{"id":4068,"date":"2022-09-22T13:21:40","date_gmt":"2022-09-22T05:21:40","guid":{"rendered":"https:\/\/fanyuzhao.com\/?p=4068"},"modified":"2022-09-28T16:00:23","modified_gmt":"2022-09-28T08:00:23","slug":"review-of-maths","status":"publish","type":"post","link":"https:\/\/fanyuzhao.com\/?p=4068","title":{"rendered":"Calculus"},"content":{"rendered":"\n

For the preparation of Quants<\/h2>\n\n\n\n

1. Functions Definition<\/h3>\n\n\n\n

1.1 Each x<\/span> has only one y<\/span><\/h4>\n\n\n\n

A function denoted f (x)<\/span> of a single variable x<\/span> is a rule that assigns each<\/strong> element of a set X<\/span> ( written x \\in X<\/span> ) to exactly one element y<\/span><\/strong> of a set Y ( y\\in Y<\/span>) :
$$
y=f(x)\\quad or \\quad x\\rightarrow f(x)
$$<\/p>\n\n\n\n

1.2 Domain of f<\/span><\/h4>\n\n\n\n

$Dom f$ Domain of Function<\/p>\n\n\n\n

$Im f$ Image of Function<\/p>\n\n\n\n

For a given value of x<\/span>, there should be at most one value<\/strong> of y<\/span>.<\/p>\n\n\n\n

1.3 Implicit Form f(x,y)=0<\/span><\/h4>\n\n\n\n

For example,
$$
4y^4-2y^2x^2-yx^2+x^2+3=0
$$<\/p>\n\n\n\n

1.4 Polynomials<\/h4>\n\n\n\n

$$
y=f(x)=a_0+a_1x+a_2x^2+\u2026+a_nx^n
$$<\/p>\n\n\n\n

2. Implicit Differentiation<\/h3>\n\n\n\n

For example,
$$
y=a^x
$$
Mainly two ways to take derivatives,
$$
ln(y)=ln(a^x)=xln(a) \\
\\frac{1}{y}\\frac{dy}{dx}=ln(a)\\quad\\text{by taking derivatives to x<\/span>}\\
\\Rightarrow \\frac{dy}{dx}=y\\cdot ln(a) \\
$$
and plug y=a^x<\/span> inside
$$
\\frac{dy}{dx}=a^x\\cdot ln(a)
$$
Or, simply we apply the exponential transformation, and take deriviatives later.
$$
y=e^{ln(a^x)}=e^{x\\cdot ln(a)}
$$
However, for a polynomial, we normally have to do the implicit differentiation.
$$
4y^4-2y^2x^2-yx^2+x^2+3=0 \\\\
16y^3y’-(4y’yx^2+4y^2x)-(y’x^2+2yx)+2x=0 $$
$$(16y^3-2yx^2-x^2)y’=-2x+4y^2x+2xy \\\\
\\Rightarrow y’=\\frac{-2x+4y^2x+2xy}{16y^3-2yx^2-x^2}
$$<\/p>\n\n\n\n

3. L ‘Hospital’s Rule & Limitations<\/h3>\n\n\n\n

If there is a limitation (, which is called as the inderterminate form<\/em>),
$$
\\lim_{x \\rightarrow a} \\frac{f(x)}{g(x)}\\equiv \\frac{0}{0} \\ or \\ \\frac{\\infty}{\\infty}
$$
then, it could be calculated as,
$$
\\lim_{x \\rightarrow a} \\frac{f(x)}{g(x)}=\\lim_{x \\rightarrow a} \\frac{f'(x)}{g'(x)}=\\lim_{x \\rightarrow a} \\frac{f”(x)}{g”(x)}=\u2026=\\lim_{x \\rightarrow a} \\frac{f^{(n)}(x)}{g^{(n)}(x)}
$$
For example, \\frac{sin(x)}{x}<\/span>, at x \\rightarrow 0<\/span>.<\/p>\n\n\n\n

4. Taylor Series<\/h3>\n\n\n\n

Approximate a function a certain point, by a series of terms.(detailing explaination sees Blog Section 6<\/a> )<\/p>\n\n\n\n

We use the 1st, 2nd, 3rd, 4th, \u2026 n^th derivatives, etc, to approximate the function at a certain value.
$$
f(x)\\approx f(x_0)+(x-x_0)f'(x)|_{x=x_0}+\\frac{1}{2}(x-x_0)f”(x)|_{x=x_0}+\u2026+\\frac{1}{n!}f^{(n)}(x)|_{x=x_0}(x-x_0)^n
$$<\/p>\n\n\n\n

For example, e^x<\/span> at x=0<\/span>.
$$
e^x=1+x+\\frac{x^2}{2!}+\\frac{x^3}{3!}+\u2026+\\frac{x^n}{n!}
$$<\/p>\n\n\n\n

5. Integration<\/h3>\n\n\n\n

5.1 Intergration by Parts<\/h4>\n\n\n\n

$$
y=u(x)v(x) \\
y’=u’\\cdot v +u\\cdot v’ \\
u’v=y’-uv’ \\
$$<\/p>\n\n\n\n

and then integrate from both sides,
$$
\\int u’v dx=\\int y’ dx-\\int uv’ dx
$$
as \\int y’ dx = y+C<\/span>, so we would get,
$$
\\int u’v\\cdot dx=\\int v\\cdot du=y-\\int u\\cdot dv +C
$$
For example,
$$
\\int xe^x\\cdot dx=\\int x\\cdot de^x \\
=xe^x-\\int e^x\\cdot dx +(C) \\
=xe^x-e^x+(C)
$$<\/p>\n\n\n\n

5.2 Reduction Formula<\/h4>\n\n\n\n

We define a integral, (I_n<\/span> is called Gamma Function<\/em>)
$$
\\int_0^{\\infty}e^{-t}t^n\\cdot dt= I_n
$$
$n$ is determined as the subscript of $I_n$, and could be treated as a constant in that integral.<\/p>\n\n\n\n

We integrate that formula, and would get,
$$
n\\int_0^{\\infty}e^{-t}t^{n-1}dt=I_n \\
n\\cdot I_{n-1}=I_n
$$
If we keep doing that, we would get,
$$
I_n= n\\cdot I_{n-1}=n(n-1)I_{n-2}=\u2026=n!\\cdot I_0
$$
where,
$$
I_0=\\int_0^{\\infty}e^{-t}\\cdot dt=1
$$
so we get,
$$
I_n=n!\\cdot I_0=n!
$$<\/p>\n\n\n\n

5.3 Other Tips<\/h4>\n\n\n\n
5.3.1 ln|f(x)|<\/span><\/h5>\n\n\n\n

$$
\\int \\frac{f'(x)}{f(x)}=ln|x|+C
$$<\/p>\n\n\n\n

For example,
$$
\\int \\frac{x}{1+x^2}dx\\
=\\frac{1}{2}\\int\\frac{1}{1+x^2}dx^2=\\frac{1}{2}\\int\\frac{1}{1+x^2}d(1+x^2) \\
=\\frac{1}{2}ln|1+x^2|+C
$$<\/p>\n\n\n\n

5.3.2 Decompose the Fraction – Factorisation<\/h5>\n\n\n\n

For example,
$$
\\frac{1}{(x-2)(x+3)}=\\frac{A}{x-2}+\\frac{B}{x+3}\\
A=\\frac{1}{5},\\quad B=-\\frac{1}{5}
$$
The further implication is that.<\/p>\n\n\n\n

Any rational expression \\frac{f(x)}{g(x)}<\/span>, ( with degree of f(x)<\/span> < degree of g(x)<\/span>), could be rewritten as.
$$
\\frac{f(x)}{g(x)}\\equiv F_1+F_2 +\u2026+F_k
$$
, where each F_i<\/span> Is,
$$
F_i=\\frac{A}{(px+q)^m}\\quad or\\quad \\frac{Ax+B}{(px+q)^m}
$$<\/p>\n\n\n\n

6. Complex Number – i<\/span><\/h3>\n\n\n\n
6.1 Definition<\/h5>\n\n\n\n

$$
z=x+iy\\
i=\\sqrt{-i}\\quad, i^2=-1
$$<\/p>\n\n\n\n

and z<\/span> could be expressed in polar co-ordinate form as,
$$
z=r(cos \\theta+i\\ sin\\theta)
$$
, where
$$
x=r\\ cos\\theta \\quad, y=r\\ sin\\theta
$$
The set of all complex numbers is denoted \\mathbb{C}<\/span>; and for any complex number z<\/span>, we could write z \\in \\mathbb{C}<\/span>. ( \\mathbb{R} \\subset \\mathbb{C}<\/span> ).<\/p>\n\n\n\n

6.2 Modulus<\/h5>\n\n\n\n

The modulus<\/em> of z<\/span> donates |z|<\/span> is defined as,
$$
|z|=r=\\sqrt{x^2+y^2}
$$
<\/p>\n\n\n

\n
\"\"
Modulus<\/figcaption><\/figure><\/div>\n\n\n
6.3 Complex Conjugate<\/h5>\n\n\n\n

$$
\\bar{z}=x-iy
$$<\/p>\n\n\n\n

For example, if z=x+iy<\/span>, then \\bar{z}=x-iy<\/span>.<\/p>\n\n\n\n

6.4 Polar Form<\/h5>\n\n\n\n

$$
z=r(cos\\ \\theta+i\\ sin \\ \\theta)=re^{i\\theta}
$$<\/p>\n\n\n\n

by Euler’s Identity<\/em><\/strong>,
$$
e^{i\\theta}=cos\\ \\theta+i\\ \\sin\\ \\theta \\
e^{-i\\theta}=cos\\ \\theta-i\\ \\sin\\ \\theta \\
|z|=r,\\quad arg\\ z=\\theta
$$<\/p>\n\n\n\n

6.5 Euler’s Formula<\/h5>\n\n\n\n

The Euler’s Identity<\/em> is shown as, by applying Taylor’s Expansion<\/em> and by i^2=-1<\/span>,
$$
e^{i\\theta}=1+i\\theta+\\frac{(i\\theta)^2}{2!}+\u2026+\\frac{(i\\theta)^n}{n!}\\
=(1-\\frac{\\theta^2}{2}+\\frac{\\theta^4}{4!}+\u2026)+i\\times(\\theta-\\frac{theta^3}{3!}+\\frac{\\theta^5}{5!}+\u2026) $$
$$=cos\\ \\theta +i\\ \\sin\\ \\theta$$
Plug \\(\\theta = \\pi\\) into Euler’s Formula,
$$
e^{i\\pi}=cos\\ \\pi+ sin\\ \\pi\\
e^{i\\pi}=-1
$$<\/p>\n\n\n\n

7.Higher Derivatives<\/h4>\n\n\n\n

$$
\\frac{\\partial^2 f}{\\partial x^2}=f_{xx}=\\frac{\\partial}{\\partial x}(\\frac{\\partial f}{\\partial x}) \\ \\
\\frac{\\partial^2 f}{\\partial x \\partial y}=f_{xy}=\\frac{\\partial}{\\partial y}(\\frac{\\partial f}{\\partial x}), $$
\\(f_{xy}=f_{yx}\\) Sequence no matters if 2nd derivatives exist and continuous.<\/p>\n\n\n\n

Reference<\/h3>\n\n\n\n
HTML – Pre-Calculus<\/a>Download<\/a><\/div>\n\n\n\n