http://mytechmemo.blogspot.com.au/2012/02/how-to-write-math-formulas-in-blogger.html

MathJax Examples

Note:

- I had to hunt for the "HTML/Javascript" gadget, down the list aways.
- I ended up putting the gadget in as a
*footer.* - You'll have to add that gadget to all blogs you want it to work for.
- Preview and Edit mode don't compute the TeX. You need to save the doc, then view the post.
- In compose "Options", "Line Breaks", I'm using 'Press "Enter" for line breaks.
- The "MyTechMemo" author doesn't use the exact code he suggests, though it works for me. His actual gadget is:

```
```` Powered by <a href="http://www.mathjax.org/docs/1.1/start.html">MathJax</a>`

<script type="text/javascript" src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML">

</script>

Alternate Hub Config in gadget, replace just first line.MathJax.Hub.Config({ TeX: { equationNumbers: { autoNumber: "AMS" } }, tex2jax: { inlineMath: [ ['$','$'], ["\\(","\\)"] ], displayMath: [ ['$$','$$'], ["\\[","\\]"] ], processEscapes: true } });

Using "all", numbers all equations.

"AMS" numbers only specified equations.

```
````<script type="text/x-mathjax-config">`

MathJax.Hub.Config({

TeX: { equationNumbers: {autoNumber: "all"} }

});

</script>

Equation Examples:

http://cdn.mathjax.org/mathjax/latest/test/sample-eqnum.html

Numbered Equn:

\begin{equation}

E = mc^2

\end{equation}

Align:

\begin{align}

a_1& =b_1+c_1\\

a_2& =b_2+c_2-d_2+e_2

\end{align}

Split:

\begin{equation}

\begin{split}

a& =b+c-d\\

& \quad +e-f\\

& =g+h\\

& =i

\end{split}

\end{equation}

http://cdn.mathjax.org/mathjax/latest/test/sample.html

**The probability of getting \(k\) heads when flipping \(n\) coins is:**

\[P(E) = {n \choose k} p^k (1-p)^{ n-k} \]

Raw Code:

\begin{equation} E = mc^2 \end{equation} \begin{align} a_1& =b_1+c_1\\ a_2& =b_2+c_2-d_2+e_2 \end{align} \begin{equation} \begin{split} a& =b+c-d\\ & \quad +e-f\\ & =g+h\\ & =i \end{split} \end{equation} \[P(E) = {n \choose k} p^k (1-p)^{ n-k} \]

Using AsciiMath (with TeX and MathML), http://www1.chapman.edu/~jipsen/mathml/asciimathsyntax.html

<script type="text/javascript" src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-MML-AM_HTMLorMML">

</script>

Or, without Tex:

<script type="text/javascript" src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=AM_HTMLorMML">

</script>

`x = (-b +- sqrt(b^2-4ac))/(2a) .`

Raw Code (back ticks as start/end tokens):

`x = (-b +- sqrt(b^2-4ac))/(2a) .`

Alternate start/end tokens, part of main package, doesn't work in Mathjax:

amath x^2 or a_(m n) or a_{m n} or (x+1)/y or sqrtx endamath

`x^2 or a_(m n) or a_{m n} or (x+1)/y or sqrtx`

Raw Code:

amath x^2 or a_(m n) or a_{m n} or (x+1)/y or sqrtx endamath

TeX graph example from AsciiMath, doesn't work in MathJax.

`\begin{graph} width=300; height=200; xmin=-6.3; xmax=6.3; xscl=1; plot(cos(log(x))); plot(2.2sin(x)) \end{graph}`

\begin{graph} width=300; height=200; xmin=-6.3; xmax=6.3; xscl=1; plot(cos(log(x))); plot(2.2sin(x)) \end{graph}

Example 1:

http://mytechmemo.blogspot.com.au/2012/12/determine-whether-ellipse-intersect.html

Assume an ellipse of width \(\sigma\) and length \(\kappa \sigma\) is centered at \((x_0, y_0)\), and has angle \(\theta_0\) with the \(x\)-axis. How do we determine whether it intersects a horizontal line or a vertical line?

It turns out the criteria is very simple:

- The ellipse intersects a horizontal line \(y = y_1\) if and only if the following equation holds: \[ \triangle_1 = \sigma^2 \left(\cos^2(\theta_0) + \kappa^2 \sin^2(\theta_0)\right) - (y_1-y_0)^2 \geq 0 \]
- The ellipse intersects a vertical line \(x = x_1\) if and only if the following equation holds:

\[ \triangle_2 = \sigma^2 \left(\sin^2(\theta_0) + \kappa^2 \cos^2(\theta_0)\right) - (x_1-x_0)^2 \geq 0 \]

Raw code:

Assume an ellipse of width \(\sigma\) and length \(\kappa \sigma\) is centered at \((x_0, y_0)\), and has angle \(\theta_0\) with the \(x\)-axis. How do we determine whether it intersects a horizontal line or a vertical line? It turns out the criteria is very simple:

- The ellipse intersects a horizontal line \(y = y_1\) if and only if the following equation holds: \[ \triangle_1 = \sigma^2 \left(\cos^2(\theta_0) + \kappa^2 \sin^2(\theta_0)\right) - (y_1-y_0)^2 \geq 0 \]
- The ellipse intersects a vertical line \(x = x_1\) if and only if the following equation holds: \[ \triangle_2 = \sigma^2 \left(\sin^2(\theta_0) + \kappa^2 \cos^2(\theta_0)\right) - (x_1-x_0)^2 \geq 0 \]

Example 2:

http://mytechmemo.blogspot.com.au/2012/04/color-balls.html

Thus if we let \(\mu_i\) be the expected number of steps getting to one color when the bag has i distinct colors, then we have

\[

\mu_i = 1 + \frac{i(i-1)}{n(n-1)} \mu_{i-1}+\left(1-\frac{i(i-1)}{n(n-1)}\right) \mu_i.

\]

From this we can use induction to prove that

\[

\mu_i = \frac{i-1}{i} n(n-1).

\]

Therefore we have

\[

\mu_n = (n-1)^2.

\]

Raw code:

Thus if we let \(\mu_i\) be the expected number of steps getting to one color when the bag has i distinct colors, then we have \[ \mu_i = 1 + \frac{i(i-1)}{n(n-1)} \mu_{i-1}+\left(1-\frac{i(i-1)}{n(n-1)}\right) \mu_i. \] From this we can use induction to prove that \[ \mu_i = \frac{i-1}{i} n(n-1). \] Therefore we have \[ \mu_n = (n-1)^2. \]

Example 3:

http://mytechmemo.blogspot.com.au/2011/02/nodes-of-height-h-in-n-element-heap.html

This is exercise 6.3-3 of CLRS version 2.

**Question:**Show that there are at most \(\lceil n/2^{h+1} \rceil\) nodes of height \(h\) in any \(n\)-element heap.

**Solution:**First some facts

- According to exercise 6.1-7, an \(n\)-element heap has exactly \(\lceil n/2 \rceil\) leaves.
- Notice that the nodes with height \(i\) become leaves after deleting all the nodes with height \(0,\cdots, i-1\).

Let \(y_i\) be the total number of elements of the new tree after deleting all the nodes with height \(0,\cdots, i-1\), and let \(x_i\) be the number of leaves of the new tree. Then we have \(x_i = \lceil y_i/2 \rceil\) by fact #1 and \(y_{i+1}=y_i - x_i\) by fact #2. Thus we have \(y_{i+1} \leq y_i/2\), and this leads to

$$y_i \leq n/2^i, \quad \forall n=0,1,\cdots.$$

The final conclusion follows from the relation \(x_i = \lceil y_i/2 \rceil\).

Raw code:

This is exercise 6.3-3 of CLRS version 2.Question:Show that there are at most \(\lceil n/2^{h+1} \rceil\) nodes of height \(h\) in any \(n\)-element heap.Solution:First some factsLet \(y_i\) be the total number of elements of the new tree after deleting all the nodes with height \(0,\cdots, i-1\), and let \(x_i\) be the number of leaves of the new tree. Then we have \(x_i = \lceil y_i/2 \rceil\) by fact #1 and \(y_{i+1}=y_i - x_i\) by fact #2. Thus we have \(y_{i+1} \leq y_i/2\), and this leads to $$y_i \leq n/2^i, \quad \forall n=0,1,\cdots.$$ The final conclusion follows from the relation \(x_i = \lceil y_i/2 \rceil\).

- According to exercise 6.1-7, an \(n\)-element heap has exactly \(\lceil n/2 \rceil\) leaves.
- Notice that the nodes with height \(i\) become leaves after deleting all the nodes with height \(0,\cdots, i-1\).

Example 4:

http://mytechmemo.blogspot.com.au/2010/12/gift-distribution-problem.html

There are 20 person in a Christmas party. What is the probability that at least one of them get his own gift?

Solution:

Let \(A_i\)=person i get his own gift

Then we want \(P(A_1 \cup A_2 \cup \cdots \cup A_n)\),

which is equal to

$$

\sum P(A_i)-\sum P(A_i \cap A_j)

+ \sum P(A_i \cap A_j \cap A_k) - \cdots

+ (-1)^n \sum P(A_1 \cap A_2 \cap \cdots \cap A_n)

$$

The general term is

$$

C(n, k)\cdot \frac{(n-k)!}{n!} = \frac{1}{k!}

$$

Therefore the answer is

$$

1-\frac{1}{2!}+\frac{1}{3!}-\cdots+\frac{1}{20!}

$$

Raw code:

There are 20 person in a Christmas party. What is the probability that at least one of them get his own gift? Solution: Let \(A_i\)=person i get his own gift Then we want \(P(A_1 \cup A_2 \cup \cdots \cup A_n)\), which is equal to $$ \sum P(A_i)-\sum P(A_i \cap A_j) + \sum P(A_i \cap A_j \cap A_k) - \cdots + (-1)^n \sum P(A_1 \cap A_2 \cap \cdots \cap A_n) $$ The general term is $$ C(n, k)\cdot \frac{(n-k)!}{n!} = \frac{1}{k!} $$ Therefore the answer is $$ 1-\frac{1}{2!}+\frac{1}{3!}-\cdots+\frac{1}{20!} $$

...

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