[b]Course 1 - Averbach & Mehta. Questions.[/b]

[florida]

I will appreciate any help with regard to the following:

Problem 6-23 (Calculus): My answer is 2 and not ( pi + 8 )/4. I think there is a mistake in the book.

Example 7-6 (Calculus): From where they conclude that:
cos x = SUM 0-inf ((-1)^n x^2n)/(2n)! and why? c sub (2n+1) = 0


Thanks

[Alya]

I will appreciate any help with regard to the following:

Problem 6-23 (Calculus): My answer is 2 and not ( pi + 8 )/4. I think there is a mistake in the book.

Example 7-6 (Calculus): From where they conclude that:
cos x = SUM 0-inf ((-1)^n x^2n)/(2n)! and why? c sub (2n+1) = 0


Thanks
I don't know what the problems are, but cos x = SUM 0-inf ((-1)^n x^2n)/(2n)! is a Taylor Series for cos(x)

[MathGuy]

(pi + 8)/4 = 2 for very small values of pi.

[retaker]

This is not why Taylor's Theorem works, but to help you remember, you might recall that cosine is an even function: cos(-x) = cos(x) \/x

If the Taylor expansion for cosine had any odd power terms, then cosine wouldn't have the above property.

[retaker]

math guy,

are you telling us that you like to: 8) for pie, in order 2 ....

[Alya]

This is not why Taylor's Theorem works, but to help you remember, you might recall that cosine is an even function: cos(-x) = cos(x) \/x

If the Taylor expansion for cosine had any odd power terms, then cosine wouldn't have the above property.

Well, he probably will be even better off knowing the general definition of the Taylor Theorem and just calculating first couple terms (cos(0) = 1, so have to start with 1, cos'(0) = 0, so no x^1, ...). Are people supposed to know how to do this for C1?

Florida,
If you need help, post the problems - there are enough people around here who probably could solve it, but don't have your book.

Wild guess -- is c_n an n-th element in the summation? If yes, then for cos(x) it is zero because odd derivatives of cos at 0 = sin(0) = 0.

[retaker]

I am not sure if they are supposed to know this, but they should be required to learn it.

..not trying to start anymore fights... :oops:

[florida]

Thanks to Alya , retaker, and math guy. I got the Example 7-6. I overlooked the definition of the cosine series. Unfortunatelly, I can't write down the problem 6-23 because it is based on a graph. But basically it says that if we have an area defined by a cardioid r=1+cos(theta) and we substract a circle with center on (0,0) of r=1, what is left? :)

[Gandalf]

Consider Quadrant1: Double integral, theta = 0 to pi/2, r = 1 to 1 + cos(theta), of r dr d theta.

= Inside integral is r^2 / 2, evaluating at limits gives [2 cos(theta) + (cos^2)(theta)] / 2.

Replace (cos^2)(theta) = (cos(2 theta) + 1)/2

So integral from 0 to pi/2 of cos(theta) + cos(2 theta)/4 + 1/4.

Integral is sin(theta) + sin(2 theta) / 8 + theta / 4

Evaluating at endpoints gives 1 + pi/8.

Since there is an equal area in quadrant 4, total is 2 + pi / 4 = (pi + 8 )/4.

[Alya]

Thanks to Alya , retaker, and math guy. I got the Example 7-6. I overlooked the definition of the cosine series. Unfortunatelly, I can't write down the problem 6-23 because it is based on a graph. But basically it says that if we have an area defined by a cardioid r=1+cos(theta) and we substract a circle with center on (0,0) of r=1, what is left? :)

i will try to explain.
(1) area of the cardoid = 3 pi/2
(2) area of the right hand side of the circle = pi / 2
(3) area of the left hand side of the cardoid = 3 pi / 4 - 2
(1) - (2) - (3) = pi/4 + 2

Now how to get (3). Area of the whole thing is 1/2 of the integral of r^2 from 0 to 2pi. Since you want only left hand side ("negative x") you integrate from pi/2 to 3pi/2. To check that your integral
is correct see http://mathworld.wolfram.com/Cardioid.html
You should get 1/2(9pi/4 - 4 - 3pi/4) = 3 pi / 4 - 2

[florida]

Thanks Gandalf. You are correct. The area remaining is on the first and fourth quadrant.

The book set the problem in the following way:

By symmetry, Area = A_2 - A_1 = 2[(1/2)Int(from pi/2 to 0) ((1+cos(theta))^2) d(theta)] - pi/2

I think it is correct except that the last pi/2 shoulb be within the brackets or outside of the brackets if it is multiplying by 2 so it will be

.........] - 2(pi/2)

Thanks again.

[florida]

I got it!!! Thank you very much Alya and Gandalf. I was making a very basic mistake!!!. It seems to me that you are well prepared for this C1 exam. Thanks for your help.

[Alya]

Alya and Gandalf...It seems to me that you are well prepared for this C1 exam
I am not taking C1 and neither does Gandalf (i think) :)

[Gandalf]

True, Alya (already FSA). And I would not have been prepared for this problem on the exam. My calculus kept producing answers not matching yours or Temple's. An Excel approximation matched Temple's, so I kept plugging.