Assignment 14

course Mth 174

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Physics II

03-25-2007

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16:42:24

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 -->

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16:50:25

what are the degree 2 and degree 3 Taylor polynomials?

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RESPONSE -->

Degree 2 = g(5) + g'(5)*(x-a) + [g""(5)*(x-a)^2]/2!

= 3 - 2(x-5) + [(x-5)^2]/2!

= 3 - 2(x-5) + [(x-5)^2]/2

Degree 3 = g(5) + g'(5)*(x-a) + [g""(5)*(x-a)^2]/2! + [g'""(5)*(x-a)^3]/3!

= 3 - 2(x-5) + [(x-5)^2]/2! + [-3(x-5)^3]/3!

= 3 - 2(x-5) + [(x-5)^2]/2 - [3(x-5)^3]/6

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16:58:13

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

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RESPONSE -->

2nd Degree: g(4.9) = 3 - 2(4.9-5) + [(4.9-5)^2]/2

= 3 - 2(-0.1) + (-0.1)^2/2

= 3.205

3rd Degree: g(4.9) = 3.205 - [3(4.9-5)^3]/6

= 3.205 - (-0.1)^3/6

= 3.2055

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17:11:47

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 -->

Using the slope formula:

m = (y2-y1)/(x2-x1)

-2 = (3 - y)/(5-4.9)

y = 2.8

The first degree Taylor polynomial is the tangent line at x=5 (in this case). Each additional degree adds a term with an additional power of x--by accounting for the change in the slope of g(x) near x and so on. Thus it produces a curve that is closer to g(x) near x=5, and each additional degree expands the interval in which the Taylor polynomial closely approximates g(x).

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17:12:46

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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17:22:46

what is your degree 3 approximation?

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RESPONSE -->

g(t) = sin(t) g(0) = 0

g'(t) = cos(t) g'(0) = 1

g""(t) = -sin(t) g""(0) = 0

g'""(t) = -cos(t) g'""(0) = -1

P3(t) = 0 + 1*t + [0*t^2]/2! - [1*t^3]/3! = 0 + t - (t^3)/6

Substituting P(3)t for sin(t) in the integral:

Integral (evaluated from 0 to 1) sin(t)/t dt

= Integral (evaluated from 0 to 1) [t-(t^3)/6]/t dt

= Integral (evaluated from 0 to 1) 1 - (t^2)/6 dt

= t - (t^3)/18 evaluated from 0 to 1

= 1 - (1/18)

= 0.94444....

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17:31:09

what is your degree 5 approximation?

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RESPONSE -->

Adding 2 terms to P3(t)

g4(t) = sin(t) g4(0) = 0

g5(t) = cos(t) g5(0) = 1

P5(t) = t - (t^3)/6 + 0*(t^4)/4! + 1*(t^5)/5!

P5(t) = t - (t^3)/6 - (t^5)/120

Substituting in the integral:

= Integral (evaluated from 0 to 1) [t - (t^3)/6 - (t^5)/120]/t dt

= Integral (evaluated from 0 to 1) 1 - (t^2)/6 + (t^4)/120 dt

= t - (t^3)/18 + (t^5)/600, evaluated from 0 to 1

= 1 - (1/18) + (1/600)

= 0.946111.....

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17:31:22

What is your Taylor polynomial?

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RESPONSE -->

See previous responses.

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17:38:00

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

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RESPONSE -->

A trapezoidal approximation takes the average of left and right Riemann sums. The right Riemann sums can be calculated for any n because the function is defined for the right side of any interval we choose. However, the left sums cannot be calculated becauxe no matter what n we choose, the function is not defined at x=0, which must be the function value for the first interval in any sum.

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17:41:24

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

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17:47:34

show how you obtained the series by taking derivatives

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RESPONSE -->

f(x) = ln(1+2x) f(0) = ln(1+0) = 0

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

f""(x) = -4/(1+2x)^2 f""(0) = -4

f'""(X) = 16/(1+2x)^3 f'""(0) = 16

f4(x) = -96/(1+2x)^4 f4(0) = -96

P(x) = 0 + 2x - 4x^2/2! + 16x^3/3! - 96x^4/4! +....

P(x) = 2x - 2x^2 +8x^3/3 - 4x^4 + ...

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17:49:46

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 -->

This series can be obtained from the series for ln(1+x) with each term multiplied by 2^n for n>=1 or or by substituting 2x for x in each term of the ln(1+x) series.

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17:56:04

What is your expected interval of convergence?

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RESPONSE -->

The interval of convergence for ln(1+x) is -1

ln(1 + 2x) = ln( 2 ( x + 1/2) ).

ln(1 + x) is defined for x > -1, while ln(1 + 2x) is defined for x > -1/2.

** Very good. Alternatively the ratio test gives us the following:

For your function we have | a(n) | = 2^n / n!, so | a(n+1) / a(n) | = 2^(n+1) / (n + 1)! / ( 2^n / n!) = 2 n / (n + 1). As n -> infinity this approaches 2. So by the ratio test the radius of convergence is therefore 1 / 2. &&

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17:58:57

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

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RESPONSE -->

I found this topic to be very interesting.

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Assignment 14

ln(1 + 2x) = ln( 2 ( x + 1/2) ). ln(1 + x) is defined for x > -1, while ln(1 + 2x) is defined for x > -1/2.

** Very good. Alternatively the ratio test gives us the following: For your function we have | a(n) | = 2^n / n!, so | a(n+1) / a(n) | = 2^(n+1) / (n + 1)! / ( 2^n / n!) = 2 n / (n + 1). As n -> infinity this approaches 2. So by the ratio test the radius of convergence is therefore 1 / 2. *&*&

Good work. See my notes and let me know if you have questions.

ln(1 + 2x) = ln( 2 ( x + 1/2) ).

ln(1 + x) is defined for x > -1, while ln(1 + 2x) is defined for x > -1/2.

** Very good. Alternatively the ratio test gives us the following:

For your function we have | a(n) | = 2^n / n!, so | a(n+1) / a(n) | = 2^(n+1) / (n + 1)! / ( 2^n / n!) = 2 n / (n + 1). As n -> infinity this approaches 2. So by the ratio test the radius of convergence is therefore 1 / 2. &&