Thursday
Wednesday
There will be a review session on Thursday night from 6-8pm in Denny 216. I will also be in the Math Study center on Friday night to answer any last minute questions.
Final Exam Study Guide
- Review Sheet (click here)
- Links to old final exams (click here)
- The worksheet from class today (great review for final click here)
The solutions to the quiz will be posted as soon as I get a chance. Also, the homeworks are in a box outside of my office. I will return what you haven't dropped by to pick up on Thursday. Only 4 more days left, and then you will all be finished with Math 126!
Wednesday
Quiz 4 Study Guide
Covers sections 14.3 and 14.4
Good problems to look at:
Good problems to look at:
Section 14.3 Read page 912
Examples 1,3,6,8,9
Section 14.4 Read page 923 Example 1
Section 14.4 Read page 923 Example 1
Friday
Homework Hints:
Section 14.1 #60
Begin by writing the
function x2 + 3y2 + 5z2 = k where k is
some constant that you can change.
Divide both sides by k, and you get an ellipsoid with a = k, b = k/3, and c = k/5. Now you can sketch the ellipsoid. I used matlab to make the following figure. It is available in the Math Sciences Computing Center. I've also included a copy of my code (this is just incase you wanted to play around with the program).
Divide both sides by k, and you get an ellipsoid with a = k, b = k/3, and c = k/5. Now you can sketch the ellipsoid. I used matlab to make the following figure. It is available in the Math Sciences Computing Center. I've also included a copy of my code (this is just incase you wanted to play around with the program).
Wednesday
Monday
HAPPY PRESIDENT'S DAY!
Problem #38 section 12.5
In order to find the
equation of the line where the two planes intersect, you have to take
the cross product of the normals. This will give you a direction
vector for your line. Solving this, you should get that
v1 = <1,-2,1>
The parameteric equation of the line then becomes
x = t + x0
y = -2t + y0
z = t + z0
In order to find x0,y0, and z0, find where the planes intersect if z = 0. This case is simple because you are left with x0 = 1 and y0 = 3.
x = t + 1
y = -2t + 3
z = t
Now, you want the plane to be perpendidular to the other plane with normal vector n = <1,1,-2>. If a plane is perpendicular to another plane, then the planes contain the normal vector. Therefore, for this problem v2 = <1,1,-2> (does this make sense, if not email me with a question).
Anyway, you have two vector that lie in the plane v1 and v2. So the normal to the plane is v1 x v2 = <3,3,3>.
Now you just have to pick a point (one on the line), and plug it in. For simplicity, let t = 0
3(x-1) + 3(y-3) + 3(z-0) = 0 ----> 3x + 3y + 3z = 12
v1 = <1,-2,1>
The parameteric equation of the line then becomes
x = t + x0
y = -2t + y0
z = t + z0
In order to find x0,y0, and z0, find where the planes intersect if z = 0. This case is simple because you are left with x0 = 1 and y0 = 3.
x = t + 1
y = -2t + 3
z = t
Now, you want the plane to be perpendidular to the other plane with normal vector n = <1,1,-2>. If a plane is perpendicular to another plane, then the planes contain the normal vector. Therefore, for this problem v2 = <1,1,-2> (does this make sense, if not email me with a question).
Anyway, you have two vector that lie in the plane v1 and v2. So the normal to the plane is v1 x v2 = <3,3,3>.
Now you just have to pick a point (one on the line), and plug it in. For simplicity, let t = 0
3(x-1) + 3(y-3) + 3(z-0) = 0 ----> 3x + 3y + 3z = 12
Saturday
Also, you can get a copy of the worksheet and the solutions to what we didn't cover in section here --> Worksheet 8 (original)
Friday
Thursday
Midterm 2 Study Guide
Dr.
Perkins review guide (click
here)
Dr. Perkins old exam from 2004 (click here)
Also, I think that it is a good idea to review quiz #3 as well as the old worksheets.
Additional Office Hours:Dr. Perkins old exam from 2004 (click here)
Also, I think that it is a good idea to review quiz #3 as well as the old worksheets.
3:00 - 5:00 pm Monday Feb. 21
GUG 405d (office) or GUG 411 (computer lab)
Since Monday is a holiday, the MSC will be closed, and I will not be holding office hours from 5 - 6.
GUG 405d (office) or GUG 411 (computer lab)
Since Monday is a holiday, the MSC will be closed, and I will not be holding office hours from 5 - 6.
Saturday
Thursday
Wednesday
In general, the quizzes cover material from the previous Friday and Monday lectures. This information is detailed on the Math 126A website. You can find it here.
Quiz 3 Study Guide
Covers section 12.5, 10.1, and 10.2
Good problems to look at:
Section 12.5 Examples 4 - 7
Section 10.1 Examples 2 - 4
Section 10.2 Examples 1 - 2
Just so you guys are all in the loop, University of
Washington recently had their Vigre grant
renewed. This is a really big deal as it provides funding for
graduate students AS WELL AS UNDERGRADUATE students interested in pure
math, applied math, or statistics. Part of the funding goes
towards putting on cool workshops. The next one is called
"Origami and Math". They will present three origami models (two
modular, one a single sheet), help people make some cool stuff, and
then talk about the mathematics behind the structures, and what courses
people can consider if they would like to learn more. This event will be
from 3:45 until 4:45 in THO 234
on Friday the 18th (a week from this coming Friday). Also, there
are free snacks and soda! Section 10.1 Examples 2 - 4
Section 10.2 Examples 1 - 2
Saturday
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Section 12.5 #17
If you take the a,b,c values for the two planes, they give you the normal direction. Their cross product should give you the direction of the line of intersection
<1,1,-1> x <2, -1, 3> = <2,-5,-3>
Now, you can find a point on the line by setting z = 0 for both of the planes. You then get two equations, and two unknowns (x and y). Solve for x and y and you get x = 1, y = 1 (remember that you already set z = 0).
Together with the point you got (1,1,0), you can get the equation of the line of intersection
x = 2t + 1
y = -5t + 1
z = -3t
If you read in the book, you can't determine the equation for a plane by a line in the plane. You have to determine a normal to the plane. If we had two vectors in the plane, we could take the cross product to get a normal to the plane. One of the vectors in the plane is <2,-5,-3>. You can find a second vector from the point (-1,2,1) (the point given to be in the plane), and to a point on the line (1,1,0).
<2,-5,-3> and <1-(-1),1-2,0-1> = <2,-1,-1>
The equation of a plane is determined by normal vector and a point in the plane. So, just take the cross product of the two vectors. This gives you the values for a,b,and c. The rest follows the definition on pg 826.
If you take the a,b,c values for the two planes, they give you the normal direction. Their cross product should give you the direction of the line of intersection
<1,1,-1> x <2, -1, 3> = <2,-5,-3>
Now, you can find a point on the line by setting z = 0 for both of the planes. You then get two equations, and two unknowns (x and y). Solve for x and y and you get x = 1, y = 1 (remember that you already set z = 0).
Together with the point you got (1,1,0), you can get the equation of the line of intersection
x = 2t + 1
y = -5t + 1
z = -3t
If you read in the book, you can't determine the equation for a plane by a line in the plane. You have to determine a normal to the plane. If we had two vectors in the plane, we could take the cross product to get a normal to the plane. One of the vectors in the plane is <2,-5,-3>. You can find a second vector from the point (-1,2,1) (the point given to be in the plane), and to a point on the line (1,1,0).
<2,-5,-3> and <1-(-1),1-2,0-1> = <2,-1,-1>
The equation of a plane is determined by normal vector and a point in the plane. So, just take the cross product of the two vectors. This gives you the values for a,b,and c. The rest follows the definition on pg 826.
Thrusday
Homework Hints:
Section 12.5 #56
If you think about the two points that they give you P1(-4,2,1) and P2(2,-4,3), then the plane containing all points equidistant to P1 and P2, should cut directly inbetween the two points. The midpoint of P1 and P2 is simply
P1 + P2
midpoint = -------
2
This gives you the point (x0,y0,z0). Now all you need is a normal vector. But this should just be the vector that goes from P1 to P2. This will give you the a,b, and c values. The rest follows directly from the equation of the plane.
n . (r - r0) = 0
If you think about the two points that they give you P1(-4,2,1) and P2(2,-4,3), then the plane containing all points equidistant to P1 and P2, should cut directly inbetween the two points. The midpoint of P1 and P2 is simply
P1 + P2
midpoint = -------
2
This gives you the point (x0,y0,z0). Now all you need is a normal vector. But this should just be the vector that goes from P1 to P2. This will give you the a,b, and c values. The rest follows directly from the equation of the plane.
n . (r - r0) = 0
Wednesday
There is a study guide below. Make sure to memorize the formulas that you might need. I didn't get a chance to post the solutions to the back side of the worksheet, but this is identical to examples 3 and 4 from section 12.4. We will go over this before the quiz.
Tuesday
Quiz 2 Study Guide
Covers sections 12.3 and 12.4
Good problems to look:
The worksheet
Examples 4-7 from section 12.3
Examples 3,4, and 6 section 12.4
Friday
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--------------------------------------
If you have any homework questions, please email me.
Here are the stats for the midterm.
Midterm Stats:
Hi 50 (3 people)
Med 37
Mean 39
Problem 2 was the weakest: average grade 54%
Problems 1 and 5 were also weak, about 70% each.
Med 37
Mean 39
Problem 2 was the weakest: average grade 54%
Problems 1 and 5 were also weak, about 70% each.
Wednesday
Homework Hints:
11.12 #31
Since you didn't cover the
Binomial theorem in class, you should do a Taylor Series
approximation. It might be best to rewrite it as
q q/D2
----- - -----------
D2 (1 + d/D)2
and then use the geometric series to get a series representation for the 2nd term (in terms of d/D). Once you do that, write out the frist 2 or 3 terms to see what cancels. You should end up with something proportional to 1/D3.
11.12 #26q q/D2
----- - -----------
D2 (1 + d/D)2
and then use the geometric series to get a series representation for the 2nd term (in terms of d/D). Once you do that, write out the frist 2 or 3 terms to see what cancels. You should end up with something proportional to 1/D3.
Don't use Taylor's inequality. Instead, use the Alternating Series estimation theorem.
Thursday
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Study materials for midterm #1
- Dr. Perkins review guide (click here)
- Dr. Perkins old exam from 2004 (click here)
Thursday
Please note that there has been a change to the syllabus. Please see the new version here
Saturday
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I've gotten a couple of email questions, so I thought that I would share them and the response with all!
For 11.9 #13 on part b I came to the answer that (-1/2) times sigma n=0 to infinity (-1)^n(n+1)nx^(n-1) which was the derivative I found. But from that I don't understand how to rewrite that to equal
(1/2)Sigma n=0 to infinity (-1)^n(n+2)(n+1)x^n
This is just a re-indexing of the summation.
For example, if you write out the first 3 or so terms from your series, you get
S = -1/2*(0 - 2x + 6*x2 ...)
which is also
S = 1/2*(2x - 6x2 ...) (distributing the negative sign
and this is the same thing as (1/2)Sigma n=0 to infinity (-1)n*(n+2)*(n+1)*xn
It is typical convention to rewrite a summation so that the first term is not equal to zero. In your original summation, the first term was 0, but re-written it is not.
Formally, to do this, you would take your original series and set it from n = 1 to infinity (because at n=0, a_n = 0). Then you would reindex by introducing a new variable m = n-1. Solving from n, your get that n = m+1. So everywhere you see a n, you replace it with m+1. (Note that when n = 1, then m = 0. This is how the bottom part of the sum changes)
inf
---
(-1/2) * > (-1)n (n+1)*n*x(n-1)
---
n=1
inf
---
(-1/2) * > (-1)(m+1)*(m+2)*(m+1)*xm
---
m=0
Then you just bring the '-' sign into the summation.
----------
For 11.9 #15 how does it go from
C - (1/5) * sigma n=0 to infinity x(n+1) divided by (5n * (n+1))
to this
C - sigma n=1 to infinty xn divided by (n*5n)
Same issue as above. Try to re-index with m = n+1, and bringing the 1/5 into the summation.
Hopefully the above should help on this. The re-indexing is convention that follows 2 rules...
1. The first term in the series should not equal zero
2. The index should start from 0
If you apply rule 1 first, usually the index will start with something like n=1. Then you should apply rule 2 so that you start from something like m = 0. (Remember that the variables m,n are not important)
Friday
To make up for all of this, I will be in my office (Guggenheim 411 or 405D) on Monday from 1-5. There is only one problem, the building is locked on holidays. I'll prop one of the back doors to the building open so that you can get in.
If you have any questions, please email me.
Thursday
Since you all studied so hard for the quiz, reward yourself by procrastinating. Watch StrongBad read his email.
Wednesday
Problem 11.9 #13
The following hint might
get you going. I will work this one in section tomorrow.
Tuesday
Will be posted tonight below.
Homework Hints:
Problem 11.8 #30 (I'm doing this off the top of my head, with no book. If I mess up a number or something, please email me so that I can update it).
The first thing that you
want to note is that the radius of convergence is centered at 0.
This is because the general form is b_n * (x-a)^2, and in this
case, a = 0.
So, if the center of the interval is at x = 0, and the series is convergent for x = -4, then the radius of convergence is at least 4. But the series is divergent when x = 6, so at most the radius of convergence is 6. So, for part a, when x = 1, should the series be convergent? The answer is yes (explain why).
What about something like x = -5? Can you tell anything about that? If you are stuck send me an email.
So, if the center of the interval is at x = 0, and the series is convergent for x = -4, then the radius of convergence is at least 4. But the series is divergent when x = 6, so at most the radius of convergence is 6. So, for part a, when x = 1, should the series be convergent? The answer is yes (explain why).
What about something like x = -5? Can you tell anything about that? If you are stuck send me an email.
Sunday
Also, PLEASE note the change in Monday office hours. Due to a scheduling conflict, I had to push back my office hours on Monday to 5:00 - 6:00 pm
I reserved a cubicle next to the "126" table, so if you don't see me, check in there!
Thursday
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Also, here is the link to the solutions for the worksheet from today
http://www.amath.washington.edu/~oliveras/math126/WS2.pdf
As always, if you see any typos, please email me.
Homework Hints:
Problem 11.2 #17
It might help to think of
4^(n) as 4*4^(n-1)
That way, you can rewrite
/ -3^(n-1)\ 1 / -3 \ ^(n-1)
a_n = | --------- | = --- |-----|
\ 4^n / 4 \ 4 /
Then this just becomes a geometric series.
That way, you can rewrite
/ -3^(n-1)\ 1 / -3 \ ^(n-1)
a_n = | --------- | = --- |-----|
\ 4^n / 4 \ 4 /
Then this just becomes a geometric series.
Problem 11.1 #25
Start by recalling the
definition of a factorial.
n! = n*(n-1)*(n-2)*...*2*1
also note that
(n-1)! = (n-1)*(n-2)*...*2*1
So we can rewrite n! as follows
n! = n*(n-1)*(n-2)*...*2*1 = n*(n-1)! = n*(n-1)*(n-2)! and so on
Now you can apply this to show that (2n+1)! = (2n+1)*(2n)*(2n-1)!
n! = n*(n-1)*(n-2)*...*2*1
also note that
(n-1)! = (n-1)*(n-2)*...*2*1
So we can rewrite n! as follows
n! = n*(n-1)*(n-2)*...*2*1 = n*(n-1)! = n*(n-1)*(n-2)! and so on
Now you can apply this to show that (2n+1)! = (2n+1)*(2n)*(2n-1)!
Problem 11.5 #26
There will be a hint
available online for this problem tomorrow (Friday)
Tuesday
http://www.amath.washington.edu/~oliveras/math126/WS1.pdf
If you see any typos, please email me.