Assignments 15  16

course Mth 174

Like i said in the work i sent to you just now i could not find the notes on any of the sections that i just worked on the CDs. I dont know if they didnt send them to me or what but they all work just not the right notes. I havent had any problems before but since i used the first two disks full of notes now i can not seem to find any CAL II notes on any of the disks.

There should be relevant notes on CDs #4 and #5. Check out the CD Contents link from your Assignments Page (look at about the fifth line of the Assignments Page).

See my email about the disks.

??n??????g???~w???Physics II

11-07-2007

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23:45:52

Query 10.4.8 (was 10.4.6 3d edition) (was p. 487 problem 6) magnitude of error in estimating .5^(1/3) with a third-degree Taylor polynomial about 0.

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well for starters we need to find the 4th derivative

f^4 (x) = 0.5^(1/3)

Abs (f^4(x)) less than 0.5^(1/3) = sqart 0.5 is less than 2 for -0.5 less than x less than 0.5

M = 2

abs(E3) = abs(f(x) - P3(x)) less than 2/4(x)^4

0.5^(1/3) = 1 - x^(1/3) + x^(1.5) + x^(2.1) + ........

>You have the right idea but you don't have the correct derivatives for the 1/3 power.

** The maximum possible error of the degree-3 polynomial is based on the fourth derivative and is equal to the maximum possible value of the n = 4 term.

The function is x^(1/3).

Its derivatives are

f'(x) = 1/3 x^(-2/3),

f''(x) = -2/9 x^(-5/3),

f'''(x) = 10/27 x^(-8/3),

f''''(x) = -80/81 x^(-11/3).

We think in terms of expanding about x = 1, since all these derivatives are undefined at x = 0, and since it is easy to substitute 1.

The maximum possible absolute value of the fourth derivative f''''(x) = -80/81 x^(-11/3) between x = .5 and x = 1 occurs at x = .5, where we obtain | f '''' (x) | = 80/81 * (.5^-(11/3) ) = 12.54 approx; to be sure we have a valid limit on the error we will use the slight overestimate M = 13.

So the maximum possible error is | 13 / 4! * (.5 - 1)^4 |, or about .034. **

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23:46:21

What error did you estimate?

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i estimated an error of 2

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23:46:32

What function did you compute the Taylor polynomial of?

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in the very first screen this is shown

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23:46:40

11-07-2007 23:46:40

What expression did you use in finding the error limit, and how did you use it?

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NOTES -------> in the very frist screen

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00:05:00

explain how you proved the result.

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I just followed the cos(x) therom that was shown in the book but just changed the signs cause sin is just the opposite of cos so that is all i did. Found that the series ended up being sin(x) = 1 + x^2/2 - x^4/4 + x^6/6 - x^8/8 + .......... for all x.

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00:05:53

What is the error term for the degree n Taylor polynomial?

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am very confused on this problem because i tried to get the notes of the CDs that were sent to me by the book store and i can not find these notes on any of the CDs. i mean all the CDs work but cant find the right notes.

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00:05:56

Can you prove that the error term approaches 0 as n -> infinity?

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** For even n the nth derivative is sin(x); when expanding about 0 this will result in terms of the form 0 * x^n / n!, or just 0.

If n is odd, the nth derivative is +- cos(x). Expanding about 0 this derivative has magnitude 1 for all n. So the nth term, for n odd, is just 1 / n! * x^n.

In terms of the theorem for the error term, we see that none of the coefficients exceeds the maximum magnitude M = 1.

For any x, lim(n -> infinity} (x^n / n!) = 0, because lim { n -> infinity} ( [ x^(n+1) / (n+1)! ] / [ x^n / n! ] ) = lim (n -> infinity) ( x / n ) = 0; the limit is zero since x is fixed and n increases without bound.

This needs to be put together formally in terms of the definition of the error term.

We get En(x) = M / (n+1)! * x^(n+1) = 1 / (n+1)! * x^(n+1) with M + 1, which as n -> infinity approaches zero. Since the error term approaches zero as n -> infinity, the series converges for all x. **

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00:05:58

What do you know about M in the expression for the error term?

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00:05:59

How do you know that the error term must be < | x | ^ n / ( n+1)! ?

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00:06:01

How you know that the limit of | x | ^ n / ( n+1)! is 0?

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00:13:38

what are the first four nonzero terms of the series?

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for 1 / sqrt (1 + theta) is

1/ sqrt(1 + theta) = (1 + theta)^(-1/2) = 1 - (1/2)theta + ((-1/2)(-3/2)/2) * theta^2 + ((-1/2)(-3/2)(-5/2)/3) * theta^3 + ........

= 1 - (1/2)theta + (3/8)theta^2 - (5/16)theta^3 + ......

Now we substitute in sin(theta) in for theta

1/ sqrt(1 + sin(theta) = theta - (1/2)(-theta^3/3) + (3/8)(theta^5/5) - (5/16)(-theta^7/7) + ...........so it ends up being 1 + theta

We substitute the Taylor expansion of sin(x) into the Taylor expansion of 1 / sqrt(u) and ignore terms with powers exceeding 4:

Expanding y = sqrt(x) about x = 1 we get derivatives

y ' = 1/2 x^(-1/2)

y '' = -1/4 x^(-3/2)

y ''' = 3/8 x^(-5/2)

y '''' = -5/16 x^(-7/2).

Substituting 1 we get values 1, 1/2, -1/4, 3/8, -5/16.

The degree-4 Taylor polynomial is therefore

sqrt(x) = 1 + 1/2 (x-1) - 1/4 (x-1)^2 / 2 + 3/8 (x-1)^3/3! - 5/16 (x-1)^4 / 4!.

It follows that the polynomial for sqrt(1 + x) is

sqrt( 1 + x ) = 1 + 1/2 ( 1 + x - 1) - 1/4 (1 + x-1)^2 / 2 + 3/8 (1 + x-1)^3/3! - 5/16 (1 + x-1)^4 / 4!

= 1 + 1/2 (x) - 1/4 (x)^2 / 2 + 3/8 (x)^3/3! - 5/16 (x)^4 / 4!.

Now the first four nonzeo terms of the expansion of sin(theta) give us the polynomial

sin(theta) = theta-theta^3/3!+theta^5/5!-theta^7/7!.

We note that since sin(theta) contains theta as a term, the first four powers of sin(theta) will contain theta, theta^2, theta^3 and theta^4, so our expansion will ultimately contain these powers of theta.

We will expand the expressions for sin^2(theta), sin^3(theta) and sin^4(theta), but since we are looking for only the first four nonzero terms of the expansion we will not include powers of theta which exceed 4:

sin^2(theta) = (theta-theta^3/3!+theta^5/5!-theta^7/7!)^2 = theta^2 - 2 theta^4 / 3! + powers of theta exceeding 4

sin^3(theta) = (theta-theta^3/3!+theta^5/5!-theta^7/7!)^3 = theta^3 + powers of theta exceeding 4

sin^4(theta) = (theta-theta^3/3!+theta^5/5!-theta^7/7!)^4 = theta^4 + powers of theta exceeding 4

Substituting sin(theta) for x in the expansion of sqrt(1 + x) we obtain

sqrt(1 + sin(theta)) = 1 + 1/2 (sin(theta)) - 1/4 (sin(theta))^2 / 2 + 3/8 (sin(theta))^3/3! - 5/16 (sin(theta))^4 / 4!

= 1 + 1/2 (theta-theta^3/3! + powers of theta exceeding 4) - 1/4 (theta^2 - 2 theta^4 / 3! + powers of theta exceeding 4)/2 + 3/8 ( theta^3 + powers of theta exceeding 4) / 3! - 5/16 (theta^4 + powers of theta exceeding 4)/4!

= 1 + 1/2 theta - 1/4 theta^2/2 - 1/2 theta^3 / 3! + 3/8 theta^3 / 3! + 2 / ( 4 * 3! * 2) theta ^4 - 5/16 theta^4 / 4!

= 1 + 1/2 theta - 1/8 theta^2 + 1/16 theta^3 - 5/128 theta^4.

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00:13:43

Explain how you obtained these terms.

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just did

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00:13:51

What is the Taylor series for `sqrt(z)?

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in first screen

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00:13:59

What is the Taylor series for 1+sin(`theta)?

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previous screen

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00:14:03

How are the two series combined to obtain the desired series?

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00:15:09

Query Add comments on any surprises or insights you experienced

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there were many surprises in this section. For one like mentioned before i could not find the notes for these sections. Havent had any problem before buti had to switch to another disk and i looked through all them and could not find the right notes for this. I dont know where they are or if i even have the right disk so i am kinda lost now.

????x???o????Cy????assignment #016

??n??????g???~w???Physics II

11-08-2007

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00:25:08

what is the fourth degree Fourier polynomial?

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to look at the fourier polynomial in the fourth degree we look at these equations

a4 = 1/pi f(x)cos(4x)dx from -pi to pi = 1/pi 1 cos(4x)dx from 0 to pi = 0

b4 = 1/pi f(x) sin(4x)dx from -pi to pi = 1/pi 1 sin(x)dx from 0 to pi = 2/4pi so the approximation is given by

f(x) = F4(x) = (1/2) + 2pi(sin(x)) + (2/3pi)sin(3x) + (2/4pi)sin(4x)

We first shift the interval (- pi , pi) to the interval (0, 1).

To shift the interval (-pi, pi) to (-1/2, 1/2) you would replace x by 2 pi x. To then shift the interval to (0, 1) you would in addition substitute x-1/2 for x. The function sin(k x) would become sin(2 pi k ( x - 1/2) ) = sin(2 pi kx - k pi). For even k this would be just sin(2 pi k x); for odd k this would be -sin(2 pi k x).

The integral of the function itself over the interval is 1, so you would have a0 = 1/2.

The integral of x * sin(2 pi k x) from 0 to 1 is easily found by integration by parts to be 1 / (2 pi k) (details of the integration: let u = x, dv = sin(2 pi k x) dx; v = -1 / (2 pi k) cos(2 pi k x), so the integral of v du is a multiple of sin(2 pi k x) and hence yields 0 at both limits; the u v term is x cos(2 pi k x) / (2 pi k), which yields 0 at the left limit and -1 / (2 pi k) at the right limit).

It follows that the first five terms of the series would be 1/2, -1/(2 pi), -1/(4 pi), -1 / (6 pi) and -1 / (8 pi).

This yields the Fourier polynomial

1/2 - (1/pi)sin(2pix) - 1/(2pi)sin(4pix) - 1/(3pi)sin(6pix) - 1/(4pi)sin(8pix).

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00:25:11

Describe the graph of this polynomial on [0,1).

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00:25:18

What substitution did you use to compensate for the fact that the period of the function is not 2 `pi?

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shown in the first screen

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00:33:32

Query problem 10.5.24 (3d edition 10.5.24) (was 9.5.24) integral of cos^2(mx) from -`pi to `pi is `pi

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for starters we say that m does not = 0 and that k and m do not equal each other

we start out by multiplying the Fourier series by cos(mx) where m is any positive integer

f(x) cos(mx) = a0cos(mx) + sum (ak)cos(kx)cos(mx) + sum (bk)sin(kx)cos(mx)

now we integrate it between -pi and pi

it is the same above equation with just the integrals in there from -pi to pi and with dx and the end of all the sections of the equation

so m does not = 0 the integral - pi to pic cos(mx)dx = 0

so we end up with

f(x)cos(mx)dx from -pi to pi = am cos(mx)cos(mx)dx from -pi to pi = pi(am)

m = 1,2,3........ in the following formula

am = 1/pi f(x)cos(mx)dx from -pi to pi

to get the bk value we multiple through with sin(mx) instead of the cos(mx) and come up with the equation bm = 1/pi f(x)sin(mx)dx from -pi to pi

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00:33:40

which formula from the table did you used to establish your result and what substitution did you use?

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on the previous screen

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00:34:08

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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Not to many surprises in this section but still didnt have the notes but i learned most of it from the book and i felt pretty good about this section.

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See my notes and be sure to let me know if you have additional questions.