michalemas math part II

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Last updated 1:30 PM on 5/31/26
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41 Terms

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delta function property (the important one)

∫ δ(x − ξ)f(x)dx = f(ξ) (∞ to −∞)

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differential operator L

L = d²/dx² + p(x) d/dx + q(x)

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The Wronskian

W = y₁y₂' -y₂y₁'

y₁, y₂ are solutions to an ODE and are independent for W≠0

<p>W = y₁y₂' -y₂y₁'</p><p>y₁, y₂ are solutions to an ODE and are independent for W≠0</p>
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Two Properties Green's Functions that allow to solve the DEs:

at the boundary:

∂G⁺/∂x - ∂G⁻/∂x =1

G⁺= G⁻

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solution for Ly(x) = f(x) using G(x; ζ)

y(x) = ∫ G(x; ζ)f(ζ) dζ from a to b (the given boundry points)

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steps for solving seperable linear PDEs

(i) inhomogeneous equation, find a particular solution.

(ii) Seek separable solutions to the homogeneous equation.

(iii) In the case of inhomogeneous boundary conditions (non constant - eg one wall is cold, one hot in the end) we split the solution into two: steady state and homogenous solution. Find the steady-state solution by guessing (eg linear gradient between C1 and C2 along x results in (C2-C1)/L x).

(iv) solve the homogeneous case (zero at the boundaries). using separable function. The BC will determine wether λ>,<, = 0. Find the eigenvalues, eg kₙ = nπ/L and the eigenfunctions Xₙ = Γₙ sin(kₙ x). the solution is the sum of eigenfunctions from n=1 to ∞.

(vi) add the solutions from (i) - (iv). Determine unknown constants using the boundary conditions.

note: The idea is to handle sources (particular integral), handle boundary conditions (if they're inhomogeneous) , then solve the homogeneous "leftover" problem using separation of variables. So the most complicated it will get is y(x,t)=v(x,t)+w(x)+p(x,t).

<p>(i) inhomogeneous equation, find a particular solution.</p><p>(ii) Seek separable solutions to the homogeneous equation.</p><p>(iii) In the case of inhomogeneous boundary conditions (non constant - eg one wall is cold, one hot in the end) we split the solution into two: steady state and homogenous solution. Find the steady-state solution by guessing (eg linear gradient between C1 and C2 along x results in (C2-C1)/L x).</p><p>(iv) solve the homogeneous case (zero at the boundaries). using separable function. The BC will determine wether λ&gt;,&lt;, = 0. Find the eigenvalues, eg kₙ = nπ/L and the eigenfunctions Xₙ = Γₙ sin(kₙ x). the solution is the sum of eigenfunctions from n=1 to ∞.</p><p>(vi) add the solutions from (i) - (iv). Determine unknown constants using the boundary conditions.</p><p>note: The idea is to handle sources (particular integral), handle boundary conditions (if they're inhomogeneous) , then solve the homogeneous "leftover" problem using separation of variables. So the most complicated it will get is y(x,t)=v(x,t)+w(x)+p(x,t).</p>
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connection between a transformation matrix and the transformed basis

Each column tells you where one original basis vector ends up after the transformation. The j-th column is the transformed version of the j-th basis vector

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symmetric and antisymmetric matrices

A^T = A

A^T = -A

<p>A^T = A</p><p>A^T = -A</p>
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definitions and properties of:

Normal matrices

Orthogonal matrices

unitary matrices

normal matrix: AA† = A†A

Orthogonal matrix: Q⊤Q=QQ⊤=I

Unitary matrix: U*U=UU* = I

all three have orthonormal eigenvectors.

Orthogonal (real) and unitary (complex) matrices are special cases of normal matrices. They preserve lengths and angles and can be thought of as a "rotation" in complex space.

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The Hermitian conjugate of a product of matrices

hermitian and anti hemitian matrix

(AB) † = B †A †

(ABCx) † = x †C †B †A †

note: end-to-start

hermitian matrix: A † = A

anti hermitian matrix: A † = - A

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Positive definiteness

v †Av ⩾ 0 , with equality iff v = 0

Positive definiteness means a matrix always gives a strictly positive value when used to measure the "squared length" x⊤Ax of any nonzero vector. This guarantees real, positive lengths/energies, and well-behaved geometry.

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⟨ u | v ⟩ ≡ u · v : Scalar/Inner Product important properties

Hermitian symmetry- u · v = (v · u)*

positive definite: v · v ⩾ 0, |v| = 0 ⇒ v = 0

'anti-linear': (au1 + bu2) · v = a^∗ (u1 · v) + b^∗ (u2 · v)

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Schwarz's Inequality

The Triangle Inequality

|⟨ u | v ⟩| ⩽ ∥u∥ ∥v∥ ∀ u, v,

∥u + v∥ ⩽ ∥u∥ + ∥v∥

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metric:

mathematical definition

meaning

3 key properties

expression for metric in terms of a change of basis matrix A.

G_ij = u_i · u_j

Encodes how lengths, angles, and distances are measured in a (possibly non-orthonormal) basis, making geometric and physical quantities independent of the coordinates. (G_ij is constructed per base).

- Hermitian(G †)ij = (G)ij

- v · w = v †G w for two random vectors v, w. (note: is orthonormal, v · w = v_i* w_i the sum of component products, and G= identity. But in non othonormal, we need to take G into account).

- positive definite: v †Gv ⩾ 0

G' = A†GA

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eigenvectors:

steps to find

significance

distinct?

1. use det(M − λI) = 0 to find the eigenvalues λ_i

2. plug them into (M − λI)x = 0 and obtain x_i.

3. normalize if necessary.

Eigenvectors are the special directions in which a linear transformation acts by simply scaling rather than rotating or mixing components, revealing the "natural" axes of the system and simplifying analysis of its behavior.

each distince eigenvalue => distinct eigenvector. if an eigenvalue repeats n times there will be 1-n eigenvectors corresponding to it.

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Diagonal matrix:

how to diagonalize M

eigenvalues of a diagonalized matrix

relation to quadric forms

4 useful properties relating Λ and M

X⁻¹MX = Λ where X is the matrix of eigenvectors in the columns ( n × n matrix is diagonalizable iff it has n linearly-independent eigenvectors.)

Diagonalization doesn't change original eigenvalues. and it has them across its diagonal.

Diagonalizing a matrix transforms it into a basis of its eigenvectors (principal axes), which helps quadric form recognition and graphing.

- Powers: Mⁿ=XDⁿX⁻¹

- det(M) = det(Λ) = product of eigenvalues

- tr(M) = dum of eigenvalues

- tr(Mⁿ) = tr(Λⁿ)

<p>X⁻¹MX = Λ where X is the matrix of eigenvectors in the columns ( n × n matrix is diagonalizable iff it has n linearly-independent eigenvectors.)</p><p>Diagonalization doesn't change original eigenvalues. and it has them across its diagonal.</p><p>Diagonalizing a matrix transforms it into a basis of its eigenvectors (principal axes), which helps quadric form recognition and graphing.</p><p>- Powers: Mⁿ=XDⁿX⁻¹</p><p>- det(M) = det(Λ) = product of eigenvalues</p><p>- tr(M) = dum of eigenvalues</p><p>- tr(Mⁿ) = tr(Λⁿ)</p>
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Properties of the Eigenvalues and Eigenvectors of an Hermitian Matrix

- The eigenvalues of an Hermitian matrix are real.

- The eigenvectors of an Hermitian matrix with distinct eigenvalues are orthogonal.

- An Hermitian matrix has n orthonormal eigenvectors

to show the first two:

Hx = λx , Hy = µy for two eigenvecs+vals.

y †H = µ ∗ y †

=> y †Hx = λy † x = µ ∗ y † x

=> y †Hx = λy † x = µ ∗ y † x

=> (λ − µ ∗ )y † x = 0

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Property of the Eigenvalues and Eigenvectors of

anti Hermitian Matrices

unitary matrices

imaginary

of unit modulus

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Normal matrix eigenvector properties

the eigenvectors of normal matrices corresponding to distinct eigenvalues are orthogonal. and a repeated eigenvalue will have an eigenvector for each repeat. its eigenvectors form an orthonormal basis.

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what matrices are diagonalizable?

how to diagonalize a matrix?

There are n linearly independent eigenvectors. satisfied automatically for normal (including anti/Hermitian, Unitary, Orthogonal).

use: X †HX = Λ

note: all normal matrices unitarily diagonalizable - X is unitary - all columns (the eigenvectors) are orthonormal.

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the eigenvalues of a metric are...

the eigenvectors of a metric are...

strictly positive

the new basis vectors of the diagonalized metric

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quadric form:

expression in terms of x

expression in terms of x' (principle basis)

what does quadric form = constant tell us?

, F(x) = x^T S x for real symmetric S and real x

F(x) = x′T Λ x′ = ∑=λₙxₙ² where Λ is the diagonalized S (all cross terms vanish. eg 2x²+3y²+z² = C is an ellipsoid).

we get an equation like in the photo and this can be:

Ellipsoid, Hyperboloids (sand-clock, attached or detached), Paraboloids, Elliptical conical surfaces, planes, cylinders.

<p>, F(x) = x^T S x for real symmetric S and real x</p><p>F(x) = x′T Λ x′ = ∑=λₙxₙ² where Λ is the diagonalized S (all cross terms vanish. eg 2x²+3y²+z² = C is an ellipsoid).</p><p>we get an equation like in the photo and this can be:</p><p>Ellipsoid, Hyperboloids (sand-clock, attached or detached), Paraboloids, Elliptical conical surfaces, planes, cylinders.</p>
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the Rayleigh quotient: definition and relation to eigenvalues.

λ(x) = (x^T S x)/(x^T x)

inverse of the distance from a point on a form to the origin, normalized to remove depndence of the constant to which the form was equal.

It is proven that for an hermitian matrix, the values of λ(x) = x †Hx/x†x at its stationary point are the eigenvalues.

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Cauchy's principle of convergence

sufficient condition for the sequence sₙ to converge:

for any positive number ε, | s_(n+m) − s_n | < ε for all positive integers m, for sufficiently large n.

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Bounded sequences definition

he sequence sn is bounded as n → ∞ if there exists a positive number K such that |sₙ| < K for sufficiently large n

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Convergent series

as n → ∞, sₙ tends to a finite limit s. sₙ here is the partial sum sₙ = Σu_r from r₀ to n.

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necessary condition for convergence

u_r → 0 as r → ∞

not a sufficient condition for convergence!

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Absolute and conditional convergence definitions

absolute:

Σ |u_r| from 1 to ∞ converges.

this implies Σ u_r from 1 to ∞ converges.

conditional: Σ |u_r| diverges and Σ u_r converges.

example: harmonic vs oscillating harmonics

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log(1+x), sin(x), cos(x), exp(x) taylor expansions

knowt flashcard image
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D'Alembert's ratio test

if ϱ = lim u_(r+1) / u_r as r → ∞

Σ u_r converges if ϱ < 1 and diverges for ϱ>1

<p>if ϱ = lim u_(r+1) / u_r as r → ∞</p><p>Σ u_r converges if ϱ &lt; 1 and diverges for ϱ&gt;1</p>
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Cauchy's test

lim u_r ^(1/r) = ϱ as r→∞

Σ u_r converges if ϱ < 1 and diverges for ϱ>1

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The O notation

f(z)/g(z) is bounded as z→z0, f(z) = O(g(z)) as z→z0

f(z)/g(z)→0 as z→z0, f(z) = o(g(z)) as z→z0

g(z)/f(z)→1 as z→z0, f(z) ∼ g(z) as z→z0

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The Cauchy-Riemann equations (complex derivatives) and concequences

for f(z) = u(x, y) + iv(x, y) , necessary conditions for convergence are -

∂u/∂x = ∂v/∂y , ∂v/∂x = − ∂u/∂y

consequences:

- If we know the real/imaginary part of an analytic function we can find its imaginary/real part up to an additive constant by integrating the Cauchy-Riemann equations.

- If a function of a complex variable is analytic in a region R it is differentiable any number of times in R.

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Analytic function

Entire function

examples

Analytic: have a complex derivative at every point z in a region R of the complex plane → analytic in R

entire: analytic in the whole complex plane

examples (entire): any combination of exponentials, polynomials, trig functions. Basically all functions that can be composed of power series, which are all "nice" functions (for example, excluding f(z) = z* )

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definitions:

zeros

Poles

connection between zeros and poles

essential singularities

zeros: points where f(z0) = 0 . has order N if all derivatives up to the N-1 give 0: f(z0) = f ′ (z0) = f ′′(z0) =...= f (ⁿ⁻¹)(z0) = 0 . simple zero means N=1.

poles: f(z) has a pole of order n if f(z) = (z − z0)⁻ⁿ g(z) . simple pole means N=1.

If f(z) has a zero of order N at z = z0, then 1/f(z) has a pole of order N there, and vice versa.

essential singularities: if a power series of a function around z0 requires infinite negative powers, z0 is an essential singulaity (unlike removable, which can be removed by multiplying the function by (z-z0)ⁿ.

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Radius of convergence:

mathematical definition

physical meaning

how to find

Σ(z − z0)^r converges for |z − z0| < R and diverges for |z − z0| > R

it is equal to the distance of the nearest singular point of the function f(z) from z0.

to find, use D'Alembert's ratio test (if the limit exists)

lim | a_(r+1) / a_r | = 1/R as r→∞

or cauchy's test:

lim | u_r |^(1/r) = 1/R as r→∞

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second solution to an ODE using the wronskian and y1

expression for the wronskian

y2 = y1 ∫W(η)/[y1(η)]² dη upper limit: x

W = k exp(-∫p(ζ) dζ ) where p is the coefficient of y' (when the coefficient of y'' is 1).

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ordinary, regular singular and irregular singular points of a complex ODE

for a differential equation y ′′(z) + p(z)y ′ (z) + q(z)y(z) = 0 :

ordinary point: if p(z) and q(z) are both analytic at z = z0, then z = z0 is called an of the ODE.

regular singular points: (z − z0)p(z) and (z − z0) 2 q(z) are both analytic at z = z0.

irregular singular points: else.

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solving a complex homogenous ODE as a power series around an ORDINARY point

1. assume y = Σaₙzⁿ from 0 to ∞ (origin shift if needed). find y' and y''.

2. if p and q are rational functions (i.e. ratios of polynomials) it is a much better idea to multiply the equation through by a suitable factor to clear denominators.

3. substitute y, y', y'' into the equation.

4. change the sums by substituting r = n-1 or n-2 , to obtain one big sum.

5. since each power coefficient should be zero, obtain a recurrence relation.

6. if needed, find a nicer relation (eg odd and even decoupling) in terms of a0 and a1, and substitute back into y = Σaₙzⁿ.

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solving a complex homogenous ODE as a power series around an REGULAR SINGULAR point

1. assume y = Σaₙz^(n+σ) from n=0 to ∞ (origin shift if needed).

2. find s = zp(z) and t = z²q(z) in terms of polynomials or infinite sums.

3. find σ₁ , σ₂ using σ(σ-1)+σs₀+t₀ = 0

4. for each σ, use Σ[ (σ + n)(σ + n − 1) + (σ + n)s(z) + t(z) ]aₙzⁿ =0 to obtain recurrence relation for aₙ.

5. for σ₁ ≠ σ₂ and σ₁ - σ₂ ≠ N , y₁ = z^σ₁ Σaₙzⁿ and y₂ = z^σ₂ Σaₙzⁿ. This also works for σ₁ - σ₂ is an odd integer specificaly for bessel.

6. for σ₁ ≠ σ₂ and σ₁ - σ₂ is an even integer, y₁ = z^σ₁ Σaₙzⁿ for the larger σ , and y₂ = z^σ₂ Σaₙzⁿ + Ky₁ln(z) . This also works for a repeated root.

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Variation of Parameters: when is it useful, steps.

Solves any nonhomogeneous ODE (sometimes can't obtain closed form). For y ′′(z) + p(z)y′(z) + q(z)y(z) = f(x):

1. solve the homogenous case with f(x) = 0, and denote the result y_c(x) = c₁y₁(x)+c₂y₂(x). Hence y_p(x) = u₁(x)y₁(x)+u₂(x)y₂(x)

2. compute the wronskian W = y₁y₂′​−y₁′​y₂

3. compute u₁ and u₂ using:

u₁(x) = -∫​y₂(x)f(x)/W(x) dx

u₂(x) = ∫​y₁(x)f(x)/W(x) dx

4. plug u₁(x) , u₂(x) into y_p(x). The final solution to the equation is y_c(x) +y_p(x).