For the thought experiment, instead of flashcards, suppose you have a stack of cheatsheets. You wanna memorize them. You go and put the cheatsheets in a memory castle.

In mindspace, you can alter the cheatsheets – the words are no longer confined to A4 sheets, and art can be three-dimensional and include the sense of touch.

Obama's insight about blue suits can be represented by Obama wearing a blue suit, e.g. (Though the presence of a person would be overpowering, so persons should only be used for important insights. A framed Obama?)

For advanced potential, perhaps add smell to different rooms – if you focus inwardly on really feeling that smell, you kind of will smell it, regardless of what your flesh nose is sensing. And then, turn your attention to the room, and it should come back more easily. To mint new memory objects, it may help to go in meatspace to smell the appropriate thing for that room.

For example… an Indian-styled room, smelling of chai tea. Go smell some meatspace chai before even trying to envision it (because memory retcons itself, you may cause details to disappear forever).

Instead of words, you can have objects. Suppose you've read a book and written notes from it. Eventually these notes become a box of objects in your memory castle.

Now make a small change to the thought experiment. Instead of representing what were originally paper cheatsheets, your memory objects represent what were originally flashcards. These flashcards still exist in your SRS, so they are now effectively duplicated.

Now start adding a new class of flashcard to the same SRS deck.

For this new class, you can put pictures of your memory rooms. Eventually you can even draw the objects you've created and put those drawings in SRS – so your SRS contains both the original flashcard and another flashcard that refers to the first one! (Option: attach a random image to every flashcard, well before turning them into memory objects. That way you can reuse those images)

Take care, don't create too many objects. You don't need as many flashcards as you think.

Memory objects can themselves hold memory objects. For example, in one room, you might find a bicycle (a Ribbing bike). On its saddle, you might find a Buddha statue. On his lap, you might find a flower. On the petals, you might find a fly. On the fly, you might find a crown. On the surface of the crown (really zoomed in), you might find a golden castle on a flat field of gold. Inside the castle, you might find a banquet. In the banquet, you will NOT find another Ribbing bike – don't reuse objects.

For a grace period, you practice on flashcards both on actual paper/SRS, and in the mindworld. Eventually, you may detach from the software and have it all in your head.

For true spaced repetition, try a Leitner system (inside the mindworld!). Arrange your objects. Hold them, smell them – every time you handle an object. Arrange them in a system. Not necessarily numeric. Just "on that desk" is recent stuff, "upstairs on that bookshelf" is old flashcards, "in the wine cellar behind that scary vampire nest and in front of the liliputian" is some old but important stuff. Re-use things from your life. The amphitheatre, some notes on love and relationships in Momo's room, with whom you can have a conversation while you're at it. You would of course draw your imagination of Momo's room often.

Sketch, sketch, sketch. No need to sketch the memory objects in use (you can if you want), just sketch the rooms, the 'bases'.

Use emotion. Fear (the scary vampire nest) is one. A narrow bridge…

Flashcard front: Buddha Flashcard back: Buddha with rose on his lap

Flashcard front: Rose Flashcard back: Rose with a fly

etc.

• Ritual of Changing My Mind
• Flashcards

We learn this this high school: a function xⁿ has the derivative nx^n-1.

A function f(x) that composes two functions g(x) and h(x) such that one is inside the other like: g(h(x)), will have the following derivative: g'(h(x)) * h'(x).

Note that in the first factor you neglect modifying the inner function.

Example: the derivative of sin(2x) is 2cos(2x), because you are deriving two functions: sin(something), and 2x. The derivative of 2x is just 2.

Aside: As you know, the sine of x derives to the cosine of x, but there are two functions in sine of x: the sine, and the x itself. What gives? Indeed you should derive both. But x derives to 1, which you just don't need to write.

A product of two functions f(x) g(x) has the derivative f'(x) g(x) + f(x) g'(x).

For a product of three functions, derive "one at a time", leaving the rest untouched, for each term.

Take f(x).

• Reflect across the x-axis: -f(x). This should be intuitive.
• Reflect across the y-axis: f(-x)
• Reflect across both axes: -f(-x)

An even function is where f(-x) = f(x), i.e. symmetric, reflected across y-axis. An odd function is where -f(-x) = f(x). i.e. that if you reflect it across both axes, you end up with the same graph.

If there is a constant — an y-intersect offset from origin — it will fail to be odd.

A function is neither even nor odd when f(x) doesn't tell you anything about f(-x), i.e. when it isn't perfectly reflected across axes.

The terminology came from exponents of some functions, but the exponents don't necessarily determine it.

When the highest-degree exponent is odd, the function may be odd. When the highest-degree exponent is even, the function may be even.

All these are functions:

• sin x
• x²
• 2x
• x
• ln x
• e^x
• 2^x

A function evaluates to only one value for a given x.

△ABC ≅ △DEF means triangle ABC is congruent to (has the same measurements as) triangle DEF.

Compare two triangles:

1. Two pairs of sides being proportional and the angle between them the same => similar triangles.
2. Any pairs of sides being proportional => similar. (??)
3. Any pair of angles the same => similar. (??)

A way to change a rational expression into multiple terms, which is then easier for some purposes – like taking the derivative or integral, or doing a Laplace transform.

Part of elementary school in some countries…

Take an example. $\dfrac{10x^2 + 12x + 20}{x^3 - 8}$

First check to see what degree these polynomials are in. If the denominator has a lower degree, then you do a polynomial division, and then do the partial fraction expansion, okay? The technique is for stuff that you can't divide, like the remainder left by a polynomial division.

Now observe the denominator. What x makes it reach zero? That's 2. So you can factor out (x - 2), right? (Scribble a polynomial division of $x^3-8$ with $x-2$ )

We get an altered denominator: $\dfrac{10x^2 + 12x + 20}{(x - 2)(x^2 + 2x + 4)}$

You could also try to factor the second-degree expression we just got, but doing it in our head tells us it will not be possible in the reals. Skip it.

Now the aim of the technique is to set variables A and B, sometimes C etcetera:

$$\frac{10x^2 + 12x + 20}{(x - 2)(x^2 + 2x + 4)} = \frac{A}{x -2} + \frac{Bx + C}{x^2 + 2x + 4}$$

A is a constant because the denominator is first-degree. You perceive the rule: the numerator should be one degree less.

To put the right-hand side over the same denominator, expand the fractions. You know how…

$$\frac{10x^2 + 12x + 20}{(x - 2)(x^2 + 2x + 4)} = \frac{A(x^2 + 2x + 4)}{(x -2)(x^2 + 2x + 4)} + \frac{(Bx + C)(x -2)}{(x -2)(x^2 + 2x + 4)}$$

Now you can just remove the denominator from both sides. Unfactor the RHS.

Addition: lg 5 + lg 5 = lg 25 Subtraction: lg 10 - lg 5 = lg 10/5 = lg 2

Exponents: lg 2³ = 3 lg 2

Change of base formula

$$\log_2 x = \frac{\log_{10} x}{\log_{10} 2}$$

$$\log_2 x = \frac{\ln_{} x}{\ln_{} 2}$$

"What is the limit of $n \cos \frac1n \sin \frac1n$ when n goes to infinity?"

Solution: we want to do a variable substitution on the function inputs. We'll have to change the n out front so that it is something something 1/n. The expression above is equal to $\frac{\cos \frac1n \sin \frac1n}{\frac1n}$ .

Thereafter, subbing 1/n for x, we ask the limit of x as it goes to zero – because 1/infty goes to zero. We get the expression $\lim_{x\to 0} \frac{\sin x}{x} \cos x$

What happens here? The solution lies in Standard limits. The limit goes to 1 * 1 = 1.

$$\begin{array}{rrr} x \to +\infty & x \to 0 & x \to - \infty \\ (a > 1) & \frac{x^n}{a^x} \to 0 & \frac{\sin x}{x} \to 1 \end{array}$$

Differential equations involve a function and its derivative, sort of. In a "first-order" differential equation, the highest order of derivative is one. In an "ordinary" differential equation, the function is of one variable.

If there is a requirement that the solution take on a particular value at a particular input, the exercise is called a `begynnelsevärdesproblem'.

You can plot the solutions to differential equations, in the form of a grid of 'directions' in a coordinate system: the tangent that a solution function would have at a given (x,y). `Riktningsfält'.

Now, by a linear differential equation, this is meant:

Take a generic equation $y' + g(x)y = h(x)$ , and name the LHS $L(y)$ . Linearity will imply that $L(y_1 + y_2) = L(y_1) + L(y_2)$ and that $L(ky) = kL(y)$ . This is similar to linearity in other fields of math, no?

Proves the useful formula of Insattningsformeln (PB 298), which is just the basic rule of how to integrate a defined interval.

Can also be used for (PB 298)

Finding the "best" solution to a problem, like the maximum payoff or minimal material use. Involves finding asymptotes, zeros of derivatives, etc.

A function is continuous around a point if it isn't undefined nor jumps to some unexpected value at exactly that point. Mathematically, the limit from both sides of that point is equal to the value of the function at that point.

The derivatives of common functions are as follows

When you work with unknown functions and use their derivatives, it's called implicit differentiation. Nothing to be scared of.

• |xy| = |x||y| Proof:
• For |x + y|, see Triangle inequality

Taking the absolute value of a complex number lets you get rid of the "i".

Consider the graph of an absolute value linear function like |x - 1|. It has two distinct segments, and no other notation would give it that shape. It's not a smooth function, it's two functions, depending on which x you're at.

So to rewrite |x - 1| without bar notation, make cases for different x. The quick way is to see where it becomes zero, which is at x=1. For the case of x>1, you can just remove the bars. For the case of x<1, also flip the signs.

It's easier with complex numbers. Consider the case of having a complex number inside bar notation, like |a + bi|. Rewriting it as a vector gives us a circle around origin, any point on which may be touched by the vector, but it's irrelevant - we want only the radius r in re^iθ{}.

Now consider the case of some unknown complex number z, plus other stuff: |z + 1 + i|. The origin of the "vector" z will now be at -1, -i, so this circle is centred off-origin.

Example solving the above: Set z := x + yi. Say its absolute value is 2. Write the equation.

$$2 = |z + 1 + i| = |(x + 1) + (y + 1)i| = \sqrt{(x+1)^2 + (y+1)^2}$$

The last form can be recognized as Pythagoras.

After squaring, the equation resembles that of a circle:

$$\iff 4 = (x + 1)^2 + (y+1)^2$$

[2024-04-22 Mon] Created Maclaurin/Taylor series

In these cases, θ is a number between 0 and 1 that depends on x and n.

$$\begin{equation} e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots + \frac{x^n}{n!} + R_{n+1}(x) \end{equation}$$ \hfill where $\displaystyle R_{n+1}(x) = \frac{e^{\theta x}}{(n+1!)} x^{n+1}$

$$\begin{equation} \ln(1 + x) \end{equation}$$

$$\begin{equation} (1 + x)^{\alpha} \end{equation}$$

$$\begin{equation} \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots + (-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!} + R_{2n+1}(x) \end{equation}$$ \hfill where $\displaystyle R =$

$$\begin{equation} \cos x \end{equation}$$

$$\begin{equation} \arctan x \end{equation}$$

• Math

$${\sin x = \frac{1}{\csc x}} \text{\ \ \ \ \ } {\csc x = \frac{1}{\sin x}}$$
$${\cos x = \frac{1}{\sec x}} \text{\ \ \ \ \ } {\sec x = \frac{1}{\cos x}}$$

Secant and cosecant not taught in Sweden.

$${\tan x = \frac{1}{\cot x}} \text{\ \ \ \ \ } {\cot x = \frac{1}{\tan x}}$$

If you prefer expressing them in sine and cosine:

$${\tan x = \frac{\sin x}{\cos x}} \text{\ \ \ \ \ } {\cot x = \frac{\cos x}{\sin x}}$$

Interestingly, note that tan is 1/cotan, but sine is not 1/cosine.

$$\sin \theta = \cos(\frac{\pi}{2} - \theta)$$

Imagine the unit circle. Instead of starting at zero, you start at pi/2. Then, due to subtraction, you walk clockwise, not counterclockwise.

There are different definitions of the domain of inverse trig functions. Multi-valued arcsin is a thing. Mathematica in particular might define the cotangent differently from Sweden. But here are the rules for basic-uni Swedish math.

$\arcsin x: -\pi/2 \le x \le \pi/2, -1 \le y \le 1$

We can use it to find stuff about nonright triangles.

Angle A, opposing side a;

$$\frac{\sin(A)}{a} = \frac{\sin(B)}{b}$$

Num and denom can be swapped.

Can tell you stuff about nonright triangles. Like with the Law of Sines, the capital letters denote angle opposing a side denoted in lower case.

$$c^2 = a^2 + b^2 - 2ab cosC$$

The Pythagorean Theorem is a special case of this Law.

In Swedish, it's called trigonometriska ettan. cos²x + sin²x = 1²

Should be obvious if you look at the unit circle and remember the Pythagorean Theorem.

sin(2x) = 2 sin(x) cos(x)

General form: e^iθ = cos(θ) + isin(θ)

Insert π and check the value. It gives you zero sin, -1 cos. So: e^iπ = -1. You quickly realize that inputting any multiple of pi will give you a real number.

Trig angle-addition formulas can be derived: substitute (a + b) for θ. Convert to cis(a + b) form. Perform the complex multiplication. The terms with i will correspond to the angle addition formula for sin, since the sine axis in the unit circle and the imaginary axis are both vertical. The terms without i then, of course, correspond to the angle addition formula for cos.

De Moivre's formula follows from Euler's formula. (cosθ + isinθ)ⁿ is (e^iθ)ⁿ, which by simple exponent rules is e^iθn, which by definition is cos(nθ) + isin(nθ).

(cosθ + isinθ)ⁿ is equal to cos(nθ) + isin(nθ). You can state the cisθ in terms of Euler's formula to prove it.

To get the formula for tan, merely do the angle addition for sin, divided by angle addition for cos. Tan is sin/cos.

$$\frac{\sin(x) \cos(y) + \cos(x) \sin(y)}{\cos(x) \cos(y) - \sin(x) \sin(y)}$$

Multiply both numerator and denominator by $\frac{1}{\cos(x) \cos(y)}$ and you will get a simplified form.

$$\frac{\tan(x) + \tan(y)}{1 - \tan(x) \tan(y)}$$

• Math

See 9. Urval med återläggning

The basis for all combinatorics.

An array of numbers.

The "sum" of an array of numbers.

You can expand (x + y)ⁿ as follows.

${n\choose0} x^n y^0 + {n\choose1} x^{n-1} y^1 + {n\choose2} x^{n-2} y^2 + \dots + {n\choose{n-1}} x^1 y^{n-1} + {n\choose n} x^0 y^n$

This is not the same thing as (xⁿ + yⁿ)! That is a sum of e.g. cubes like x³ + 1, where the exponents are "inside the parentheses", for which there is a different formula (Sum/difference of cubes).

    1       zeroth row -- a constant: (x + y)^0
   1 1      first row --- a binomial
  1 2 1     second row -- a binomial to the second degree
 1 3 3 1    third row
1 4 6 4 1   fourth row

Finding combinations of k items from n, $${n\choose k},$$ "n choose k", is $${ n!\over{k!(n - k)!} }.$$

For example, $${10 \choose 2} = { 10!\over{2!8!}}.$$

A startling symmetry is that $${10 \choose 8} = { 10!\over{8!2!}},$$ which is the same as the previous identity. This symmetry, that ${n \choose{k}} = {n \choose{n-k}},$ works for any numbers $n$ and $k$ you want.

A combination lock should be called a permutation lock. The order matters, 2-7-1 isn't 2-1-7.

There are more permutations than combinations.

Note that 10! / 8! is the same as 10 * 9. If it's not obvious why, think.

• Math