Introduction to LaTeX: Formatting and Math

INTRODUCTION TO LATEX 1. Format and Font You can just start typing normally. You can center something. You can use bold. You can use italics. You can even underline. This document uses a 12 point font size, but you can write smaller or smaller or smaller. You can also write larger and You can larger and larger and larger . . . combine these styles . . . 2. Horizontal Space Leaving more than one space between words does not change the output. But you can create a little space like this. You can create a little more like this. You can have as much space as you like. 3. Vertical Space A new line in the source document does not change the output. A new line in the source document does not change the output. A new line in the source document does not change the output. Leaving a line blank starts a new paragraph. Double backslash starts a new line. Double backslash starts a new line. You can use bigskip to skip a line. You can use vspace to create as much space as you want between lines. You can start a new page like this. 4. Lists Now I want to make a list. • first thing on my list – list within a list – another thing on my sublist • second thing on my main list Now I want to number my list. (1) yellow (a) flowers (i) sunflowers (ii) chamomile (iii) arnica (b) my highlighter (2) blue I can specify my own labels on a list. A. is for Allison B. Ball 5. Math 5.1. Math environments. We can use dollar signs to write mathematical expressions a + b < 2 in the middle of a line (inline). (If you ever want to type a dollar sign, do this $.) Some expressions, like limx→1 f (x), don’t always look so great when you write then inline. You can see if displaystyle, like lim f (x), makes them look x→1 better. You can use double dollar signs to write equations centered on their own line. x2 + y 2 ≤ 9 Backslash square brackets around an expression does the exact same thing. x2 + y 2 ≤ 9 You can give an equation a number using the equation environment Z b 5 (5.1) f (x)dx = 9 a or use the * to take the number away if you change your mind. Z b 5 f (x)dx = 9 a You can label a numbered equation and refer back to it later. (Sometimes you have to typeset twice before the numbers will show up.) eiπ + 1 = 0 (5.2) Equation (5.2) is Euler’s equation. You can use the align environment for longer arguments. (5.3) (5.4) |x − y| = |x − a + a − y| ≤ |x − a| + |a − y| Use align* to take all the numbering away. Use nonumber to remove the numbering from a single line. |x − y| = |x − a + a − y| (5.5) ≤ |x − a| + |a − y| 5.2. Theorems, proofs, and stuff like that. I can write a theorem. Theorem 5.1. Everything I say is true. Proof. This is true because I said so.  Ok, that’s not really a proof. I’ll change it to a claim. Claim 1. Everything I say is true. Ok, here is something that is true. Notation. We write dy = f 0 (x). dx Notice the different ways these are labeled and look at how they are defined in the preamble. You can change these settings if you want. 5.3. More examples. Here are some examples from other stuff I have written. I can cite them like this [1]. My references appear at the end of the document. (Sometimes you have to typeset twice to get citations to work.)     −1 1 Exercise 1. Let ~u = and ~v = . Find the projection proj~v ~u of ~u onto 2 −1 ~v . Draw each of these vectors in the same coordinate plane. Solution. ~v · ~u −3 proj~v ~u = ~v = 2 |~v | 2     1 −3/2 = −1 3/2 Exercise 2. Let z = x + iy be a complex number (i.e. z ∈ C) with x, y ∈ R and i2 = −1. Define the modulus |z| and complex conjugate z̄ of z, respectively, by |z| = (x2 + y 2 )1/2 and z̄ = x − iy (1) Show that |λz| = |λ||z| if λ ∈ R. Solution. √ p p |λz| = (λx)2 + (λy)2 = λ2 x2 + y 2 = |λ||z| (2) Show that |z1 z2 | = |z1 ||z2 | if z1 , z2 ∈ C. Solution. |z1 z2 | = |(x1 x2 − y1 y2 ) + i(x1 y2 + x2 y1 )| p = (x1 x2 − y1 y2 )2 + (x1 y2 + x2 y1 )2 q = x21 x22 + y12 y22 + x21 y22 + x22 y12 q = (x21 + y12 )(x22 + y22 ) = |z1 z2 | (3) Show that |z|2 = z z̄. Solution. z z̄ = (x + iy)(x − iy) = x2 + y 2 = |z|2 Exercise 3. For z ∈ C a complex number, define the exponential function z e = ∞ X zn n=0 n! Show that this series converges absolutely for all z, i.e. show that Solution. We apply the ratio test to the series ∞ X zn converges. n! n=0 ∞ X |z|n n=0 n! an+1 |z|n+1 n! |z| = lim = lim =0 n n→∞ an n→∞ (n + 1)! |z| n→∞ n + 1 lim Thus, the series converges for all z. Exercise 4. Suppose that f (x) is periodic with period 2π. Show that for any number a we have Z π Z a+π f (x) dx = f (x) dx −π a−π Solution. Z π Z a+π Z −π Z a+π −π π f (x) dx f (x) dx + f (x) dx = a+π Z −π f (x) dx + = −π Z a+π = f (u + 2π) du a−π Z −π f (x) dx + −π Z a+π f (u) du a−π f (x) dx = a−π Exercise 5. Consider the steady-state heat equation (Laplace’s equation) in the upper half-plane.   uxx + uyy = 0 −∞ < x < ∞, y > 0  u(x, 0) = f (x) x ∈ R Use the Fourier transform to show that the bounded solution is u(x, y) = (f ∗ Py )(x), 1 y where Py is the Poisson kernel for the upper half-plane, Py (x) = . 2 π x + y2 Solution: Taking the Fourier transform in x, we have −ξ 2 u b(ξ, y) + ∂ 2u b (ξ, y) = 0 ∂y 2 For fixed ξ, this is an ODE for u b(ξ, ·) and the solutions are u b(ξ, y) = A(ξ)e−|ξ|y + B(ξ)e|ξ|y Since the second term is unbounded on the upper half-plane, we take B(ξ) = 0. So u b(ξ, y) = A(ξ)e−|ξ|y . From the boundary condition we have u b(ξ, 0) = fb(ξ). Thus, u b(ξ, y) = fb(ξ)e−|ξ|y . √ py (ξ), Now if pby (ξ) = e−|ξ|y , then u b(ξ, y) = fb(ξ)b py (ξ). And since f\ ∗ py (ξ) = 2π fb(ξ)b taking inverse Fourier transforms gives u(x, y) = √12π (f ∗ py )(x). q a Note that for a > 0 the Fourier transform of g(x) = e−a|x| is gb(ξ) = π2 a2 +ξ 2 , and q y 2 ǧ(x) = gb(−x) = gb(x) since gb is even. Thus, we take py (x) = so that π y 2 +x2 pby (ξ) = e−y|ξ| as desired. Then 1 u(x, y) = √ (f ∗ py )(x) 2π r Z ∞ 1 2 y =√ f (x − t) 2 dt y + t2 2π π −∞ Z 1 ∞ y = f (x − t) 2 dt π −∞ y + t2 = (f ∗ Py )(x) where Py = √1 py . 2π Exercise 6. Consider the linear system  0    x 1 1 x = . 0 y −10 3 y One  eigenvalue  of the matrix above is λ = 2 + 3i and a corresponding eigenvector is 1 − 3i v= . Find two linearly independent real-valued solutions Y1 and Y2 of this 10 system. Solution. Left to the reader. This is getting pretty complicated. . . Take a break and enjoy this comic. Chaos