Learn Math The Easiest

My Latex is not working

June 15th, 2011

I have to apologize to those of you that use my script to render latex symbols in their blog. My website was down and then the hosting company of my site move me to another server. I didn’t remember to put everything back as it was. So the location of the script was changing. Now if you still want to use the mathtex you can use the code below. To adjust with your template change the color from black into whatever color that suitable for you (you may need to find some latex documentation to see what colors are well defined).

The weakness of the mathtex script is that there is no reliable public server for mathtex. John Forkosh doing a great job for giving this public server but sometimes if you produce a lot of math symbols from time to time there is a warning that you need to install mathtex on your own.

I will discuss how to use another method (for example mathjax that I use it myself on my question and answer forum: http://www.ask.watchmath.com)

If you still have problem with the script below please let me know.

<script src="http://www.watchmath.com/main/cgi-bin/mathtex3.js?" type="text/javascript"></script>
<script type="text/javascript">window.mathPreamble =
 '\\usepackage[usenames]{color}\\color{black} \\gammacorrection{1.3}\\png \\normal ';replaceMath( document.body );</script>
 
<center>
<a href="http://www.watchmath.com">
<img alt="" src="http://www.watchmath.com/main/images/formula.png" width="100"></a>
<br>
<a href="http://watchmath.com/vlog/?p=1461">
Install LaTeX?</a></center>

As usual here is the demo site:
http://www.watchmath.blogspot.com

Posted in Uncategorized |

15 Comments »

Fundamental Theorem of Calculus

May 12th, 2011

If you want to know for example of how to compute

$\frac{d}{dx}\int_{x^2}^{x^3}\sqrt{1+\sin t}\,dt$ is please proceed to the watch the video. If you want to understand more why the fundamental theorem is important please bear with me to read what come after this

.

First we need the statement of the fundamental theorem of calculus

FTC

Let $f$ be a continuous function on $[a,b]$ . Then the function defined on the interval $[a,b]$ by

$g(x):=\int_a^x f(t)\,dt$

is differentiable and

$g'(x)=f(x)$

Ok, so what the statement really says?

1) If $f$ is a continuous function, then $f$ has an antiderivative (on $[a,b]$ )

2) Roughly the statement says that taking integral and taking derivative cancel and give you back your original function.

Well it doesn’t sound so important! Then don’t worry about that , you can just move on with your life .

Practice Problems:

Find the derivative of

1) $\int_2^x \cos(t^2+5)\,dt$

2) $\int_{x}^5 e^{t^4+2}\,dt$

3) $\int_3^{\sin x} \sin^2(t)+t^2\,dt$

4) $\int_{e^x}^{x^3}\sqrt{\sin^2{t}+1}\,dt$

Please try to do the above problems. You may write your answer on the comment section below.

Posted in Calculus |

Tags: Calculus, Continuous function, Functions, Fundamental theorem of calculus, Riemann sum |

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The set of polynomials with integer coefficients is not a PID

April 6th, 2011

Let $I:=\{f\in \mathbb{Z}[x]\mid f(0) \text{ is even }\}$ (i.e., the set polynomials with even constant term). Show that $I$ is an ideal and show that $I\neq <g>$ for any $g\in \mathbb{Z}[x]$ .

Answer

If $f,g\in I$ then $(f+g)(0)=f(0)+g(0)=0$ . Hence $f+g\in I$ . If $r\in \mathbb{Z}$ then $rf(0)=0$ . Hence $rf\in I$ . Therefore $I$ is an ideal.

Suppose on the contrary $I=<g>$ for some $g\in\mahtbb{Z}[x]$ . Since $2\in I$ , then $2=g(x)h(x)$ for some $h(x)\in \mathbb{Z}[x]$ . By the degree consideration $g(x)$ must be a constant polynomial, and $g$ divides $2$ . It follows that $g=\pm 1$ or $g=\pm 2$ . If $g=\pm 1$ , then $I=\mathbb{Z}[x]$ which is not true since $1\not\in I$ . If $g=\pm 2$ , then $x+2\in I=<g>=<2>$ . Then $x+2=2f(x)$ for some $f\in \mathbb{Z}[x]$ which is not true since $f=x+(1/2)$ (contradiction!). Therefore $I\neq <g>$ for any $g$ .

Posted in Abstract Algebra |

1 Comment »