Assignment 14

course Mth 174

{€Šf·×©Í|ÔËا`ß¼°ÈÀ|y‚ýþÿassignment #014

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Physics II

06-23-2009

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13:00:39

query problem 10.1.23 (3d edition 10.1.20) (was 9.1.12) g continuous derivatives, g(5) = 3, g'(5) = -2, g''(5) = 1, g'''(5) = -3

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RESPONSE -->

part a)

P3 = 3 - 2(x - 5) + (1) * ((x-5)^2)/2!

= 3 - 2(x - 5) + 1/2 * (x - 5)^2

P4=3 - 2(x - 5) + (1) * ((x-5)^2)/2! - (3)* ((x - 5)^3)/3

=3 - 2(x - 5) + 1/2 * (x - 5)^2 - (x - 5)^3

part b)

P3(4.9) =3 - 2(4.9 - 5) + 1/2 * (4.9 - 5)^2= 3.205

P4(4.9)= P3(4.9) - (x-5)^3 = 3.206

The degree-n Taylor polynomial about a is

 

g(x) = g(a) + g ‘ (a) ( x – a ) + g ‘ ‘ (a) (x – a)^2 / 2! + … + g [n] (a) ( x – a)^n / n!.

 

The degree-2 polynomial is

 

g(5) + g ‘ (5) ( x – 5 ) + g ‘ ‘ (5) (x – 5)^2 / 2! =

3 – 2 ( x – 5) + 1 ( x – 5)^2 / 2!

= 3 – 2 ( x – 5) + 1 ( x – 5)^2 / 2

 

The degree-3 polynomial is

 

g(5) + g ‘ (5) ( x – 5 ) + g ‘ ‘ (5) (x – 5)^2 / 2! + g ‘ ‘ ‘ (5) (x – 5)^3 / 3!

= 3 – 2 ( x – 5) + 1 ( x – 5)^2 / 2! – 3 ( x – 5)^3 / 3!

= 3 – 2 ( x – 5) + 1 ( x – 5)^2 / 2 – 3 ( x – 5)^3 / 6

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13:00:47

what are the degree 2 and degree 3 Taylor polynomials?

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RESPONSE -->

part a)

P3 = 3 - 2(x - 5) + (1) * ((x-5)^2)/2!

= 3 - 2(x - 5) + 1/2 * (x - 5)^2

P4=3 - 2(x - 5) + (1) * ((x-5)^2)/2! - (3)* ((x - 5)^3)/3

=3 - 2(x - 5) + 1/2 * (x - 5)^2 - (x - 5)^3

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13:00:55

What is each polynomial give you for g(4.9)?

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RESPONSE -->

part b)

P3(4.9) =3 - 2(4.9 - 5) + 1/2 * (4.9 - 5)^2= 3.205

P4(4.9)= P3(4.9) - (x-5)^3 = 3.206

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13:02:54

What would g(4.9) be based on a straight-line approximation using graph point (5,3) and the slope -2? How do the Taylor polynomials modify this tangent-line estimate?

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RESPONSE -->

As you find each higher degree for the Taylor polynomial the answer gets closer and closer to the actual graph and is more convergent.

The straight-line approximation is

 

y-y1=m(x-x1); for the point (5, 3) and slope -3 this is

 

y-3=-2(x-5) which we solve for y to obtain

 

y=-2x+13. Substituting x = 4.9 we obtain

 

y = -2(4.9)+13

 

=-9.8+13

 

=3.2

 

The degree-2 Taylor polynomial differs from this by .05, which is a small modification for the curvature of the graph.

 

The degree-3 Taylor polynomial differs by an additional .005 and takes into account the changing curvature of the graph.

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13:19:00

query problem 10.1.35 (3d edition 10.1.33) (was 9.1.36) estimate the integral of sin(t) / t from t=0 to t=1

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RESPONSE -->

f(x) = sint

f '(x)= cost

f ''(x)= -sint

f '''(x)= -cost

f ''''(x) = sint

f '''''(x)= cost

f(0)= 0

f '(0)= 1

f ''(0)= 0

f '''(0)= -1

f ''''(0)= 0

f '''''(0)= 1

t = 0

P3= 0 + 1x + 0* x^2/2! - 1* x^3/3!

= x - x^3/3

P5=x - x^3/3 + 0 * x^4/4! + 1*x^5/5!

= x - x^3/3 + x^5/5

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13:19:16

what is your degree 3 approximation?

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RESPONSE -->

P3= 0 + 1x + 0* x^2/2! - 1* x^3/3!

= x - x^3/3

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13:19:27

what is your degree 5 approximation?

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RESPONSE -->

P5=x - x^3/3 + 0 * x^4/4! + 1*x^5/5!

= x - x^3/3 + x^5/5

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13:20:12

What is your Taylor polynomial?

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RESPONSE -->

P5=x - x^3/3 + 0 * x^4/4! + 1*x^5/5!

= x - x^3/3 + x^5/5

Good.

The degree 4 approximation of sin(t) is sin(t) = t - t^3 / 6, approx.

 

So the degree 3 approximation of sin(t) / t is P3(t) = 1 - t^2 / 6, approx.

 

The degree 6 approximations are for sin(t) is t - t^3 - 6 + t^5 / 120 approx.,

so the degree-5 approximation so sin(t) / t is P5(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. **

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13:22:30

Explain in your own words why a trapezoidal approximation will not work here.

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RESPONSE -->

A trapezoid approximation will not work here because the graph changes from increasing to decreasing, unlike the graph of f(x) = 1/x. The trapezoid approximation does not account for these fluctuations.

STUDENT ANSWER: 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.

I'm not positive about this but I think it has something to do with the fact that the integral is undefined at t = 0.

 

INSTRUCTOR RESPONSE: ** The biggest problem is the undefined value at t = 0. **

 

The first trapezoid will always begin with an undefined side. So the trapezoidal approximation on any interval including t = 0 cannot be defined.

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13:47:06

Query problem 10.2.25 (3d edition 10.2.21) (was 9.2.12) Taylor series for ln(1+2x)

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RESPONSE -->

I checked the answer in the book and I didn't arrive at the same one. This is what I did:

f(x) = ln (1 + 2x)

f '(x)= 1/ 1+ 2x

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

f '''(x) = 2 (1 + 2x)^-3

f ''''(x) = -6 (1 + 2)^-4

f(0)= 0

f '(0)= 1

f ''(0) = -1

f '''(0)= 2

f ''''90) = -6

P5 = 0 + 1*x - x^2/2! + 2x^3/3! - 6x^4/4!

= x - x^2/2! + 2x^3/3! - 3x^4/2!

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 + ...

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13:47:09

show how you obtained the series by taking derivatives

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RESPONSE -->

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13:51:52

how does this series compare to that for ln(1+x) and how could you have obtained the above result from the series for ln(1+x)?

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RESPONSE -->

The answer is

2x - 2x^2 + (8/3)x^3 +...

I understand that this is similar to the series of ln(1 + x) because each item is multiplied by 2 (n + 2)

x * 2 = 2x

2x^2 * 4 = 4x^2 = 2x

3x^3 * 8 = 24x^3 / 3 = (8/3)x^3

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 + …

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13:53:39

What is your expected interval of convergence?

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RESPONSE -->

I would expect it to converge from 0 < x < 1 because the graph has moved one unit to the left and up 2 units

For your function we have | a(n) | = 2 for all n, so | a(n+1) / a(n) | = 2 / 2 = 1 for all n. By the ratio test the interval of convergence is therefore 1 / 1 = 1.

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13:55:26

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

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&#Your work looks good. See my notes. Let me know if you have any questions. &#