N.T.Lam's blog: How to type Math formula in blog

Thứ Bảy, 16 tháng 12, 2017

How to type Math formula in blog

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For example:
(1)

\infty \sum i = 0 \frac{x^{i}}{i!} = lim n \to \infty n \prod i = 1 (1 + a_{i} x) = x \int - \infty e^{y} d y = e^{x}

$\sum_{i=0}^{\infty} \frac{x^i}{i!} = \lim\limits_{n\to\infty}\prod_{i=1}^{n}(1+a_{i}x) = \int\limits_{-\infty}^{x} e^y dy = e^x$

(2)

x \in N \Rightarrow \exists y \in N : y = x + 1

$x\in\mathbb{N}\Rightarrow\exists y\in\mathbb{N}\colon y=x+1$ (3)

\forall x \in R, \exists y \in R : x + y = 0 \exists̸ y \in Q : y^{2} = 2

$\forall x\in\mathbb{R}, \exists y\in\mathbb{R}\colon x+y=0 \quad \nexists y\in\mathbb{Q}\colon y^2=2$
(4)When

a \neq 0

$a \ne 0$ , there are two solutions to

a x^{2} + b x + c = 0

$ax^2 + bx + c = 0$ and they are

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$ Or you may love using Online LaTeX formula editor and browser-based math equation editor
http://hostmath.com/
Another source:How-do-you-use-LaTeX-mathematical-on-Quora?

Another example:
What is the solution of the differential equation

(\frac{d y}{d x})^{2} + 4 y \frac{d y}{d x} t a n x + 4 y^{2} s e c^{2} x = c o s^{6} x + 4 y^{2}

$(\frac{dy}{dx})^2 + 4y\frac{dy}{dx}tan{x} + 4y^2sec^2{x} = cos^6{x} + 4{y}^2$

Let see how the problem will be solved by Yebraw Mada, studies Mathematics and Physics at University of Bristol (2021)
What a horrific looking equation. This may take some time to solve…
After a while of pondering I’ve realized that this is actually quite a lot simpler than it looks. We just need to rearrange the fuck out of it. Here goes:

We start with

(\frac{d y}{d x})^{2} + 4 y \frac{d y}{d x} t a n x + 4 y^{2} s e c^{2} x = c o s^{6} x + 4 y^{2}

$(\frac{dy}{dx})^2+4y\frac{dy}{dx}tan{x}+4y^2sec^2{x}=cos^6{x}+4y^2$
Let’s subtract the

4 y^{2}

$4y^2$ , giving us

(\frac{d y}{d x})^{2} + 4 y \frac{d y}{d x} t a n x + 4 y^{2} s e c^{2} x - 4 y^{2} = c o s^{6} x

$(\frac{dy}{dx})^2+4y\frac{dy}{dx}tan{x}+4y^2sec^2{x}-4y^2=cos^6{x}$
Now factorizing

(\frac{d y}{d x})^{2} + 4 y \frac{d y}{d x} t a n x + 4 y^{2} (s e c^{2} x - 1) = c o s^{6} x

$(\frac{dy}{dx})^2+4y\frac{dy}{dx}tan{x}+4y^2(sec^2{x}-1)=cos^6{x}$
Now remembering our trig identities this becomes

(\frac{d y}{d x})^{2} + 4 y \frac{d y}{d x} t a n x + 4 y^{2} t a n^{2} x = c o s^{6} x

$(\frac{dy}{dx})^2+4y\frac{dy}{dx}tan{x}+4y^2tan^2{x}=cos^6{x}$
The next step is to multiply by

c o s^{2} x

$cos^2{x}$

c o s^{2} x (\frac{d y}{d x})^{2} + 4 y \frac{d y}{d x} s i n x c o s x + 4 y^{2} s i n^{2} x = c o s^{8} x

$cos^2{x}(\frac{dy}{dx})^2+4y\frac{dy}{dx}sin{x}cos{x}+4y^2sin^2{x}=cos^8{x}$
It is not so obvious to spot the next step, but notice that the LHS factorizes perfectly.

(\frac{d y}{d x} c o s x + 2 y s i n x)^{2} = (c o s^{4} x)^{2}

$(\frac{dy}{dx}cos{⁡x}+2ysin{⁡x})^2=(cos^4{⁡x})^2$
Note that here I will just focus on the positive square root of each side. The next method will apply for the negative values too.

\frac{d y}{d x} c o s x + 2 y s i n x = c o s^{4} x

$\frac{dy}{dx}cos{x}+2ysin{x}=cos^4{x}$
Dividing through by

c o s x

$cos{x}$

\frac{d y}{d x} + 2 y t a n x = c o s^{3} x

$\frac{dy}{dx}+2ytan{x}=cos^3{x}$
Now we solve using an integrating factor. Here it is

e^{\int 2 t a n x d x}

$e^{\int{2tan⁡{x}d{x}}}$

e^{2 l n s e c x}

$e^{2lnsec{x}}$

s e c^{2} x

$sec^2{x}$
Now multiplying by this integrating factor we get

\frac{d y}{d x} s e c^{2} x + 2 y s e c^{2} x t a n x = s e c^{2} x c o s^{3} x

$\frac{dy}{dx}sec^2{x}+2ysec^2{x}tan{x}=sec^2{x}cos^3{x}$

\frac{d}{d x} (y s e c^{2} x) = c o s x

$\frac{d}{dx}(ysec^2{x})=cos{x}$
We eventually end up with

y = c o s^{2} x (s i n x + C)

$y=cos^2{x}(sinx+C)$
Phew. That was tiring. This solution can then be checked by substituting it back into the original differential equation.
Thanks for the A2A!

N.T.Lam's blog

Thứ Bảy, 16 tháng 12, 2017

How to type Math formula in blog

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