Fundamental Mathematics/Calculus

Calculus

Mathematical operations that perform on function, equation

Calculus mathematics

Change in variables

For any function f(x) . Over the interval of $x$  to $x+\Delta x$

Change in variable x

$\Delta x=(x+\Delta x)-x$

Change in function f(x)

$\Delta f(x)=f(x+\Delta x)-f(x)$

Rate of change

Rate of change

${\frac {\Delta f(x)}{\Delta x}}={\frac {f(x+\Delta x)-f(x)}{\Delta x}}$

Limit

$\lim _{x\to a}f(x)$

Finite Limit We call $L$  the limit of $f(x)$  as $x$  approaches $c$  if $f(x)$  becomes arbitrarily close to $L$  whenever $x$  is sufficiently close (and not equal) to $c$  .

When this holds we write

$\lim _{x\to c}f(x)=L$

or

$f(x)\to L\quad {\mbox{as}}\quad x\to c$

Infinite Limit We call $L$  the limit of $f(x)$  as $x$  approaches infinity if $f(x)$  becomes arbitrarily close to $L$  whenever $x$  is sufficiently large.

When this holds we write

$\lim _{x\to \infty }f(x)=L$

or

$f(x)\to L\quad {\mbox{as}}\quad x\to \infty$

Similarly, we call $L$  the limit of $f(x)$  as $x$  approaches negative infinity if $f(x)$  becomes arbitrarily close to $L$  whenever $x$  is sufficiently negative.

When this holds we write

$\lim _{x\to -\infty }f(x)=L$

or

$f(x)\to L\quad {\mbox{as}}\quad x\to -\infty$

Differentiation

Let $f(x)$  be a function. Then

${\frac {d}{dt}}f(t)=f'(x)=\sum \lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}$  wherever this limit exists.

In this case we say that $f$  is differentiable at $x$  and its derivative at $x$  is $f'(x)$  .

Integration

Mathematics operation on a continuous function to find its area under graph . There are 2 types of integration

$\int f(x)dx=F(x)+C$

Where $F$  satisfies $F'(x)=f(x)$

Suppose $f$  is a continuous function on $[a,b]$  and $\Delta x={\frac {b-a}{n}}$  . Then the definite integral of $f$  between $a$  and $b$  is

$\int \limits _{a}^{b}f(x)dx=\lim _{n\to \infty }A_{n}=\lim _{n\to \infty }\sum _{i=1}^{n}f(x_{i}^{*})\Delta x$

Where $x_{i}^{*}$  are any sample points in the interval $[x_{i-1},x_{i}]$  and $x_{k}=a+k\cdot \Delta x$  for $k=0,\dots ,n$  .}}

Solving differential equations

Given

${\frac {d}{dt}}f(t)=-sf(t)$
${\frac {df(t)}{f(t)}}=-sdt$
$\int {\frac {df(t)}{f(t)}}=\int -sdt$
$Lnf(t)=-st+c$
$f(t)=e^{-st+c}=Ae^{-st}$
$A=e^{c}$

In summary

 Ordered differential equation Equation of the form Root of equation 1st ordered differential equation ${\frac {d}{dt}}f(t)=-sf(t)$ $f(t)=Ae^{-st}$ 2nd ordered differential equation ${\frac {d^{2}}{dt^{2}}}f(t)=-{\sqrt {s}}f(t)$ $f(t)=Ae^{\pm j{\sqrt {s}}t}$ $n$ th ordered differential equation ${\frac {d^{n}}{dt^{n}}}f(t)=-n{\sqrt {s}}f(t)$ $f(t)=Ae^{\pm jn{\sqrt {s}}t}$ Solving Ordered Differential Equations

1st ordered differential equation

Equation of general form

$A{\frac {d}{dx}}f(x)+Bf(x)=0$

After arrangement, equation above becomes

${\frac {d}{dx}}f(x)=-sf(x)$  Where $s={\frac {B}{A}}$

Equation has a root

$f(x)=Ae^{-st}$

2nd ordered differential equation
$A{\frac {d^{2}}{dx^{2}}}f(x)+B{\frac {d}{dx}}f(x)+C=0$
${\frac {d^{2}}{dx^{2}}}f(x)+{\frac {B}{A}}{\frac {d}{dx}}f(x)+{\frac {C}{A}}=0$
$s^{2}f(x)+2\alpha sf(x)+\beta f(x)=0$
$s^{2}f(x)=-2\alpha sf(x)-\beta f(x)$

The solution of the 2nd ordered polynomial equation above

 One real root $s=-\alpha$ $i(t)=Ae^{-\alpha t}$ Two real roots $s=-\alpha \pm {\sqrt {\beta -\alpha }}$ $i(t)=Ae^{(-\alpha \pm {\sqrt {\alpha -\beta }})t}$ One complex root $s=-\alpha \pm j{\sqrt {\beta -\alpha }}$ $i(t)=Ae^{(-\alpha \pm j{\sqrt {\beta -\alpha )}}t}$ With

$A(\alpha )=Ae^{-\alpha t}$
$\omega ={\sqrt {\beta -\alpha }}$
$\beta ={\frac {1}{T}}={\frac {1}{LC}}$
$\alpha =\gamma \beta$
$T=LC$
$\gamma =RC$

Partial Differential Equations

${\frac {\partial }{\partial t}}f(t)=-sf(t)$

Integral Transformation

Any function f(t) can be transform into Laplace function or Fourier function by using Laplace transform or Fourier transform

 $f(t)$ $F(s)$ $F(j\omega )$ $f(t)$ $\int f(t)e^{-st}dt$ $\int f(t)e^{-j\omega t}dt$ ${\frac {d}{dt}}$ $s$ $j\omega$ $\int dt$ ${\frac {1}{s}}$ ${\frac {1}{j\omega }}$ Example

 $f(t)$ $F(s)$ $F(j\omega )$ $L{\frac {d}{dt}}$ $sL$ $j\omega L$ ${\frac {1}{L}}\int dt$ ${\frac {1}{sL}}$ ${\frac {1}{j\omega L}}$ Go to the School of Mathematics