Assignments 13  14

course Mth 174

I am trying my best as these assignments they are just hard to understand these but i am trying to do better so i can get my C in this course.

ˆ³r¬æR˜¾±Ñ‡ˆ¦—|ÁšÐÌ£|assignment #013

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Physics II

10-30-2007

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15:37:37

With what known series did you compare this series, and how did you show that the comparison was valid?

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RESPONSE -->

with the known series of n greater or equal to 1 we know that n^3 less than equal to n^3 so

0 less than equal to 1/3^n+1 less than equal to 1/n^3

so every term here in the series is less than or equal to the corresponding term of 1/n^3

Your solution is OK, but should include the reasons why you believe 1 / n^3 to be convergent.

GOOD STUDENT SOLUTION WITH COMMENT: For large values of n, this series is similar to 1 / 3^n. As n approaches infinity, 1 / 3^n approaches 0. A larger number in the denominator means that the value of the function will be smaller. So, (1 / 3^n) > (1 / (3^n + 1)). We know that since 1 / 3^n converges, so does 1 / (3^n + 1).

COMMENT: We know that 1 / (3^n) converges by the ratio test: The limit of a(n+1) / a(n) is (1 / 3^(n+1) ) / (1 / 3^n) = 1/3, so the series converges.

We can also determine this from an integral test. The integral of b^x, from x = 0 to infinity, converges whenever b is less than 1 (antiderivative is 1 / ln(b) * b^x; as x -> infinity the expression b^x approaches zero, as long as b < 1).

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15:42:34

Query problem 9.4.40 (3d edition 9.3.18) (was 9.2.24) partial sums of 1-.1+.01-.001 ... o what does the series converge?

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RESPONSE -->

Ok problem 10

the radius of convergence here in the series of 1 / (2n)! is shown by using the ratio test which tells us that whater the radius is it is two times the n so we will have to use the ratio test to establish our answer. To us the ration test we suppose the sequence of rations abs(an + 1 / an) which will have a limit of lim as n goes to infinity so abs(an+1) / (an) = L

*&*& The ratio test takes the limit as n -> infinity of a(n+1) / a(n). If the limit is less than 1 then the series converges in much the same way as a geometric series sum(r^n), with r equal to the limiting ratio.

In this case a(n+1) = 1 / (2n + 2)! and a(n) = 1 / (2n) ! so

a(n+1) / a(n)

= 1 / (2n+2) ! / [ 1 / (2n) ! ]

= (2n) ! / (2n + 2) !

= [ 2n * (2n-1) * (2n-2) * (2n - 3) * ... * 1 ] / [ (2n + 2) * (2n + 1) * (2n) * (2n - 1) * ... * 1 ]

= 1 / [ (2n+2) ( 2n+1) ].

As n -> infinity this result approaches zero. Thus the series converges for all values of n, and the radius of convergence is infinite. *&*&

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15:43:44

What are the first five partial sums of the series?

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RESPONSE -->

ok now #40

in the series they are just moving the decimal place one time and making the number smaller and smaller and just switching the sign from positive to negative every time.

This is an alternating series with | a(n) | = .1^n, for n = 0, 1, 2, ... .

Thus limit{n->infinity}(a(n)) = 0.

An alternating series for which | a(n) | -> 0 is convergent.

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15:57:26

Query 9.5.6. What is your expression for the general term of the series p x + p(p-1) / 2! * x^2 + p(p-1)(p-2) * x^3 + …?

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RESPONSE -->

the expression for the general term of this series is explained in an equation but first we have to look at the series itself. we must write it in the sum notation. the series is about p - n. We use (p-n)^n * x^n and we begin with n = 0 then going to n = 1. The series alternates thought is it keeps adding p - n every time it goes to the next series. p - n we divide by 1. Another way to write this is series is

sum (1)^n (p - n)^n / n^1

Good attempt. There's a little more to it but you're definitely on the right track here.

** The general term is the coefficient of x^n.

In this case you would have the factors of x^2, x^3 etc. go down to p-1, p-2 etc.; the last factor is p minus one greater than the exponent of x. So the factor of x^n would go down to p - n + 1.

This factor would therefore be p ( p - 1) ( p - 2) ... ( p - n + 1).

This expression can be written as p ! / (p-n) !. All the terms of p ! after (p - n + 1) will divide out, leaving you the desired expression p ( p - 1) ( p - 2) ... ( p - n + 1).

The nth term is therefore (p ! / (p - n) !) / n! * x^n, of the form a(n) x^2 with a(n) = p ! / (n ! * (p - n) ! )

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16:01:28

Query 9.5.18. What is the radius of convergence of the series x / 3 + 2 x^2 / 5 + 3 x^2 / 7 + …?

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RESPONSE -->

the radius of convergence of this series can be found by evaluating the series in question. so we first look at the series which is started in the question. The general term of the series is x^n / n so if n is odd and -x^n / n is even so Cn = (-x)n-1 / n then we can use the ratio test. using the ration test we will use

abs(an+1) / abs(an) for an = Cn (x-a)^n

To find the radius of convergence you first find the limit of the ratio | a(n+1) / a(n) | as n -> infinity. The radius of convergence is the reciprocal of this limit.

a(n+1) / a(n) = ( n / (2n+1) ) / (n+1 / (2(n+1) + 1) ) = (2n + 3) / (2n + 1) * n / (n+1).

(2n + 3) / (2n + 1) = ( 1 + 3 / (2n) ) / (1 + 1 / (2n) ), obtained by dividing both numerator and denominator by 2n. In this form we see that as n -> infinity, this expression approaches ( 1 + 0) / ( 1 + 0) = 1.

Similarly n / (n+1) = 1 / (1 + 1/n), which also approaches 1.

Thus (2n + 3) / (2n + 1) * n / (n+1) approaches 1 * 1 = 1, and the limit of a(n+1) / a(n) is therefore 1.

The radius of convergence is the reciprocal of this ratio, which is 1.

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16:01:42

What is your expression for the general term of this series, and how did you use this expression to find the radius of convergence?

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RESPONSE -->

it is showed in the previous screen.

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16:03:56

Query 9.5.28 (3d edition 9.4.24). What is the radius of convergence of the series p x + p(p-1) / 2! * x^2 + p(p-1)(p-2) * x^3 + … and how did you obtain your result?

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RESPONSE -->

we must look at the series first and evaulate it and urse the ratio test first and we see that every time the series increases there is a (p - n) added to the top plus on the top the x value is always increasing so we see an x^n on top as well and we see on the bottom just a n because it is just going up by one everytime. so another way we could right this is

1 + (p - n)^n * x^n / 1^n

Once more, good attempt.

*&*& As seen in 9.4.6 we have

a(n) = p ! / (n ! * (p - n) ! ) so

a(n+1) = p ! / [ (n+1) ! * ( p - (n+1) ) ! ] and

a(n+1) / a(n) = { p ! / [ (n+1) ! * ( p - (n+1) ) ! ] } / {p ! / (n ! * (p - n) ! ) }

= (n ! * ( p - n) !) / {(n+1)! * (p - n - 1) ! }

= (p - n) / (n + 1).

This expression can be written as

(p / n - 1) / (1/n + 1). As n -> infinity both p/n and 1/n approach zero so our limit is -1 / 1 = -1.

Thus the limiting value of | a(n+1) / a(n) | is 1 and the radius of convergence is 1/1 = 1. *&*&

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assignment #014

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Physics II

10-30-2007

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17:14:47

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 -->

we have to find the function first and then we will approximate the true value of f. so we have here

3 -2 (x -5)

+ 0.5(x - 5)^2

-0.5(x - 5)^3

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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17:14:57

what are the degree 2 and degree 3 Taylor polynomials?

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RESPONSE -->

shown in the previous screen

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17:15:13

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

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RESPONSE -->

shown in previous screen

Substituting x = 4.9 into the degree-2 polynomial results in value 3.205.

Substituting x = 4.9 into the degree-3 polynomial results in value 3.2055.

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17:15:16

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 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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17:19:58

what is your degree 3 approximation?

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RESPONSE -->

so if we take it to degree 3 we end up with

f(x) = sint/t

f'(x) = cos(t)

f''(x) = - sin(t)

f'''(x) = -cos(t)

after that we will take it into Taylor polynomial approx.

so we get 0 + 1 * t + 0 * t^2 / 2

What you give is the degree 2 approximation to sin(t).

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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17:22:24

what is your degree 5 approximation?

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RESPONSE -->

we do the same thing here we take

f(t) = sin(t) / t

f'(t) = cos(t)

f''(t) = -sin(t)

f'''(t) = -cos(t)

f^4 = sin(t)

f^5 = cos(t)

so we would get using the Taylor approx here

0 + 1 * t + 0 * t^2/2 - 1 * x^3/3 + 0 * x^4/4

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17:22:33

What is your Taylor polynomial?

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RESPONSE -->

shown in the previous screen

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17:22:46

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

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RESPONSE -->

do not know how to explain this

** The problem is the undefined value at t = 0. A trapezoidal approximation requires values at the two endpoints of the interval, and you have no value for the t = 0 endpoint.**

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17:25:22

show how you obtained the series by taking derivatives

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RESPONSE -->

to find the taylor series we take derivatives here as follows

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

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

f''(x) = 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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17:25:46

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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it would be the same series there as well and would come up with almost the same answer

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17:25:49

What is your expected interval of convergence?

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17:26:37

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

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i really didnt know how to do the last problem except for taking the derviatives there and i am trying my best to do these but it is hard to learn math online here and by teaching yourself. I am just trying to get a C or better in this course so i can graduate.

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You made creditable attempts on most of these problems and showed reasonable detail on most, though still more detail might be helpful. Be sure you review my notes and let me know what questions you have about the given solutions, etc.