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 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 \(x_i = \lceil y_i/2 \rceil\).
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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