0 2026/notes/miscHi! These notes are part of this larger collection of notes from my freshman year at UW.
Matthew Conroy, Spring 2026.
Field axioms and Order axioms: Reals are an ordered field.
Absolute value.
\forall x, |x| \ge 0|x y| = |x| |y||x + y| \le |x| + |y|If \forall x \in \mathbb R, f(f(x)) = f(x), then f is idempotent. E.g. \lfloor x \rfloor, |x|, x, 1.
If S is finite and nonempty, both \min S and \max S exist.
Completeness axiom: Bounded above and nonempty implies supremum exists in \mathbb R, but not in \mathbb Q.
“Bounded” = bounded both above and below.
If S is bounded below, then \{-x \mid x \in S\} is bounded above.
Archimedean property is a consequence of the completeness axiom: For all x,y \in \mathbb R with x>0, there exists n \in \mathbb N such that n x > y. There are no infinitely big or small reals.
x>0, it suffices to show that \exists n, n > y/x. Assume FTSOC that \forall n \in \mathbb N, n \le y/x. Then \mathbb N \subseteq \mathbb R is bounded above and this has a supremum by the completeness axiom. Then \sup \mathbb N - 1 is not a natural bound, so let n'\in\mathbb N be a natural such that n' > \sup \mathbb N - 1. Adding 1 to both sides, we get n' + 1 > \sup \mathbb N, contradicting the fact that it is an upper bound. Our assumption was false and there is such an n that n > y/x.All nonempty finite sets have a maximum (Induction on set cardinality).
A set S\subseteq \mathbb R is dense if for all x \in \mathbb R and all \varepsilon > 0, there exists s \in S such that |x -s| < \varepsilon.
Lemma: Let a,b \in \mathbb R be reals with b - a > 1. Then \exists n \in \mathbb Z, n \in (a, b).
a > 1. Let S = \{n \in \mathbb N : n < a\}, and this must be nonempty since 1 \in S. Again since a > 1, by the archimedean property we have m \in \mathbb N where m > a. Thus, S \subseteq \mathbb N_{<m}, so S is finite. Thus, S has a max M = \max S, and M \in S with M < a. And then M + 1 \not\in S so M + 1 \ge a. Transitive property: M < a, and < + 1 < a + 1 < b. Thus, a \le M + 1 < b. Suppose that M + 1 = a. Then a < a + 1 \le M + 2 < b. Thus, we either have M + 1 or M + 2 in the interval (a, b).a \le 1: then 1 - a \ge 0, so by the archimedean principle we have n \in \mathbb N such that n > 1 - a. So a + n > 1. Now we apply our weaker version to obtain m \in \mathbb N in the interval (a +n, b+ n), and thus m - n \in (a, b) and m - n \in \mathbb Z.Theorem: \mathbb Q is dense in the reals. Let a,b\in\mathbb R with a < b, then \exists q \in \mathbb Q with a < q < b.
b-a>0, so by the archimedean principle we can find n \in \mathbb N with n(b-a) > 1. Thus, (na, nb) is an interval of length >1, so there must exist m \in \mathbb Z in that interval by the previous lemma. Thus, a < \frac{m}{n} < b.A sequence a_n : \mathbb N \to \mathbb R has a limit L \in \mathbb R if
\forall \varepsilon > 0, \exists N \in \mathbb N, \forall n > N, |a_n - L | < \varepsilon.
Irrationals are dense in the reals: Suppose \forall x \in (a,b), x \in \mathbb Q. Then we have an injection (a,b) \to \mathbb Q, so (a,b) is countable. This is a contradiction with the fact that (a,b) is uncountable.
If (a_n) is a convergent sequence, then (a_n) is bounded: Choose some \varepsilon, and for n > N \varepsilon is the bound, but n \le N is a finite set and is this bounded.
(a_n) is not bounded, it does not converge.Abuse of notation:
\forall B>0, \exists N such that \forall n > N, a_n > B, then (a_n) does not converge (it is unbounded) and we write \lim a_n = +\infty.\forall B<0, \exists N, \forall n>N, a_n < B, then (a_n) divergens and \lim a_n = -\infty.Theorem: If (a_n) and (b_n) both converge, then
c_n = a_n + b_n also converges with \lim c_n = \lim a_n + \lim b_n.\forall s \in \mathbb R, the sequence c_n = s a_n converges with \lim c_n = s \lim a_n.c_n = a_n \cdot b_n converges with \lim c_n = \lim a_n \cdot \lim b_n.Squeeze theorem: If for all n \in \mathbb N, a_n \le b_n \le c_n and \lim a_n = \lim b_n=L, then \lim c_n = L.
Let (S_n) be a sequence where \forall n, S_n \ne 0 and \lim S_n \ne 0. Then \exists m > 0 such that \forall n, |S_n| > m. We say S_n is bounded away from zero.
Let (S_n) be a sequence where L = \lim S_n \ne 0 and \forall n, S_n \ne 0. Then the sequence \left(\frac{1}{S_n}\right) also converges to 1/L.
Let (a_n) and (b_n) be convergent sequences with \lim b_n \ne 0. Then
\lim \left(\frac{a_n}{b_n}\right) = \frac{\lim a_n}{\lim b_n}.
Monotone Sequences: A sequence (a_n) is increasing or non-decreasing if \forall n, a_{n+1} \ge a_n and decreasing or non-increasing if \forall n, a_{n+1} \le a_n. These sequences are monotone.
(a_n) consider the set A = \{a_n : n \in \mathbb N \}. WLOG take that a_n is increasing and bounded above. Then by the completeness axiom A has a supremum, and \lim a_n = \sup A by definition of supremum (for any \varepsilon we can find some a_N \in A where a_N > \sup A - \varepsilon, and since the sequence is monotone we get all a_n after this one are also greater and thus \forall n > N, |a_n - \sup A| < \varepsilon).
a_{n+1} = \frac{1}{2} a_n + \frac{1}{a_n} with a_1 = 2 is monotone and converges to \sqrt{2}. First show that a_n > \sqrt{2} implies thatAnd since\begin{align}\frac{1}{2} a_n + \frac{1}{a_n} &> \sqrt{2} \\ \frac{1}{4} a_n^2 + \frac{1}{a_n^2} + 1&>2 \\ \frac{1}{4}a_n^2 + \frac{1}{a_n^2}&>1 \\ \frac{1}{4}a_n^4 + 1&>a_n^2 \\ \left(\frac{1}{4}a_n^2 -1\right)a_n^2&>-1\end{align}
a_n > \sqrt{2} we know a_n^2 > 2 so \frac{1}{4} a_n^4 > 1 and so \frac{1}{4} a_n^2 - 1 > 0 and we get the desired result. So a_n is bounded below by \sqrt{2}, and we can show it is decreasing by evaluating \frac{1}{2} a_n + \frac{1}{a_n} \le a_n. Then we can say \lim a_{n+1} - \lim \frac{a_n}{2} - \lim \frac{1}{a_n} = 0, and then solve for \lim a_n! This sequence actually comes from doing newtons method on x^2 - 2.A sequence (x_n) is Cauchy iff for all \varepsilon > 0, there exists some N \in \mathbb N such that for all n,m > N, we have |x_m - x_n| < \varepsilon.
A sequence converges if and only if it is Cauchy.
Let (a_n) be a sequence, and let n_k be a strictly increasing sequence of positive integers. Then (a_{n_k}) is a subsequence of (a_n).
(a_n) converges to L, every subsequence also converges to L. The contrapositive is very useful to prove non convergence.Comparison test: Suppose \sum a_n and \sum c_n are series.
|a_n| < c_n for n \ge N_0 for N_0 \in \mathbb N and \sum c_n converges, then \sum a_n converges.a_n \ge c_n \ge 0 for n \ge N_0 for N_0 \in \mathbb N and \sum c_n diverges, then \sum a_n diverges.Cauchy condensation test: Suppose \sum a_n is a decreasing positive term series (\forall n, a_n \ge 0\land a_{n+1}\le a_n). Then \sum a_n converges if and only if S=\sum_{i=0}^\infty 2^i a_{2^i} converges.
\frac{a_1 + S}{2} =a_1 + \sum_{i=0}^\infty 2^i a_{2^{i+1}}\le \sum a_n\le \sum_{i=0}^\infty 2^i a_{2^i} = S
Easy way to prove convergence/divergence of p-series 1/n^p.
If \sum a_n converges, then \lim a_n = 0 (Cauchy criterion it).
Absolute convergence test: If \sum |a_n| converges, then \sum a_n also converges.
\sum |a_n| converges we have some N \in \mathbb N where for all \varepsilon > 0 and m,k > N, we get
so we converge by Cauchy criterion.\left|\sum_{i=m}^k a_i\right|\le\sum_{i=m}^k |a_i| = \left|\sum_{i=m}^k |a_i|\right| < \varepsilon
Limit comparison test: Let \sum a_n and \sum b_n be positive term series (a_n, b_n > 0). Suppose that \lim \frac{a_n}{b_n} = L and L > 0. Then \sum a_n converges if and only if \sum b_n converges.
\varepsilon < L and get L - \varepsilon < \frac{a_n}{b_n} < L + \varepsilon so b_n(L-\varepsilon) < a_n < b_n (L + \varepsilon). If \sum a_n converges then \sum b_n(L-\varepsilon) converges by comparison test, and if \sum a_n diverges, then \sum b_n (L+\varepsilon) diverges, so \sum b_n must diverge because otherwise \sum b_n(L+\varepsilon) = (L + \varepsilon)\sum b_n. The backward direction of the iff is true since \lim \frac{a_n}{b_n} = L \implies \lim \frac{b_n}{a_n} = L^{-1} (positive term series).Ratio test: Suppose we have a series \sum a_n where \lim \left|\frac{a_{n+1}}{a_n}\right|= L. Then \sum a_n converges if L < 1 and diverges if L > 1. N.B. that L=1 tells us nothing.
L < 1 we find |a_{n+1}| \le r |a_n| for all n > N for some N and r<1. Then |a_n| < C r^n for some C and all n, so we have the absolute sum is bounded above by a geometric series that converges, so \sum a_n converges.L > 1, we find |a_{n+1}| \ge r |a_n| for all n > N for some N and r > 1. Then |a_n| > C r^n for some C and all n, so the absolute sequence diverges and the sum definitely diverges.Root test: Let (a_n) be a sequence with L = \lim |a_n|^{1/n}. Then \sum |a_n| converges if L < 1 and diverges if L > 1. N.B. If L does not exist/converge or if L=1 this test tells us nothing.
Alternating Series: Let (a_n) be a sequence of non-negative terms. Then we say \sum (-1)^n a_n an alternating series
Alternating series test: Let (a_n) be a decreasing sequence of non-negative reals, and suppose \lim a_n = 0. Then \sum (-1)^{n+1} a_n converges.
S_k = \sum\limits_{i=1}^k (-1)^{i+1}a_i. First consider S_{2n+3} - S_{2n+1} = a_{2n+3}-a_{2n+2} \le 0 (summing the sequence 2 terms at a time, odd terms. since a_n is decreasing). Now consider the even terms S_{2n+2} - S_{2n} = a_{2n} - a_{2n+2}\ge 0. Then S_{2n} is an increasing sequence, and S_{2n+1} is a decreasing sequence. Then, since S_{2m+1} = S_{2m} + a_{2m+1}, note that we have
Thus,S_2\le S_{2m} \le S_{2m+1}\le S_1.
S_{2n} and S_{2n+1} are bounded and monotonic, so they converge. Then \lim S_{2m+1} = \lim S_{2m} + \lim a_{2m+1}, and so \lim S_{2m} = \lim S_{2m+1}. We can now prove this is \lim S_m too pretty easily.Sequence definition: A function f : U \to \mathbb R is continuous at x_0 \in U if and only if for every sequence (x_n) : \mathbb N \to U with \lim x_n = x_0, we have f(x_0) = \lim f(x_n).
\varepsilon-\delta definition: A function f : U \to \mathbb R is continuous at x_0 \in U if and only if
\forall \varepsilon > 0, \exists \delta>0,\forall x\in U, |x-x_0|<\delta \implies |f(x) - f(x_0)| < \varepsilon.
These two defs are equivalent. Sequence \implies \varepsilon-\delta is a proof by contradiction, as not \varepsilon-\delta lets us build a sequence that breaks the sequence definition. \varepsilon-\delta \implies Sequence
Pathological function: Thomae’s function is continuous on all irrationals and discontinuous on all rationals.
f(x)=\begin{cases}1/q & x \in \mathbb Q \text{ with } x = p/q \text{ and } p,q \text{ relatively prime.} \\ 0 & \text{otherwise}\end{cases}
The set of discontinuities of any function must be F_{\sigma}, the countable union of closed sets.
f(x) = \begin{cases} -x & x \in \mathbb Q \\ x & x \not \in \mathbb Q.\end{cases}
is continuous at 0 and discontinuous everywhere else.
Every bounded sequence has a convergent subsequence. Rephrasing:
A subset of \mathbb R^n is sequentially compact if and only if it is closed and bounded.
(a_n) is a bounded sequence with a_n \in [a,b] for all n. Then we can binary search as either [a,m] or [m, b] must have infinitely many terms with m \in [a,b]. Then apply Cauchy.Every convergent sequence contained in a closed interval has a limit in the interval.
All subsequences of a convergent sequence a_n converge to \lim a_n.
Extreme value theorem: If f is continuous on [a,b], then f is bounded on [a,b] and the image of [a,b] has a max and min.
f is not bounded. Define a sequence x_n \in \mathbb [a,b] where |f(x_n)| > n. By Bolzano Weierstrass, there is some convergent subsequence y_n. Then \lim y_n \in [a,b], but clearly \lim f(y_n) does not converge. Thus, f(\lim y_n) \neq \lim f(y_n) so our function f is not continuous at \lim y_n \in [a,b], a contradiction.M = \sup f([a,b]). Then we can define a sequence x_n \in [a,b] with M - f(x_n) < 1/n. Again by Bolzano Weierstrass there is some convergent subsequence y_n, and \lim y_n \in [a,b]. Since f is continuous at \lim y_n, we know M = \lim f(y_n) = f(\lim y_n), so we attain the sup. Same logic for inf.A function f:A \to \mathbb R is uniformly continuous if and only if
\forall \varepsilon > 0, \exists \delta > 0, \forall x,y \in A, |x-y|<\delta \implies |f(x)-f(y)| < \varepsilon.
Theorem: If f is continuous on a closed interval [a,b], then f is uniformly continuous on [a,b].
f is not uniformly continuous. Let \varepsilon > 0 be such that for all \delta > 0, \exists x,y \in [a,b],|x-y|<\delta \land |f(x)-f(y)|\ge \varepsilon. Define a sequence \delta_n = \frac{1}{n}, and we have sequences x_n and y_n. Since x_n, y_n \in [a,b], by Bolzano Weierstrass there is a convergent subsequence x_{n(j)}. Note that since |x_n-y_n| < \delta_n, we must have that y_{n(j)} also converges with \lim x_{n(j)} = \lim y_{n(j)} \in [a,b]. Since f must be continuous at this point, f(\lim x_{n(j)}) = f(\lim y_{n(j)}) so \lim |f(\lim x_{n(j)}) - f(\lim y_{n(j)})| = 0. This means that f(x_{n(j)}) and f(y_{n(j)}) must get within \varepsilon of each other for big enough n (small enough \delta), contradicting the construction of x_n and y_n. Our assumption was false, and f is uniformly continuous.Farbod Shokrieh, Winter 2026.
Statement/sentence
Remark: P \wedge Q is not a statement, but a “statement form”. We only get a statement when we substitute statements for P and Q like (1+1=2)\wedge(x=x).
Axioms of Set theory
X be a set and P : X \to \text{Prop} be a proposition on X. Then \exists(A \subseteq X), A = \{x \in X \mid P(x)\}.
X is required, and we can only define subsets. Otherwise we have Russel’s paradoxQuantifiers: \exists! = exists unique, etc etc
\begin{align}A = B &\iff \forall x, (x \in A \iff x \in B) \\ &\iff A \subseteq B \land B\subseteq A\end{align}
Def Subset: A \subseteq B \iff \forall x, (x\in A \implies x \in B)
Union, intersection, setminus:
A \cup B = \{x \mid x \in A \lor x \in B\},
A \cap B = \{x \mid x \in A \land x \in B\},
A \setminus B = \{x \mid x \in A \land x \not\in B\}.
Theorems on cartesian product.
(A \cup B) \times C = (A \times C) \cup (B \times C)(A \cap B) \times C = (A \times C)\cup (B \times C)(A \setminus B)\times C = (A\times C) \setminus (B\times C)\varnothing \times A = \varnothingA,B\ne \varnothing, then A\times B = B\times A \iff A = B.U_A \subseteq A and U_B \subseteq B, then U_A \times U_B \subseteq A \times B.Let S \ne \varnothing. A partition \Pi of S is a pairwise disjoint collection of subsets of S that cover S. I.e. \Pi = \{ A_i \in \mathcal P(S) \}_{i \in I} such that A_i \ne \varnothing, \forall (i \ne j \in I), A_i \cap A_j = \varnothing, and \bigcup A_i = S.
A relation R is a set R \subseteq S \times S. The inverse relation R^{-1} = \{(x,y)\mid (y,x) \in R\} (sends \ge\;\to\;\le). The complement is the negation of the relation (\ge\;\to\;\lt).
Properties of relations (\forall x,y,z \in S):
x R x.x R y \implies y R x.x R y \land y R z \implies x R z.x R y \land y R x \implies x=y.x R y \lor y R x.Hasse diagrams for orderings: a total ordering’s diagram will just be a line
Given a partition \Pi=\{A_i\}_{i\in I} on S, there is an induced equivalence relation R where xRy \iff \exists(i \in I), x,y \in A_i. In fact, all equivalence relations are induced by a partition.
\sim be an equivalence relation on S. Define the equivalence class of x\in S to be [x] = \{y \in S \mid x \sim y\}. Then the set \bigcup_{x\in S} \{[x]\} is a partition.A function f : A \to B is a relation f \subseteq A \times B such that \forall (x \in A), \exists! (y \in B), (x, y) \in f.
Examples of functions: Inclusion map \iota : A \xhookrightarrow{} B for A \subseteq B. Being a subset is the same thing as the existence of an inclusion map. Constant maps. Identity maps.
Equality of functions: f = g if they are equal as sets (i.e. domains are equal and \forall x in the domain, we have f(x) = g(x)).
f : \mathbb Z \to \mathbb Z and g : \mathbb R \to \mathbb R are both given by x \mapsto x + 1, then f \ne g, but g \mid_\mathbb{Z} = f and f is an (non-unique) extension of g from \mathbb{Z} to \mathbb R.f : \mathbb R \to \mathbb R to be equal to g : \mathbb R \to \mathbb C if they agree on all inputs.consider the image of f : A \to B, given by f(A) = \{f(x) \mid x \in A\}. We say f is surjective if f(A) = B. We say that f is injective if \forall (x,y \in A), f(x) = f(y) \implies x = y. A function f is bijective (one-to-one correspondence) if it is both injective and surjective.
We use \tan : (-\pi / 2, \pi/2) \to \mathbb R as a bijection, and f : (a, b) \to (0, 1) given by x \mapsto \frac{x-a}{b-a} is also a bijection. Thus, R \approx (a,b).
Pigeonhole principle: For n > m, there exists no injection f : \mathbb N_n \to \mathbb N_m. Proof by inducting on n, proving that \forall (m < n), \lnot \exists(\text{injective } f : \mathbb N_n \to \mathbb N_m). Take the convention that 0 \in \mathbb N, and \mathbb N_n = \{0,1,\dots, n-1\}.
n=0, there is no m<n so its trivial.\forall (m < n-1), \lnot \exists(\text{injective } f : \mathbb N_{n-1} \to \mathbb N_m). Assume FTSOC that we have m < n and injective f : \mathbb N_n \to \mathbb N_m. If m=0, we have n>0 and no function exists. Thus, WLOG assume m > 0. Consider g:\mathbb N_m \to \mathbb N_m given by
g(x) = \begin{cases}m-1 & x = f(n-1) \\ f(n-1) & x = m-1 \\ x & \text{otherwise}\end{cases}
g is clearly a bijection, and we have injective g \circ f : \mathbb N_n \to \mathbb N_m. Consider the restriction g \circ f |_{\mathbb N_{n-1}}. By the injectivity of g\circ f, there does not exist any other preimage for g(f(n-1)), so g(f(n-1)) \not\in (g\circ f)(\mathbb N_{n-1}), and m-1 is not in the image of g\circ f|_{\mathbb N_{n-1}}. Therefore, we can consider g\circ f : \mathbb N_{n-1} \to \mathbb N_{m-1} to be an injective function, contradicting the inductive hypothesis. Our assumption was false, and for all m < n there does not exist any injection f : \mathbb N_n \to \mathbb N_m.Generalized pigeonhole: If n pigeons go into m holes and n > km, then some hole will have at least k+1 pigeons. Proof: If not, every hole has at most k pigeons, so we must have had at most km pigeons, but we have n > km pigeons!
Generalization 2: Given a set A = \{a_1, \dots, a_{n+1}\} \subseteq \mathbb Z, there exist x, y \in A with x \ne y and x - y = k n for some k \in \mathbb Z (i.e. x \equiv y \mod n). Proof: Consider the remainder by n function f : A \to \mathbb N_n. If there are no two elements that share remainders, f is injective, a contradiction with pigeonhole principle.
Ramsey theory: R(3,3)=6: https://en.wikipedia.org/wiki/Ramsey%27s_theorem#R(3,_3)_=_6
Proof of PIE: |A \cup B| = |A| + |B| - |A \cap B|: do A \cup B = (A \setminus B) \cup (A \cap B) \cup (B\setminus A) as the union of disjoint sets. Then we do A = (A \setminus B) \cup (A \cap B) as the union of disjoint sets (and same for B), and we do algebra on the equations.
|A \times B| = |A| \cdot |B|: Write A \times B = \bigcup_{a \in A} \{a\}\times B as the union of pairwise disjoint sets.
Cantor’s theorem: |S| < |\mathcal P(S)|: This is true for the empty set, so only look at nonempty sets. Assume FTSOC that we have a bijection f : S \to \mathcal P(S). Consider U \subseteq S defined by U = \{x \in S \mid x \not\in f(x)\}. Then consider the element y = f^{-1}(U). We have that y \in U if and only if y \not\in f(y) by the definition of U. However, f(y) = U by the definition of y, so y \not\in f(y) if and only if y \not\in U. We have y \in U if and only if y \not\in U, a contradiction.
f:p \to (p \to \text{False}) and g : (p \to \text{False})\to p, so by LEM we either have p, in which case we use f\;p\;p to get \text{False}, or we have n : p \to \text{False} so we use n\;(g\;n) to get \text{False}. Actually, we don’t need LEM: we have h \;p := f\; p\; p so h : p \to \text{False}!!Every subset A \subseteq B of a countable set B is countable: only need to consider countably infinite A: use well-order of the naturals to repeatedly select the least element of A to put it into bijection with \mathbb N.
Miguel Morales, Spring 2026.
Simple Harmonic Motion
Energy conservation
In[14]:= x[t_] := A Cos[Sqrt[k/m] t + \[Phi]]
In[15]:= 1/2 k x[t]^2 + 1/2 m (x'[t])^2
Out[15]=
1/2 A^2 k Cos[Sqrt[k/m] t + \[Phi]]^2 +
1/2 A^2 k Sin[Sqrt[k/m] t + \[Phi]]^2Read out v_\text{max} = A \omega = A \sqrt{\frac{k}{m}}.
Vertical/external force: F = F_\text{sp} + F_g = -k (x - x_0) + F_g, can just set 0 at equilib.
Measurement
(x + \delta x)^n, ignore higher powers). Really dumb.Proportional is not the same as linear! Affine… “Unrelated” = dependent var is constant wrt. independent var.
Undamped oscillation: \omega_0 = \sqrt{k / m}.
Undamped Pendulum \omega = \sqrt{g / L}.
Underdamped oscillator: \omega = \sqrt{\omega_0^2 - \frac{b^2}{4m^2}}
Full solution is
A \exp\left({-\frac{b}{2m}t}\right) e^{i\omega t}
Wave equation
u_{tt} = \frac{F_T}{\mu} u_{xx}
or more generally
\frac{\partial^2 u}{\partial t^2} = c^2 \nabla^2 u.
Wave speed is v=\sqrt{\frac{F_T}{\mu}}. Solution is
u(x,t) = A\exp(i\omega t+ikx)+B\exp(i\omega t-ikx)
Wavenumber k = 2\pi/\lambda where \lambda is wavelength. Wave speed is v=\lambda f or \omega/k. Remember \omega = 2\pi/T where T is period, and f = 1/T.
General formula for doppler shift:
f_\text{obs} = \frac{v\pm v_\text{obs}}{v \mp v_\text{src}} f_\text{src}
where v is the speed of sound in the medium, v_\text{src} is speed of source relative to medium, and v_\text{obs} is speed of observer relative to medium. For sources moving towards each other, use +v_\text{obs}, and otherwise for sources moving away from each other use -v_\text{obs}.
Power and intensity of waves is propto amplitude^2, with proportionality constant depending on the type of wave I = \frac{P}{\text{Area}} \propto A^2
This is because we have total energy = \int_{\text{wavefront}}\frac{1}{2}kA^2\,dx where this pseudo-spring constant k has units N/m per unit area of wavefront (pressure per meter displaced?).
Decibels
\beta = 10 \log_{10}\left(\frac{I}{I_0}\right)
For waves, we take m =\text{num antinodes} - 1: we start counting from m=0,1,2,3 where m=0 is usually no wave? I think that’s how it works.
p \propto -\nabla\cdot d for some reason. Open end = pressure = 1 atm, so pressure antinode and displacement node. Closed end = 0 displacement, so displacement antinode and pressure node.Textbook convention for phase constant is \sin(\omega t + k x + \phi). Sin, + phi!!
Young’s double slit experiment and their cursed small angle approximations…
d, distance to screen L, and d <<< L.\theta from midpoint between sources, call distance from point on screen to midpoint r.\Delta r between the sources is r_2 - r_1 = d \sin\theta, right angle between the parallel lines.y along the screen is y/L = \tan \theta, and we take small angle \tan \theta = \theta.d \sin\theta = k \lambda, and we can plug in our small angle \theta and also take that \sin\theta = \theta to get d y /L = k \lambda.y = \frac{k L \lambda}{d}
For diffraction grating, the analysis is the same except you CANNOT USE THE SMALL ANGLE APPROXIMATION since d (distance between each slit) is too small. Thus, you just end up with y = L \tan \theta and d \sin \theta = k \lambda.
d. You can do spectroscopy since small changes in \lambda = relatively large changes in \theta.I_0, amplitude at interference will be N \sqrt{I_0} (cause amplitude of wave is sqrt intensity), so we have intensity I \propto N^2 I_0. With no interference (random decoherent phase shifts), we expect an intensity of I \propto N I_0 instead, so since each fringe is N times as bright they need to be 1/N the width to conserve energy? Weird ass argument.For single slit, we can use Heugen’s principle and integrate over the width of the slit a. That’s too hard though, so we can play a trick and pair each point source with a point source a/2 away from it, and notice they destructively interfere with the same conditions as the single slit case (we have an a/2 double slit, and we need the phase diff to be \lambda/2, so the two effects cancel. We can consider \frac{a}{2} \sin(\theta) = k \lambda / 2, or we can consider more pairings (maybe pair each source with one a/4 away) as \frac{a}{2k} \sin(\theta) = \lambda / 2 which is obviously the same thing).
y = \frac{k L \lambda}{a}, this time for k=\pm1,\pm2,\dots instead of starting at 0.To deal with double slit, the interference is \Delta \phi' = \frac{2\pi d}{\lambda L} y and we can use 2a\cos\left(\frac{\Delta \phi'}{2}\right)\exp(i(kx-\omega t)) where we redefine a=A_0 /\sqrt{L} or whatever to conserve energy, and we pretend the wave has the same amplitude over the entire screen? We take the intensity I \propto \text{Amplitude}^2 as
I \propto 4 |a|^2\cos^2\left(\frac{\pi d}{\lambda L}y\right)
\Delta \omega apart after a prism results in a strobing white light at \Delta \omega. Bright intensity when all lasers are in phase for constructive interference, which happens every \Delta \omega if \Delta \omega divides \omega? If \Delta \omega does not divide \omega fourier transform it. \cos\left([\omega+k\Delta \omega] \left[0 + \frac{2\pi}{\Delta \omega}m\right] \right)Guillermo Sanmarco, Winter 2026.
y'(x)+P(x)y = Q(x)
[\mu(x) y]' = \mu(x) Q(x)
\mu'(x) = \mu(x)P(x)
\mu(x) = e^{\int P(x)\,dx}
y = \frac{1}{e^{\int P(x)\,dx}}\int \mu(x) Q(x)\,dx
\sin(\theta)=-i\frac{1}{2}\left(e^{i\theta} - e^{-i\theta}\right)
\cos(\theta) = \frac{1}{2}\left(e^{i\theta} + e^{-i\theta}\right)
\therefore Ae^{iat}+Be^{-iat} \mapsto A \sin(at)+B \cos(at).
\begin{align*} A \cos(a t) + B \sin(a t) = \sqrt{A^2 + B^2} \cos(at-\varphi)\\ A= \sqrt{A^2+B^2}\cos(\phi),\qquad B=\sqrt{A^2+B^2} \sin(\phi) \end{align*}
\sin(\theta+\varphi)=\sin(\theta)\cos(\varphi)+\cos(\theta)\sin(\varphi) \qquad \cos(\theta + \varphi) = \cos(\theta) \cos(\varphi) - \sin(\theta)\sin(\varphi)
\cos(x)^2 = \frac{1}{2}
\mathcal L\{f(t)\} = s\mapsto \int_0^\infty f(t) e^{-s t}\,dt
Works for any f(t) bounded by exponential: f(t) \le e^{r t} for some r \in \mathbb R. The transformed F(s) may exist for all s or just for all s > C. The laplace transform is a bijection on the space of functions.
\mathcal L\{e^{c t}\} = \int_0^\infty e^{c t - s t}\, dt = \left[\frac{1}{c-s}e^{(c-s)t}\right]^\infty_0 = \frac{1}{s-c}
Time-domain Derivative rule (integration by parts):
\mathcal L\{f'(t)\} = \int_0^\infty f'(t)e^{-s t}\,dt = \left[f(t) e^{-s t} + s \int f(t) e^{-s t}\right]_0^\infty = s \cdot \mathcal L\{f(t)\} - f(0).
Frequency-domain Shift rule (variable substitution s \mapsto s-a):
\mathcal L\{e^{a t} f(t)\} = \int_0^\infty f(t)e^{-(s-a)t}\,dt = F(s-a)
Time-domain shift rule (u-sub, convolution theorem with dirac delta):
\mathcal L\{f(t-a)u(t-a)\} = \int_0^\infty e^{-s t} f(t-a)u(t-a)\,dt =\int_0^\infty e^{-s(t+a)}f(t)\,dt = e^{- a s} \mathcal L\{f(t)\}.
Frequency-domain derivative rule (time-domain t-multiplication): evaluate the derivative of F(s)
-\frac{d}{ds} F(s) = -\frac{d}{ds}\int_0^\infty f(t)e^{-s t}\,dt = \mathcal L\{t f(t)\}.
Gamma function:
\mathcal L \{t^n\} = \int_0^\infty t^n e^{-s t}\,dt = -\frac{1}{s}\left[e^{-s t} t^n - n\int e^{-s t} t^{n-1}\right]_0^\infty = \frac{n}{s} \mathcal L\{ t^{n-1}\} = \frac{n!}{s^{n+1}}
Convolution theorem
\mathcal L\{ f \ast g\} = F \cdot G
\mathcal L \{f \cdot g\} = F \ast G
Dirac delta \delta(t) = u'(t).
Region of convergence: In s-domain, everything to the right of the last pole converges (this is when our e^{-s t} damping can overpower our function f(t)). One sided: Initial conditions at f(0).