Assignment 14 Query

course

Sorry about all the bugs in this one. I had to restart the program twice.

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Physics II

11-13-2007

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12:04:36

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 -->

a) Find Taylor polynomial of degree 2 for x near 5:

To do this I need to take the general form of a Taylor polynomial and insert the values of the derivatives at the right points. This gives me a polynomial that looks like:

P_2(x-5) = 3 - 2(x-5) +(1/2)(x-5)^2

P_3(x-5) = 3 - 2(x-5) +(1/2)(x-5)^2 - (1/2)(x-5)^3

b) Approximate g(4.9)

To do this, all I did was plug in 4.9 for x into both equations:

P_2(4.9) = 3.205

P_3(4.9) = 3.2055

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12:04:41

what are the degree 2 and degree 3 Taylor polynomials?

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RESPONSE -->

See above.

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12:04:47

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

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RESPONSE -->

See above.

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13:00:16

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 -->

The approximation that was created was not a straight-line approximation. That would have been the tangent line approximation.

The Taylor polynomials take this tangent-line approximation and turn it into an approximation of any power. As the power increases, so does the approximation over the interval of convergence.

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assignment #014

?????????????

Physics II

11-13-2007

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14:30:57

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 -->

Had to attend another class. Restarted.

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14:31:01

what are the degree 2 and degree 3 Taylor polynomials?

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RESPONSE -->

.

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14:31:04

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

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RESPONSE -->

.

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14:31:07

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 -->

.

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15:24:45

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 -->

a) Estimate with a Taylor polynomial, n=3.

The first thing I did was find the derivatives of sin t:

f(0) = sin 0 = 0

f'(0) = cos 0 = 1

f'' = -sin 0 = 0

f''' = -cos 0 = -1

Putting these together into a Taylor polynomial results in this:

P_3 = t - t^3/3!

I then inserted this into the integral:

int( (t - t^3/3!) / t, t, 0, 1)

= int( 1 - t^2/6, t, 0, 1)

Then I found the antiderivative:

t - t^3/18

Finally, I found the overall value by using the first fundamental theorem (F(b) - F(a)):

F(1) - F(0)

= 1 - 1/18 -(0)

= 17/18

Roughly equivalent to 0.9444

b) Estimate for n=5

I did the same thing for this equation, first creating a Taylor polynomial using derivatives:

P_5 = t - t^3/3! + t^5/5! = t - t^3/6 + t^5/120

I found the integral:

int(1 - t^2/6 +t^4/120, t, 0, 1)

The antiderivative:

t - t^3/18 + t^5/600

And finally, the value using the first fundamental theorem:

1 - 1/18 + 1/600 = 0.94611

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15:24:50

what is your degree 3 approximation?

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RESPONSE -->

See above.

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15:24:55

what is your degree 5 approximation?

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RESPONSE -->

Up.

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15:27:56

What is your Taylor polynomial?

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RESPONSE -->

I believe I put it in above.

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15:50:37

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

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RESPONSE -->

Hmm. Not quite sure. Would it be because trapezoidal approximation uses the origin, 0, and therefore fails because it tries to do division by 0?

Exactly. That first trapezoid has no left-side altitude.

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15:51:23

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 -->

Ok.

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{???????w????assignment #014

?????????????

Physics II

11-13-2007

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15:57:09

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 hit the wrong button somewhere...

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16:00:36

show how you obtained the series by taking derivatives

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RESPONSE -->

After the first derivative I realized that I could use negative exponents on (1+2x), which would make it easier than dealing with the fraction line:

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

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

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

f^4 = -96(1+2x)^-4

Good. The following goes a little further:

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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16:09:55

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 value of x is doubled for the original function (ln(1+2x). In comparing the Taylor series, each part in ln(1+2x) is multiplied by 2^n.

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16:10:42

What is your expected interval of convergence?

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RESPONSE -->

I expected the interval to be half as large because the values of x were doubled, but I had no reason other than that. Any suggestions would be greatly appreciated!

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16:12:21

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

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RESPONSE -->

Series are a lot different from the previous topics I have worked with, but it is good to see that basic calculus topics such as derivation and integration apply to many different topics.

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