0 obsidian/miscnotes

Table of Contents

• Lagrange Multipliers
• Cramer’s Rule
• \alpha-\beta pruning Minimax/Negamax search
  • • • Fail hard
      • Fail soft Version
    • Null windows
• Comp Arch
  • Boolean Problem (2025-01-23)
  • Buffer
    • AND
    • OR
  • Inverter
    • NAND
    • NOR
  • Curveball 1
  • Curveball 2
• Harmonic Series Stuff
  • The Actual Stuff
  • Extending Where n\le0
  • Cases for n\gt0
• Cursed U-sub
• e^x Stuff
  • Orgo Chem 2022-09-25
  • Le Chatelier’s 2022-12-07
  • Math of Kinetics
• Test Suite for TikzJax

Lagrange Multipliers

The problem: Maximize scalar function f(\vec{v}) under the constraints c_i(\vec{v}) = 0.

Solution: Solve for \nabla f(\vec{v}) = \sum_i \lambda_i \nabla c_i(\vec{v}). The values of the \lambda_i do not matter, only the value of \vec{v}.

If the gradient of f does not lie purely in the space spanned by the gradients of the constraints, then we can always increase f by moving along the constraint manifold (moving along the constraint manifold = moving in a direction perpendicular to the gradients of the constraint functions, think about contours of constant value). Adding constraints removes constraints on the gradient of f.

Cramer’s Rule

A system can be represented as

\begin{bmatrix} i & j & k \\ l & m & n \\ p & q & r \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}

^abe765

where we are solving for x, y, z and know everything else.

Because the area of a parallelepiped is base times height, (diagram most likely needed here)

\begin{vmatrix} 1 & 0 & x \\ 0 & 1 & y \\ 0 & 0 & z \\ \end{vmatrix} = z

\begin{vmatrix} x & 0 & 0 \\ y & 1 & 0 \\ z & 0 & 1 \end{vmatrix} = x

^ab78d8

For determinants, we know

\det{(A)}\cdot\det{(B)} = \det{(AB)}

[[Cramer’s Rule#^ab78d8|Focus on x]]: The [[Cramer’s Rule#^abe765|original equation]] says \begin{bmatrix} x \\ y \\ z\end{bmatrix} transformed by the matrix must be \begin{bmatrix} a \\ b \\ c\end{bmatrix}, so we know

\begin{bmatrix} i & j & k \\ l & m & n \\ p & q & r \end{bmatrix} \begin{bmatrix} x & 0 & 0 \\ y & 1 & 0 \\ z & 0 & 1 \end{bmatrix} = \begin{bmatrix} a & j & k \\ b & m & n \\ c & q & r \end{bmatrix}

^84f58c

By taking the determinants on both sides of [[Cramer’s Rule#^84f58c|this equation]] and substituting in x, we get Cramer’s rule:

\begin{vmatrix} x & 0 & 0 \\ y & 1 & 0 \\ z & 0 & 1 \end{vmatrix} = x = { \begin{vmatrix} a & j & k \\ b & m & n \\ c & q & r \end{vmatrix} \over { \begin{vmatrix} i & j & k \\ l & m & n \\ p & q & r \end{vmatrix} } }

Similar logic can be applied for y and z.

\alpha-\beta pruning Minimax/Negamax search

Returns a lower bound on the true score of a node

**Search function returns a lower bound for the real score, unless

Fail hard

// alpha = best score we can get (from whole search up until this depth)
// beta = best score our opponent can get (from whole search up until this depth)
// fail hard: Only returns values in [alpha, beta]
int search(Position pos, int alpha, int beta) {
	for (auto move : pos.legal_moves()) {
		pos.make(move);
		
		// next beta = our best move
		// next alpha = best score opponent can get (beta)
		// negatives for perspective change
		int score = -search(pos, -beta, -alpha);
		
		pos.unmake(move);

		alpha = std::max(score, alpha);
		if (alpha >= beta) {
			// higher up, the search found a better
			// branch, so this path doesn't need to be explored further
			return beta;
		}
	}

	return alpha;
}

Fail soft Version

Can return values outside of the [\alpha, \beta] range

int search(Position pos, int alpha, int beta) {
	int val = std::numeric_limits<int>::min();
	for (auto move : pos.legal_moves()) {
		pos.make(move);
		val = std::max(val, -search(pos, -beta, -alpha));
		pos.unmake(move);

		alpha = std::max(val, alpha);
		if (alpha >= beta)
			return val;
	}

	return val;
}

Null windows

Quick boolean testing

int test = -search(pos, -alpha - 1, -alpha);

• Deeper iterations experience tons of \beta cutoffs & is super quick
• Check alpha < test && test < beta

Comp Arch

Boolean Problem (2025-01-23)

\left[\left(A^C\cap B^C \cap (A\cap B)^C\right)\cup (A \cap B)\right]\cap C \cup (A\cup B)\cap (A\cap B)^C \cap C^C

Simplify the following expression:

(a+\overline{b})(b+\overline{a})(\overline{ab}\,\overline{b}\,\overline{a}+ab)c+(a+b)\,\overline{ba}\,\overline{c}

Solution: Just expand out using DeMorgan’s + axioms and eventually you will get

\begin{align} &= \overline{a}\,\overline{b}\,c + a b c + a \overline{b}\,\overline{c} + \overline{a}\,b\,\overline{c} \\ &= \boxed{a\oplus b\oplus c} \end{align}

where \oplus denotes XOR.

Derivation: Note that

a\oplus b = \overline{a}\,b+a\,\overline{b} = (a + b)\overline{ab}

We can write a simple expression for a \oplus b \oplus c by nesting these expressions, using both versions for the a\oplus b subexpression in (a\oplus b)\oplus c:

\overline{\left(\overline{a}\,b+a\,\overline{b}\right)}\;c+\left[(a+b)\overline{ab}\right]\overline{c}

Expanding the extreme left term with de Morgan’s twice:

\overline{\left(\overline{a}\,b+a\,\overline{b}\right)}=\overline{\left[a \overline{b}\right]}\overline{\overline{a}b}=(\overline{a}+b)(a+\overline{b})

However, by the idempotency we can expand it and AND it together with itself with no effect:

= (\overline{a}+b)(a+\overline{b})(\overline{a}+b)(a+\overline{b}) = (\overline{a}+b)(a+\overline{b})(\overline{a}\,\overline{b}+ba)

By the a + b = a + \overline{a}b absorption identity, we can further do

= (\overline{a}+b)(a+\overline{b})(\overline{ab}\,\overline{a}\,\overline{b}+ba)

Substituting this into the original expression, we obtain the full problem.

Buffer

A	out = A
0	0
1	1

AND

A	B	out = AB
0	0	0
0	1	0
1	0	0
1	1	1

OR

A	B	out = A+B
0	0	0
0	1	1
1	0	1
1	1	1

Inverter

A	out = \overline{A}
0	1
1	0

NAND

A	B	out = \overline{AB}
0	0	1
0	1	1
1	0	1
1	1	0

NOR

A	B	out = \overline{A+B}
0	0	1
0	1	0
1	0	0
1	1	0

Curveball 1

A	B	C	out = A(B+C)
0	0	0	0
0	0	1	0
0	1	0	0
0	1	1	0
1	0	0	0
1	0	1	1
1	1	0	1
1	1	1	1

Curveball 2

A	B	C	out = A+BC
0	0	0	0
0	0	1	0
0	1	0	0
0	1	1	1
1	0	0	1
1	0	1	1
1	1	0	1
1	1	1	1

Harmonic Series Stuff

P_n(x) = \sum_{k=1}^x {k^n}

P_n(x+d)=P_n(x)+\sum_{k=1}^d {(x+k)^n}

P_{-n}(x+d)=P_{-n}(x)+\sum_{k=1}^d {1\over{(x+k)^n}}

H(x)=P_{-1}(x)

\lim_{x\to \infty}{[H(x+d)-H(x)]} = 0

\lim_{x \to \infty} {H'(x)} =0

The Actual Stuff

\lim_{x\to \infty}H(d+x) = \lim_{x\to \infty}H(x)

\lim_{x\to \infty}{[H(d) + \sum_{k=1}^x {1\over{d+k}}]} = \lim_{x \to \infty} {\sum_{k=1}^x {1\over{k}}}

H(d)=\lim_{x\to \infty} {\sum_{k=1}^x {({1\over{k}} - {1\over{d+k}})}}

Extending Where n\le0

\lim_{x\to \infty}P_{n}(d+x) = \lim_{x\to \infty}P_{n}(x)

\lim_{x\to\infty}[P_{n}(d)+\sum_{k=1}^x {{(d+k)^n}}] = \lim_{x\to\infty}\sum_{k=1}^x{{k^n}}

P_n(d) = \lim_{x\to\infty}\sum_{k=1}^x[{k^n}-{(d+k)^n}] \quad\rm{for\space}n\le0

Cases for n\gt0

P_n(d) = \sum_{k=1}^d {k^n} = \lim_{x\to\infty}\sum_{k=1}^x[{k^n}-{(d+k)^n}] \quad\rm{for\space}n\le0

Cursed U-sub

\int_0^1 {\sqrt{1-x^2}dx}

\sin(u)=x, {dx\over{du}}=\cos{u}

\int_0^{\pi/{2}} {\sqrt{1-\sin^2{u}}\cos(u)du}=\int_0^{\pi/2} {\cos^2(u)du}

\cos^2{u} = \cos{2u} + \sin^2{u} = \cos{2u}+1-\cos^2{u}

\cos^2{u} = {1\over{2}}\cos{2u}+{1\over{2}}

w=2u, \int_0^{\pi} ({1\over{4}}\cos{w}+{1\over{4}})dw

{1\over{4}}[\sin{w}+w|^{\pi}_{0}={\pi\over{4}}

e^x Stuff

e^x = {d\over{dx}} (e^x) = \lim_{h\to 0} {{e^{x+h}-e^x}\over{h}} = e^x \lim_{h\to 0} {{e^h-1}\over{h}}

\lim_{h\to 0} {{e^h - 1}\over h}=1

\lim_{h\to 0} {e^h\over{h}} = \lim_{h\to 0} [{h\over{h}} + {1\over{h}}]

\lim_{h\to 0} e^h = \lim_{h\to 0} (h + 1)

e = \lim_{h\to 0} (h+1)^{1\over h}

finally:

\int_0^a \int_0^x f(y)\,dy\,dx=\int_0^a (a-y)f(y)\,dy

𝟙\rm{\frac{d}{dx}} = \lim_{h\to0}\frac{e^{h \rm{\frac{d}{dx}}}-\mathbb{1}}{h}

Orgo Chem 2022-09-25

R = carbon-containing thing ()

Hybridization	End	Internal
sp^2	aldehyde (\ce{R-C(=O)H})	ketone \ce{RC(=O)R'}
sp^3	alcohol (\ce{ROH})	ether \ce{R-O-R'}
Both	(carboxyllic) acid (\ce{C(=O)OH})	ester

Primary Amines: \ce{NH_2R} Secondary Amines: \ce{NHR^1R^2} Tertiary Amines: \ce{NR^1R^2R^3}

Le Chatelier’s 2022-12-07

\ce{H2(g) + 3H2(g) <-> 2NH3(g) \Delta H = -129 kJ/mol}

Equilibrium will counteract any stress

Increase in temperature

• LCP: Reaction runs in direction that is endothermic/opposite of temp change
  • Runs forward
• Q/K Analysis: “Q stays the same,” K must increase to agree with LCP

Math of Kinetics

For the reaction \ce{aA + bB + ...-> pP + ...}

Let A(t) be the concentration of the limiting reactant in the reaction. Let a be the coefficient that reactant has in the equation. Let R(t) be the rate of the reaction. Let k be any positive real number. Let n be the order of the reaction.

R(t) = k\cdot A(t)^n

A'(t) = -a\cdot R(t) = -a\cdot k\cdot A(t)^n

For zeroth-order n=0 reactions,

A'(t) = -a\cdot k

A(t) = A(0)-a\cdot k\cdot t

For first-order n=1 reactions,

A'(t) = - a\cdot k \cdot A(t)

A(t) = A(0)\cdot e^{-a\cdot k\cdot t}

For second order n=2 reactions, A'(t)=-a\cdot k\cdot A(t)^2 A(t)=(kt+A(0)^{-1})^{-1} \therefore {1\over{A(t)}} is linear with respect to time, and its slope is the value of k.

Test Suite for TikzJax

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