intuition.tex

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\author{Or something \\  \small By Luc}
\title{Math by intuition}


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\section{Riemann sums}
Riemann sums are a way to approximate the area under a curve, 
given a high enough n the area will be closer and closer to the integral, 
see it as splitting up the area under a curve into n rectangles of width $\Delta x$ and height $f(x)$ and adding them up:

Given a function $f(x) = 6x$ on the interval $[0,4]$
\begin{equation}
\int_0^4 f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x
\end{equation}
where $\Delta x = \frac{4}{n}$ and $x_i = i \Delta x$.

Filling in the $f(x) = 6x$ we get the riemann sum:
\begin{equation}
\sum_{k=1}^{n} (6k \cdot \frac{4}{n}) \cdot \frac{4}{n} = \sum_{k=1}^n \frac{96k}{n^2}
\end{equation}

A useful property to remember is $\sum_{k=1}^n = \frac{n(n+1)}{2}$.

Remember that with summations you can take out any constants, and n is also a constant:
\begin{equation}
\sum_{k=1}^n \frac{96k}{n^2} = \frac{96}{n^2} \sum_{k=1}^n k = \frac{96}{n^2} \cdot \frac{n(n+1)}{2} = \frac{48n(n+1)}{n^2} = 48 + \frac{48}{n}
\end{equation}


\section{Fundamental theorem of calculus}
We will start with the following integral as an example: 

$\int_a^b f(x) dx$


The fundamental theorem of calculus gives us the following formula

\begin{equation}
    \int_a^b f(x) dx = F(b) - F(a)
\end{equation}

This means that the area under the curve between points a and b is equal to the area of 0 to b,
minus the area of 0 to a.

Some other useful formulas:
$\int_a^b f(x)dx = -\int_b^a f(x)dx$ A logical intuition is that the distance travelled $x$ at a positive velocity $v$ is
equal to the distance travelled at a negative velocity, but with a minus sign.

Another consequence of the fundamental theorem of calculus is that the derivative of an integral is the original function.
This is because the integral and derivative are eachother's inverse functions and the x in dx is the same as the x in the integral.
For example, $\frac{d}{dx}\int_a^x f(x)dx = f(x)$
Here the value of $a$ does not matter, the only rule is that the variable has to be at the top, but these can be switched around by using the previous formula.


In the real world, you might not have a single variable up there but a function of a variable, then you will have to use substitution.
For example, if you have $\frac{d}{dx}\int_a^{sin(x)} 3t^2dt$, you can use the substitution $u = sin(x)$, 
then you will have $\frac{d}{dx}\int_a^{u} 3t^2dt$.
Here, we are using the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ and $\frac{du}{dx} = cos(x)$:
Thus, we get $\frac{d}{du}\int_a^{u} 3t^2dt \cdot \frac{du}{dx}$.
The result will be $3u^2 \cdot cos(x) = 3sin^2(x)cos(x)$

\section{Integration By parts}
Integration by parts can be seen as the inverse of a chain rule and allows you to integrate functions that are not easily integrable.
The following is the formula for I.B.P.:
\begin{equation}
\int_a^b f(x) \cdot g\prime(x)dx = [f(x) \cdot g(x)]_a^b - \int_a^b f\prime(x) \cdot g(x)dx
\end{equation}
In textbooks you might also see these as $u$ and $v$ instead of $f()$ and $g()$.

If your function does not contain a multiplication you can still use IBP by setting either $f(x)$ or $g\prime(x)$ to 1
This can help as you know both the integral and derivative of 1.

As an example, take the following function:
\begin{equation}
something=here
\end{equation}

\section{Integration by Substitution}
When solving an integral, you can also use substitution to make the integral easier to solve.
For example, take the integral $\int x^3 \sqrt{x^2+1} dx$. This is hard to solve, so we take $u = x^2+1$. 
From this follows that $\frac{du}{dx} = 2x$ and $du = 2x dx$. so $xdx = \frac{1}{2}du$. 
We will also use $x^2 = (u-1)$.
There are no specific rules on what to take for u, 
but it is recommended to take a relatively small function or something that appears in multiple places in your formula as a derivative.
Solving this equation:
\begin{equation}
\begin{align*}
    \int x^3 \sqrt{x^2+1} dx =  \int x \cdot x^2 \sqrt{x^2+1} dx \\
    = \frac{1}{2} \int (u-1) \cdot \sqrt{u} du \\
    = \frac{1}{2} \int ( \frac{2}{5}u^{2.5} - \frac{2}{3}u^{1.5} ) du + C
\end{align*}
\end{equation}
And then finish it off by evaluating the integral and replacing all instances of $u$ with $x^2+1$.

When doing integrating with definite integrals you can also use substitution.
The only extra thing to keep in mind is that you will have to change the bounds
by evaluating the bounds with the substitution and then using the results.
For example, if you have a integral between 4 and 5, and you take $u = x^2$, then you will have to evaluate $4^2$ and $5^2$ and use those as the new bounds.

\section{Taylor Series}

\section{Complex numbers}

\section{Diffential equations}



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