Assignment 14

assignment #014

014. `query 14

Physics II

07-13-2010

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19:23:12

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 -->

The general formula for the Taylor polynomials are: P2(x)=g(5) +g'(5)(x-5)+g''(5)/2! (x-5)^2 and P3(x)= g(5)+g'(5)(x-5)+g''(5)/2!(x-5)^2+g'''(5)/3! (x-5)^3.

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19:24:39

what are the degree 2 and degree 3 Taylor polynomials?

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RESPONSE -->

Substituting in the values of g(x) makes the polynomials: P2(x)= 3-2(x-5)+1/2(x-5)^2 and P3(x)=3-2(x-5)+ 1/2(x-5)^2 - 1/2 (x-5)^3.

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19:25:24

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

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RESPONSE -->

P2(4.9)= 3.205 and P3(4.9)= 3.2055

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19:41:43

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 equation of the line using these values is: y-3=-2(x-5) and this becomes: y=-2x+13. At 4.9 this has a value of: 3.2. The Taylor polynomials 'add' onto this kind on 'honing in' on the actual value.

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19:43:47

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 -->

Here we start off by finding through the third and then fifth derivative of sin t and forming the Taylor polynomial using this information and then divide through by t.

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19:47:05

what is your degree 3 approximation?

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RESPONSE -->

Using the strategy formulated above we arrive at the third degree polynomial, which is: P3(x)=1-1/6 t^2. We use this expression for our integral and integrate over the interval of (0,1). This is: Int(1-1/6 t^2, 0,1)= t-1/18 t^3 and evaluating over the interval gives: 0.94444...

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19:49:03

what is your degree 5 approximation?

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RESPONSE -->

With the same strategy we construct the fifth degree polynomial: P5(x)= 1-t^2/6+t^4/120. The integral is then: Int(1-t^2/6+t^4/120, 0,1)= t-t^3/18+t^5/600 and evaluating this over the interval gives: 0.946111...

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19:49:23

What is your Taylor polynomial?

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RESPONSE -->

I have given the Taylor polynomials in the solutions above.

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19:50:14

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

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RESPONSE -->

The reason is because of the function being undefined at x=0 which prevents us from even forming the TRAP sum.

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20:00:21

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 -->

After a lot of derivations and then finding the general form of the derivative and plugging all of this into the polynomial equation we have: Pn(x)= 0+2x-2x^2+8/3 x^3 -4x^4+....+[(-1)^n(2^n)]/n.

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20:03:05

show how you obtained the series by taking derivatives

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RESPONSE -->

I found the first four derivatives and from those developed a general form for all the derivatives. It took a lot of trial and error to work it out but I eventually arrived at: fn(x)= [(-1)^n(2^n)(n-1)!]/(1+2x)^n. I plugged this into the the taylor polynomial formula for x=0 and this produced the result above.

One way of doing this, for comparison:

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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20:03:52

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 -->

Replacing x in the ln(1+x) polynomial with 2x would have the same results.

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20:06:11

What is your expected interval of convergence?

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RESPONSE -->

Here I used the idea above, that of comparing this function to the one in the last question, and saw that with the 2x instead x the interval would have to be divided by 2 to get: -1/2

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20:07:50

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

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RESPONSE -->

I was weary of the series section at first but really have gotten the hang of it and find it interesting how you can 'model' unknown functions behavior with known ones. This seems like it will be very useful in engineering.

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

"

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

Assignment 14

assignment #014

014. `query 14

Physics II

07-13-2010

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

19:23:12

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 -->

The general formula for the Taylor polynomials are: P2(x)=g(5) +g'(5)(x-5)+g''(5)/2! (x-5)^2 and P3(x)= g(5)+g'(5)(x-5)+g''(5)/2!(x-5)^2+g'''(5)/3! (x-5)^3.

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19:24:39

what are the degree 2 and degree 3 Taylor polynomials?

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RESPONSE -->

Substituting in the values of g(x) makes the polynomials: P2(x)= 3-2(x-5)+1/2(x-5)^2 and P3(x)=3-2(x-5)+ 1/2(x-5)^2 - 1/2 (x-5)^3.

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19:25:24

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

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RESPONSE -->

P2(4.9)= 3.205 and P3(4.9)= 3.2055

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

19:41:43

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?

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

RESPONSE -->

The equation of the line using these values is: y-3=-2(x-5) and this becomes: y=-2x+13. At 4.9 this has a value of: 3.2. The Taylor polynomials 'add' onto this kind on 'honing in' on the actual value.

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

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

19:43:47

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

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

RESPONSE -->

Here we start off by finding through the third and then fifth derivative of sin t and forming the Taylor polynomial using this information and then divide through by t.

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

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

19:47:05

what is your degree 3 approximation?

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

RESPONSE -->

Using the strategy formulated above we arrive at the third degree polynomial, which is: P3(x)=1-1/6 t^2. We use this expression for our integral and integrate over the interval of (0,1). This is: Int(1-1/6 t^2, 0,1)= t-1/18 t^3 and evaluating over the interval gives: 0.94444...

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

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

19:49:03

what is your degree 5 approximation?

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

RESPONSE -->

With the same strategy we construct the fifth degree polynomial: P5(x)= 1-t^2/6+t^4/120. The integral is then: Int(1-t^2/6+t^4/120, 0,1)= t-t^3/18+t^5/600 and evaluating this over the interval gives: 0.946111...

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

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

19:49:23

What is your Taylor polynomial?

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

RESPONSE -->

I have given the Taylor polynomials in the solutions above.

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

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

19:50:14

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

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

RESPONSE -->

The reason is because of the function being undefined at x=0 which prevents us from even forming the TRAP sum.

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

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

20:00:21

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

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

RESPONSE -->

After a lot of derivations and then finding the general form of the derivative and plugging all of this into the polynomial equation we have: Pn(x)= 0+2x-2x^2+8/3 x^3 -4x^4+....+[(-1)^n(2^n)]/n.

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

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

20:03:05

show how you obtained the series by taking derivatives

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

RESPONSE -->

I found the first four derivatives and from those developed a general form for all the derivatives. It took a lot of trial and error to work it out but I eventually arrived at: fn(x)= [(-1)^n(2^n)(n-1)!]/(1+2x)^n. I plugged this into the the taylor polynomial formula for x=0 and this produced the result above.

One way of doing this, for comparison:

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 + ...

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

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

20:03:52

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)?

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

RESPONSE -->

Replacing x in the ln(1+x) polynomial with 2x would have the same results.

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

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

20:06:11

What is your expected interval of convergence?

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

RESPONSE -->

Here I used the idea above, that of comparing this function to the one in the last question, and saw that with the 2x instead x the interval would have to be divided by 2 to get: -1/2

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

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

20:07:50

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

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

RESPONSE -->

I was weary of the series section at first but really have gotten the hang of it and find it interesting how you can 'model' unknown functions behavior with known ones. This seems like it will be very useful in engineering.

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

"

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

*&$*&$