Real Math on The Web

[Math 126] Practice Shifting - Good Examples

Prof Joe — Mon, 08 Feb 2010 13:49:52 -0500

B) Let's have this function [i]f[/i]... Graph the following...Try it! 1. $f(x-2)$ 2. $f(x)-4$ 3. $f(x+1)+2$ 4. $-f(x)$ 5. $f(-x)$ [img]http://www.joemath.com/blog/file.php?11,file=25[/img]

[Math 126] Quiz 1

Prof Joe — Fri, 05 Feb 2010 00:31:02 -0500

Here are the solutions for Sections 1.1, 1.2 [url=http://www.joemath.com/blog/file.php?11,file=23]Quiz 1[/url] as a [i][color=#FF0021] .pdf[/color][/i] file. [i][color=#0033FF]Click on link above.[/color][/i]

[Math 182] Re: Chapter 9 Practice

Prof Joe — Fri, 05 Feb 2010 00:06:12 -0500

$[[1],[2]]^T \cdot [ [1,2],[3,4]] = [[1+6],[2+8]]^T =[[7],[10]]^T $

[Math 182] Re: binomial series

Prof Joe — Wed, 16 Dec 2009 22:37:27 -0500

The binomial series should be in bag of basic math tools but I won't test it on Thursday. Make sure you know how to find the radius of convergence, interval of convergence and and how to make a taylor/mclaurin polynomial in 9.7. Nothing from 9.9.

[Math 182] binomial series

David vang — Wed, 16 Dec 2009 16:15:17 -0500

i was wondering if we need to know how to do the binomial series for the test since it was in ch9. also, did you omit 9.9, representation of functions by power series?

[Math 182] Chapter 9 Practice

Prof Joe — Wed, 02 Dec 2009 16:25:12 -0500

(:P) I made two practice exercises. [url=http://www.joemath.com/blog/file.php?0,file=18]Chapter 9 Practice I[/url] [url=http://www.joemath.com/blog/file.php?0,file=19]Chapter 9 Practice II[/url] `lim_{gpa rarr 0} text(EE) = text(HOA)`

[Math 182] Christy

Prof Joe — Sun, 01 Nov 2009 21:02:07 -0500

`\sum_{n=4}^\infty 1/n` `\int_4^infty 1/(x-3) dx = ln |x-3| = \infty - ln 1 = \infty ` `0 < a_n \le b_n` 1. if `\sum b_n` con, then `\sum a_n` does too. 2. if `\sum a_n` div, then `\sum b_n` does too. `\sum_{n=1}^{\infty} 1/(2+3^n)` `\sum_{n=1}^{\infty} a x^{n-1} = a / (1-x) if |x| < 1` `\sum_{n=1}^{10} 2^{n-1} = [1-2^10]/[1-2] = (-1023)/(-1)` `a_n = 1/[2+3^n] < 1/3^n = b_n` by the direct coparison test, since the series `1/3^n` converges, then `1/[2+3^n]` also converges `lim_{n->\00} (a_n/b_n) = L` `lim_{n->\00} ([5n+1]/[3n^2]/[1]/[n])` `lim_{n->\00} ([5n+1]/[3n^2]*[n]/[1]) =lim_{n->\00} ([5n^2+n]/[3n^2]*[n]/[1])=lim_{n->\00} ([5n^2+n]/[3n^2]) =5/3 = L` `b_n` is a divergent harmonic series. BY LCT, a_n divergeres `\sum [1]/[3n^2-4n+5]` `\sum_{n-1}^\infty [sqrt(n)]/[n^2+1]` `lim_{n->00} (b_n)/(a_n) =lim_{n->00} ([sqrt(n)]/[n^2+1])/(1/n^{3/2}) `=1, by lct, `\sum_{n=0}^\infty [2^n]/[n!]` `lim_{n->\00} [2^{n+1}]/[(n+1)!]/[2^n]/[n!]` `lim_{n->\00}[2^{n+1}n!]/[(n+1)!2^n] = [2n!]/[(n+1)n!]` `lim_{n->\00} 2/[n+1] = 0`

[Math 182] Fall 2009 Test 1Solutions

Prof Joe — Tue, 20 Oct 2009 00:01:31 -0400

Here are the solutions for test 1. read them and weep! [url=http://www.joemath.com/blog/file.php?0,file=17]Test 1 Solutions[/url] (:P)X(

[Math 182] question

amaroke — Mon, 05 Oct 2009 23:14:23 -0400

professor i was looking for the take home quiz solutions on the site, i cant find it. thanks angad

[Math 182] Re: any chance there is a missing "2" in #3?

Prof Joe — Mon, 14 Sep 2009 22:05:54 -0400

I am not sure what you are doin9. If `u = 2x-1` then `du = 2dx \rArr (du)/2 = dx` Just replace `dx` with `1/2du` Should be fine.B) angela_kallus Wrote: ------------------------------------------------------- > I have done # 3 five times and I keep getting an > extra 2 in the denominator when I differentiate > the answer. :( > > if that 2 was in the denominator in the original > integral, all would be well. > > anybody?

[Math 182] Re: any chance there is a missing "2" in #3?

Prof Joe — Mon, 14 Sep 2009 22:01:20 -0400

Make sure you put \$ signs or \` accent marks around your code. `u= sqrt(2x-1)` and `1/6u^3+1/2u+C` I also droped the () because they are not needed. David vang Wrote: ------------------------------------------------------- > what did you use for "u"? i used u= sq root(2x-1) > and ended up with . i check my > answer with the calculator and it's the same.

[Math 182] Re: any chance there is a missing "2" in #3?

David vang — Mon, 14 Sep 2009 04:22:14 -0400

what did you use for "u"? i used u= sq root(2x-1) and ended up with 1/6(u^3)+1/2(u)+C. i check my answer with the calculator and it's the same.

[Math 182] any chance there is a missing "2" in #3?

angela_kallus — Sat, 12 Sep 2009 17:16:58 -0400

Homework chapter 5/6 #3 due 9/17/09 I have done # 3 five times and I keep getting an extra 2 in the denominator when I differentiate the answer. sad smiley if that 2 was in the denominator in the original integral, all would be well. anybody?

[Math 182] any chance there is a missing "2" in #3?

angela_kallus — Sat, 12 Sep 2009 17:14:20 -0400

I have done # 3 five times and I keep getting an extra 2 in the denominator when I differentiate the answer. :( if that 2 was in the denominator in the original integral, all would be well. anybody?

[Math 126] Example of Math Stuff...

Prof Joe — Fri, 11 Sep 2009 02:46:36 -0400

Here is [url=http://www.joemath.com/blog/file.php?6,file=16]Assignment 1[/url] as a [b][i][color=#FF0021] .pdf[/color][/i][/b] file. [b]Interesting fact?[/b] [i]True or False[/i] What is $\pi$ ? I claim it is this... $\[6 \sum_{n=1}^{\infty} 1/n^2 \]^{1/2} = \[6(1+1/4+1/9+1/16+ \cdots +)\]^{1/2} = \[6+6/4+6/9+6/16+ \cdots +\]^{1/2} = \[6+3/2+2/3+3/8+ \cdots +\]^{1/2}=pi$ Go to [joemath.com] for homework problems. How did Prof Joe make links, bold text, etc...? I used [url=http://www.bbcode.org/]bbcode.org[/url]. Example: It [size=14]g[/size][size=18]r[/size][size=22]o[/size][size=26]o[/size][size=28]w[/size][size=32]s[/size]!!! It is possible to colour the text [color=red]red[/color] [color=green]green[/color] [color=blue]blue[/color] - or [color=#DB7900]whatever[/color]

[Math 182] Re: Graphing Challenge (5 points)

Prof Joe — Fri, 14 Nov 2008 02:13:30 -0500

:S [img]http://www.joemath.com/blog/file.php?7,file=15[/img] Function: `f(x)=((x-2)^2-4)/((x-2)^2+4)` x intercepts: `f(0)=f(4)=0` are given and actually work! Just set the numerator to zero. `(x-2)^2-4=0 -> (x-2)^2 =4 rArr x=2+-2 rArr x= 0,4` Now, take the first derivative:`f'(x)=(16(x-2))/([(x-2)^2+4]^2)` C = 2 is the critical point. Note: `(x-2)^2+4 != 0` since a positive (or zero) + positive > 0`. It is decreasing on `(-oo,2)` and increasing on `(2,oo)` since `f'(x) < 0` when `x < 2` and `f'(x) > 0 ` when ` x> 2`. `\:. C = 2` gives an Absolute Minimum by first derivative test, `(2, -1)`. The second derivative is more work but manageable. `f''(x)=(-16(3x^2-12x+8))/[(x-2)^2+4]^3`. Setting `3x^2-12x+8=0` gives `k = 2 +- 2/3sqrt(3)` The points of inflection are `(2 - 2/3sqrt(3), -1/2)` and `(2 + 2/3sqrt(3), -1/2)` I used `k = 1, 3` as approximations. i.e. `(1,-1/2)` and `(3,-1/2)`. From problem: ` f''(1)=f''(3)=0` and `f''(0)0,f''(4)oo)f(x)=lim_(x->oo)((x-2)^2-4)/((x-2)^2+4)=lim_(x->oo)(x^2-4x)/(x^2-4x+8)=lim_(x->oo)((x^2)/(x^2)-(4x)/(x^2))/((x^2)/(x^2)-(4x)/(x^2)+8/(x^2))=lim_(x->oo)(1-0)/(1-0+0)=1 ` `lim_(x->-oo)f(x)=lim_(x->-oo)((x-2)^2-4)/((x-2)^2+4)=lim_(x->-oo)(x^2-4x)/(x^2-4x+8)=1 ` :P

[Math 182] Graphing Challenge (5 points)

Prof Joe — Fri, 17 Oct 2008 03:12:32 -0400

Suppose we have a graph with the following characteristics: Can you graph it?? Note: I approximated the POI's for simplicity sake. `f(0)=f(4)=0, f(2)=-1` `f'(2)=0, f'(x)

[Math 181] The Graphing Challenge (5 points)

Prof Joe — Fri, 17 Oct 2008 03:10:49 -0400

Suppose we have a graph with the following characteristics: Can you graph it?? Note: I approximated the POI's for simplicity sake. `f(0)=f(4)=0, f(2)=-1` `f'(2)=0, f'(x)

[Math 181] Re: First Derivative/ Second Derivative test

Prof Joe — Tue, 14 Oct 2008 22:40:00 -0400

The second derivative finds concavity (and min/max). `f'(x) = [9-2x^2]/[(9-x^2)^(1/2)]` `f''(x) = [(9-x^2)^(1/2)(-4x)-(9-2x^2)(1/2)(9-x^2)^(1/2-1)(-2x)]/[((9-x^2)^(1/2))^2 ]` `f''(x) = [-4x(9-x^2)^(1/2)+x(9-2x^2)(9-x^2)^(-1/2)]/[9-x^2 ]= {x(9-x^2)^(-1/2)\[-4(9-x^2)^(1)+(9-2x^2)\]}/[9-x^2 ]` `f''(x) = {x(9-x^2)^(-1/2)\[-36+4x^2+9-2x^2\]}/[9-x^2 ] = {x(-27+2x^2)}/[(9-x^2)^(3/2) ]` Set `f''(x)=0` to find possible points of inflection. This means either `x = 0 ` or `27-2x^2 = 0`, but `27-2x^2 = 0 rArr x^2 = 13.5 rArr x = +- sqrt(13.5) ~~ +- 3.67` which is outside the domain. Therefore, c = 0, is the only possible point of inflection. Since `f''(-1) ={-1(-27+2(-1)^2)}/[(9-(-1)^2)^(3/2) ] = [-(-25)]/[8^(3//2)] = [25]/[8^(3//2)] > 0` Concave UP on [-3,0] Since `f''(1) ={1(-27+2(1)^2)}/[(9-(1)^2)^(3/2) ] = [-25]/[8^(3//2)] = [-25]/[8^(3//2)] < 0` Concave DOWN on [0,3] Hence, (0,0) is a point of inflection. [img]http://www.joemath.com/blog/file.php?7,file=11[/img]

[Math 181] First Derivative/ Second Derivative test

Prof Joe — Tue, 14 Oct 2008 19:31:34 -0400

Suppose `f(x)=x sqrt(9-x^2)` First, look at the domain of the graph: Since we have square root we know that the radicand, the thing under the radical has to be non-negative. $9-x^2 \ge 0$ or $9 \ge x^2 \rArr$ $x^2 \leq 9$ Take square root of both sides... $\sqrt(x^2) \leq \sqrt(9)$ $ |x| \leq 3 \rArr -3 \le x \le 3 $ which means `[-3,3]`. NOTE: `f(x)` has zeros at `x = -3, 0, 3.` i.e. x-intecepts at (-3,0), (0,0), and (3,0). `f'(x) = (1)(9-x^2)^(1/2) +x(1/2)(9-x^2)^(1/2-1)(-2x)` `f'(x) = (9-x^2)^(1/2) -x^2(9-x^2)^(-1/2)` Factor out the term with smallest exponent `f'(x) = (9-x^2)^(-1/2)\[(9-x^2)^1 -x^2\]` Multiply out to verify `f'(x) = [9-x^2-x^2]/[(9-x^2)^(1/2)] = [9-2x^2]/[\sqrt(9-x^2)]` Now set `f'(x) = 0` and solve for x. The critical occur when the numerator is zero, `9-2x^2 = 0 rArr x^2 = 9/2 rArr x = +- 3/sqrt(2) `, and when `f'(x)` is undefined at `x=+- 3`. NOTE: -3 and 3 are also endpoints, therefore possible max or min. [IMG]http://www.joemath.com/blog/file.php?7,file=12[/img] By the first Derivative test, we have a relative minimum at `-3/sqrt(2)` and a relative maximum at `3/sqrt(2)`. Both of these turn out to be an absolute minimum and an absolute maximum. Why?? Now the graph... [IMG]http://www.joemath.com/blog/file.php?7,file=11[/IMG]

[Math 181] Review for test 1

Prof Joe — Tue, 30 Sep 2008 11:51:34 -0400

B) 1.`\quad y=x/(x+1)` `\quad \quad dy/dx = [(x+1)(1)-x(1)]/[(x+1)^2] = [x+1-x]/[(x+1)^2]=[1]/[(x+1)^2]` 2. `f(x)=x^2 ` find `f'(x)` using `\quad d/dxf(x)=lim_(h->0)(f(x+h)-f(x))/h` `\quad \quad f'(x)=lim_(h->0)(f(x+h)-f(x))/h = lim_(h->0)[(x+h)^2-x^2]/h = lim_(h->0) (x^2+2hx+h^2-x^2)/h = lim_(h->0) (2hx+h^2)/h =lim_(h->0) (h(2x+h))/h =lim_(h->0) (2x+h) = 2x+0 =2x` 3. `\quad lim_(x->4) (-4x^2+2x-3) = -4(16)+2(4)-3 = -64+8-3=-59` 4. `\quad y=sec(x)`, find `(d^2y)/dx^2`. `\quad \quad dy/dx = sec(x)\quad tan(x)`, `\quad \quad \quad (d^2y)/dx^2 = [sec(x) tan(x)]tan(x) + sec(x) sec^2(x) = sec(x)tan^2(x)+sec^3(x)` 5. `\quad lim_(x->0) sin(3x)/x = lim_(x->0) 3sin(3x)/(3x) = 3lim_(x->0) sin(3x)/(3x) = 3(1) = 3` 6. `\quad x^2y-xy^2+tan(x) =10 ` Use implicit differentiation. `\quad \quad 2xy+x^2 dy/dx -((1)y^2 +x2y dy/dx) + sec^2(x)=0` `\quad \quad 2xy+x^2 dy/dx -y^2 -2xy dy/dx + sec^2(x)=0` `\quad \quad 2xy -y^2 + sec^2(x)=2xy dy/dx-x^2 dy/dx` `\quad \quad 2xy -y^2 + sec^2(x)=(2xy -x^2) dy/dx` `\quad \quad [2xy -y^2 + sec^2(x)]/(2xy -x^2)= dy/dx` 7. `\quad f(\theta) = 1/cos(\theta)` `\quad \quad f'(\theta) = sec(\theta)=[cos(\theta)(0)-(1)(-sin(\theta))]/[cos^2(\theta)] = sin(\theta) / [cos^2(\theta)]=1 /cos(\theta) \cdot sin(\theta)/cos(\theta) =sec(\theta) tan(\theta) ` 8. `\quad ` Just evaluate. `lim_(x->2^-) [x+3]/[x-3]=(2+3)/(2-3)=-5` 9. `\quad x^2 +y^2 =10`, ` (-1,3)`. `\quad \quad 2x + 2y dy/dx = 0` `\quad \quad dy/dx = -x/y = -(-1)/3=1/3` Find line: `y - 3=1/3(x-(-1))` gives ` x-3y=-10`. 10.`\quad y=2/x^3 = 2x^(-3)` `\quad \quad y' = -3(2)x^(-3-1) = -6x^(-4)`. `\quad \quad y'' = -4(-6)x^(-4-1) = 24x^(-5) = 24/x^5`.

[Math 181] Chain Rule

Prof Joe — Wed, 24 Sep 2008 00:36:01 -0400

X( Chain rule: Do you really understand the chain rule? If `y=sqrt(2x-3)`, what is `dy/dx = dy/(du) (du)/dx`? HINT: Let `u=2x-3 `, then `(du)/dy = 2`. Rewrite: `y=(2x-3)^(1//2) = u^(1//2)`, then `dy/(du)=1/2u^(1//2-1)=1/2u^(-1//2)`. `dy/dx = dy/(du) (du)/dx = 1/2 u ^(-1//2) \cdot 2 = 1/2(2x-3)^(-1//2) \cdot 2 = (2x-3)^(-1//2)=1/(2x-3)^(1//2) = 1/sqrt(2x-3)` Some observations: `(2x-3)^(-1//2)` or `1/(2x-3)^(1//2)` or `1/sqrt(2x-3)` are all acceptable answers. Most text books and software like to leave the answers in the same form as the original expression. i.e. If starts as a radical, then leave as a radical. `dy/(du)` means the derivative of `y` with respect to `u`. `(du)/dx` means the derivative of `u` with respect to `x`.

[Math 181] Review for quiz 2 solutions

Prof Joe — Mon, 22 Sep 2008 13:32:36 -0400

Quiz 2 Solutions Find the limit, if exists. You must show work!! 1. `lim_(x->0)(sinx)/(3x) =[1]/[3]\[ lim_(x->0)(sinx)/(x) \] = [1]/[3]\cdot 1 = [1]/[3]` 2. `lim_(x->3) sqrt(x-3)= text{dne}` since the limit does not exist coming from the left. 3. `lim_(x->1^-)(|x-1|)/(x-1) = lim_(x->1^-)[-(x-1)]/[x-1] =lim_(x->1^-)(-1) = -1 ` 4. `lim_(x->3)(sqrt(x-3)-5)/(x-4) =(sqrt(3-3)-5)/(3-4) = [0-5]/[-1] = 5` 5. `lim_(x->\pi/6)tan(2x) = tan(2\cdot \pi/6) = tan (\pi/3) =[sin (\pi/3) ]/[cos (\pi/3) ]= [sqrt(3)/2]/[1/2]=sqrt(3)` Find the derivative of each function. 6. `f(x)=3x^3-2x-\pi` $f(x)$ 7. `y=sqrt(x)+root3(x^2)` 8. `h(x) = 3cosx` 9. `k(x)=cotx` 10. `f(x)=ax^(n-1)` Solutions posted later....(:P)

[Math 181] Please post questions or discussion

Prof Joe — Fri, 19 Sep 2008 20:23:01 -0400

X( Please post math stuff here. Ask questions or just chat. You may answer each others questions. :S I will check it daily. Please register so I can privately email or publicly post. B) Not everything needs to be public.

[Math 182] Re: Ask Questions here.

Prof Joe — Thu, 18 Sep 2008 11:58:40 -0400

(tu) $sum_(k=0)^n((n),(k)) x^{n-k}y^k=((n),(0))x^n+((n),(1))x^{n-1}y+\cdots+((n),(n)) y^n$

[Math 182] Re: Test 2 problem solution

Prof Joe — Mon, 05 Nov 2007 21:11:32 -0500

1.` int cos^3x*sin^4x dx = int cos^2x*cos x*sin^4x dx = int (1-sin^2x)*cos x*sin^4x dx=int (1-sin^2x)*sin^4x*cos x dx` Let `u=sinx` then `du=cosx dx` `int (1-u^2)u^4 du=int (u^2-u^6) du=1/3u^3-1/7u^7+c=1/3sin^3+1/7sin^7+c` 2. `int(sinx)/(sqrt(cosx))dx ` Let `u=cosx, du=-sinx dx` `int(-1)/(sqrt(u))du=-intu^(-1/2)du=-2/1u^(1/2)+c=-2sqrt(cosx)+c` 3.`intx^3e^(2x)dx` Hint: Use the table method with `u=x^3,dv=e^(2x)dx` 4. `int1/(sqrt(25-x^2))dx` 5. `int(x^2-x+9)/((x^2+9)^2)dx` 6. `lim_(x->oo) x^2/e^x`

[Math 182] Test 2 problem

Prof Joe — Sat, 03 Nov 2007 16:41:15 -0400

(tu) Try these? Post the solutions... 1. `intcos^3x*sin^4x dx` 2. `int(sinx)/(sqrt(cosx))dx` 3.`intx^3e^(2x)dx` 4. `int1/(sqrt(25-x^2))dx` 5. `int(x^2-x+9)/((x^2+9)^2)dx` 6. `lim_(x->oo) x^2/e^x`

[Math 182] Test 1 Solutions

Prof Joe — Sat, 13 Oct 2007 02:44:24 -0400

X( Click here to get solutions to test 1. [url=http://www.joemath.com/blog/file.php?6,file=10]Test 1 Solutions[/url] Come back and post questions. (:P)

[Math 182] Ask Questions here.

Prof Joe — Sat, 13 Oct 2007 00:28:56 -0400

B) You can post complicated math symbols here. `sin^2(2x)=(1-cos[2(2x)])/(2) = (1-cos(4x))/(2) ` `tan(2x)=(tanx+tanx)/(1-tanx*tanx)=(2tanx)/(1-tan^2x)` `e^x=sum_(n=0)^oo(x^n)/(n!) ` So `e^1 = 1/(0!)+1/(1!)+1/(2!)+1/(3!)+1/(4!)+1/(5!)+cdots+1/(n!)+cdots~~ 1+1+1/2+1/6+1/24+1/120+1/720 ~~ 2.718055cdot` WE know that `e=2.18281828459045_cdots` Increase n and we get a better approximation. How cool. :P(tu)

[Statistics 152] Please Read - Homework Assignments Needed

Angelo — Thu, 22 Mar 2007 15:25:30 -0400

I have been feeling horrible this week and have been unable to attend class. If anyone could please post what has been assigned for homework for March 20 + march 22 (this tuesday and thursday) I would very much appreciate it. Thank you