assignment 14

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

Using the standard form y-y1=m(x-x1) we obtain the equation

y-3=-2(x-5), which simplifies to

y =-2x+13

Evaluating at x = 4.9 we obtain

Y =-2(4.9)+13

=-9.8+13

=3.2.

This is identical, as must be the case, with the first-degree Taylor polynomial y = 3-2(x-5), which simplifies to y = -2 x + 13.

This is close to both the second- and third-degree approximations, but not as accurate as either. The third-degree approximation is the most accurate.

The second-degree polynomial adds .05 to the linear approximation, while the third-degree polynomial adds another .005 to the second-degree approximation.

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

These are the polynomials for sin(x). To get the polynomials for sin(t) / t you would have to substitute t for x, then divide by t. The to answer the question given in the problem, you would have to integrate the resulting polynomial over the given interval.

The degree 3 approximation of sin(t) is sin(t) = t - t^3 / 6, approx.

So sin(t) / t = 1 - t^2 / 6, approx.

The degree 5 approximations are sin(t) = t - t^3 - 6 + t^5 / 120 approx., and sin(t) / t = 1 - t^2 / 6 + t^4 / 120

Antiderivatives would be

integral( sin(t) / t) = t - t^3 / 18 approx. and

integral( sin(t) / t) = t - t^3 / 18 + t^5 / 600, approx.

The definite integrals would be found using the Fund Thm. You would get

1 - 1/18 = .945 approx. and

1 - 1/18 + 1/600 = .947 approx.

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

GOOD STUDENT RESPONSE:

I think that a trapezoid approximation won`t work well because of the shape of the graph of the sin t. Also, it definently won`t work for our initial expression from the book, because the fn is undefined at t=0. The altitudes of each line go from being too small to too large and so on. Perhaps that is the reason, but I am not completely certain.

INSTRUCTOR RESPONSE: The biggest problem is the undefined value at t = 0, which you recognize in your answer. Very good.

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

ln(1 + 2x) is a composite function. Its successive derivatives are a little more complicated than the derivatives of most simple function, but are not difficult to compute, and form a pattern.

ln(1 + 2x) = f(g(x)) for f(z) = ln(z) and g(x) = 1 + 2x.

f ' (z) = 1 / z, and g ' (x) = 2. So the derivative of ln(1 + 2x) is

f ' (x) = (ln(1 + 2x) ) ' = g ' (x) * f ' ( g(x) ) = 2 * (1 / g(x) ) = 2 / ( 1 + 2x).

Then

f ''(x) = (2 / ( 1 + 2x) )' = -1 * 2 * 2 / (1 + 2x)^2 = -4 / (1 + 2x)^2.

f '''(x) = ( -4 / (1 + 2x)^2 ) ' = -2 * 2 * (-4) = 16 / (1 + 2x)^3

etc.. The numerator of every term is equal to the negative of the power in the

denominator, multiplied by 2 (for the derivative of 2 + 2x), multiplied by the previous numerator. A general expression would be

f [n] (x) = (-1)^(n-1) * (n - 1)! * 2^n / ( 1 + 2x) ^ n.

The (n - 1)! accumulates from multiplying by the power of the denominator at each step, the 2^n from the factor 2 at each step, the (-1)^n from the fact that the denominator at each step is negative.

Evaluating each derivative at x = 0 gives

f(0) = ln(1) = 0

f ' (1) = 2 / (1 + 2 * 0) = 2

f ''(1) = -4 / (1 + 2 * 0)^2 = -4

f '''(1) = 16 / (1 + 2 * 0)^2 = 16

f[n](0) = (-1)^n * (n - 1)! * 2^n / ( 1 + 2 * 0) ^ n = (-1)^n * (n-1)! * 2^n / 1^n = (-1)^n (n-1)! * 2^n.

The corresponding Taylor series coefficients are

f(0) / 0! = 0

f'(0) / 1! = 2

f''0) / 2! = -4/2 = -2

...

f[n](0) = (-1)^n (n-1)! * 2^n / n! = (-1)^n * 2^n / n

(this latter since (n-1)! / n! = 1 / n -- everything except the n in the denominator cancels).

So the Taylor series is

f(x) = 0 + 2 * (x-0) - 2 ( x - 0)^2 + 8/3 ( x - 0)^3 - ... + (-1)(^n) * 2^n /n * ( x - 0 ) ^ n

= 0 + 2 x - 2 x^2 + 8/3 x^3 - ... + (-1)(^n) * 2^n /n * x^n + ...

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

The derivatives of g(x) = ln(x) are

g'(x) = 1/x

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

g'''(x) = -2 / x^3

g''''(x) = 6 / x^4

...

g[n](x) = (-1)^n * (n-1)! / x^n

yielding the Taylor series about n = 1:

ln(x) = (x - 1) - (x - 1)^2 / 2 + (x-1)^3 / 3 - (x - 1)^4 / 4 ... + (-1)^(n-1) (x-1)^n / n + ...

To get ln(1 + x), we can just substitute 1 + 2x for x in the above. It follows that

ln(1 + 2x) = (1 + x - 1) - (1 + x - 1)^2 / 2 + (1 + x - 1)^3 / 3 - (1 + x - 1)^4 / 4 ... + (-1)^(n-1)(1 + x-1)^n / n + ...

= x � x^2 / 2 + x^3 / 3 � x^4 / 4 + � + (-1)^(n-1) x^n / n.

The function ln(1 + 2x) is obtained by just substituting 2x into the previous:

ln(1 + 2x) = 2x - (2x)^2 / 2 + (2x)^3 / 3 - (2x)^4 / 4 ... + (-1)^(n-1) ( 2x )^n / n + ...

or writing out the terms more explicitly

ln(1 + 2x) = 2 x � 2^2 x^2 / 2 + 2^3 x^3 / 3 � 2^4 x^4 / 4 + � + (-1)^(n-1) x^n / n + �

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

See my notes. It would be a very good idea for you to insert detailed self-critiques on most of these questions (denoted by *&*& so I can easily identify your self-critiques) and send me the resulting document.