + sech2x = 1Relation to Circular Functions

sinh(ıx) = ı sin x sinh x = −ı sin(ıx)cosh(ıx) = cos x cosh x = cos(ıx)tanh(ıx) = ı tan x tanh x = −ı tan(ıx)Angle Sum and Difference Identities

sinh(x ± y) = sinh x cosh y ± cosh x sinh ycosh(x ± y) = cosh x cosh y ± sinh x sinh ytanh(x ± y) = tanh x ± tanh y

1 ± tanh x tanh y =

sinh 2x ± sinh 2ycosh 2x ± cosh 2ycoth(x ± y) = 1 ± coth x coth y

coth x ± coth y =

sinh 2x ∓ sinh 2ycosh 2x − cosh 2y

Function Sum and Difference Identities

sinh x ± sinh y = 2 sinh 1

2(x ± y) cosh

1

2(x ∓ y)cosh x + cosh y = 2 cosh1

2(x + y) cosh

1

2(x − y)cosh x − cosh y = 2 sinh 1

2(x + y) sinh

1

2(x − y)tanh x ± tanh y = sinh(x ± y)

cosh x cosh ycoth x ± coth y = sinh(x ± y)

sinh x sinh y

Double Angle Identities

sinh 2x = 2 sinh x cosh x, cosh 2x = cosh2x + sinh2xHalf Angle Identities

sinh x sinh y = 1

2cosh(x + y) −

1

2cosh(x − y)cosh x cosh y = 1

2cosh(x + y) +

1

2cosh(x − y)sinh x cosh y = 1

2sinh(x + y) +

1

2sinh(x − y)See Figure M.1 for plots of the hyperbolic circular functions

-2 -1 1 2

-3 -2 -1 1 2 3

-1 -0.5

0.5 1

Figure M.1: cosh x, sinh x and then tanh x

rJν(j0ν,mr)Jν(j0ν,nr) dr = j

02 ν,n− ν2

2j02 ν,n

Jν(j0ν,n)
2δmn

Z 1 0

rJν(αmr)Jν(αnr) dr = 1

2α2 n

 a2

b2 + α2n− ν2

(Jν(αn))2δmnHere αn is the nth positive root of aJν(r) + brJ0ν(r), where a, b ∈ R

Appendix O

Formulas from Linear Algebra

Kramer’s Rule Consider the matrix equation

A~x = ~b

This equation has a unique solution if and only if det(A) 6= 0 If the determinant vanishes then there are either nosolutions or an infinite number of solutions If the determinant is nonzero, the solution for each xj can be written

xj = det Ajdet Awhere Aj is the matrix formed by replacing the jth column of A with b

Example O.0.1 The matrix equation

,has the solution

x1 =

5 2

6 4

1 2

3 4

= −9

−2 =9

2.

u · dr

Appendix Q

Partial Fractions

A proper rational function

p(x)q(x) =

p(x)(x − a)nr(x)Can be written in the form

p(x)(x − α)nr(x) =



a0(x − α)n + a1

(x − α)n−1 + · · · + an−1

x − α

+ (· · · )where the ak’s are constants and the last ellipses represents the partial fractions expansion of the roots of r(x) Thecoefficients are

ak = 1k!

dk

dxk

 p(x)r(x)



x=α

Example Q.0.2 Consider the partial fraction expansion of

1 + x + x2(x − 1)3 The expansion has the form

a0

(x − 1)3 + a1

(x − 1)2 + a2

x − 1.

The coefficients are

a0 = 10!(1 + x + x

2)|x=1 = 3,

a1 = 11!

d

dx(1 + x + x

2)|x=1 = (1 + 2x)|x=1 = 3,

a2 = 12!

1 + x + x2

x2(x − 1)2.The expansion has the form

a0

x2 +a1

x +

b0(x − 1)2 + b1

x − 1.The coefficients are

a0 = 10!

 1 + x + x2

(x − 1)2

 ...

5

6

1

3

= 8< /sup>

−2 = −4, x2 =... data-page="21">

Appendix Q

Partial Fractions

A proper rational function

p(x)q(x) =

p(x)(x − a)nr(x)Can be written in the form

p(x)(x − α)nr(x)... an−1

x − α

+ (· · · )where the ak’s are constants and the last ellipses represents the partial fractions expansion of the roots of r(x) Thecoefficients are

ak