query 15

ΒΰήέW鲩Γ°~Κδˆ…²ΪρΗ΅χψassignment #015

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Question:

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Question: Query 10.4.8 (was 10.4.1 3d edition formerly p. 487 problem 6) magnitude of error in estimating .5^(1/3) with a third-degree Taylor polynomial about 0.

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Your solution:

We want to find the magnitude of error in estimated .5x^(1/3) using the 3rd degree taylor polynomial about x=0. We begin by finding derivatives.

f(a)=.5x^(1/3)

f’(a) = (1/3)x^(-2/3)

f’’(a)= (-2/9)x^(-5/3)

f’’’(a)=(10/27)x^(-8/3)

f’’’’(a)= (-80/81)x^(-11/3)

when we try to find f(0) we see that all of the derivatives are undefined so we will need to evaluate and expand about x=1.

We also know that since we are asked to use the 3rd degree to find the error, we know that that error is found using the 4th degree. So we use

f’’’’(a)= (-80/81)x^(-11/3) to find the M

we will use .5 and 1 as values of x to see with is larger.

f’’’’(.5)= (-80/81)(.5)^(-11/3)=-12.5424

f’’’’(1)= (-80/81)1^(-11/3)=-.98765

we will use the value of 13 for M since the value of f(.5) was the largest and represents the maximum error.

So now to find the maximum possible magnitude of error we use the formula

M/4!(a-1)^4

13/24 (.5-1)^4= -.0339

confidence rating #$&*: 2

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Given Solution:

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Given Solution: ** The maximum possible error of the degree-3 Taylor polynomial is based on the fourth derivative and is equal to the maximum possible magnitude of the n = 4 term of the Taylor series.

The present function is x^(1/3). Its derivatives are

• f ' (x) = 1/3 x^(-2/3),

• f '' (x) = -2/9 x^(-5/3),

• f ''' (x) = 10/27 x^(-8/3),

• f '''' (x) = -80/81 x^(-11/3).

All these derivatives are undefined at x = 0.

Since all the derivatives are easily evaluated at x = 1, we expand about x = 1.

The maximum possible magnitude of the fourth derivative f''''(x) = -80/81 x^(-11/3) between x = .5 and x = 1 occurs at x = .5, where we obtain | f '''' (x) | = 80/81 * (.5^-(11/3) ) = 12.54 approx. We will still have a valid limit on the error if we use the slight overestimate M = 13.

So the maximum possible error is | 13 / 4! * (.5 - 1)^4 |, or about .034.

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Self-critique (if necessary):

### I was ok until I got a negative for the 4th term which was f’’’’(.5)= (-80/81)(.5)^(-11/3)=-12.5424

So I wasn’t sure what to do about that negative sign. Also how do we know what values to find the maximum possible magnitude, is it always between x=.5 and x=1?????

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Self-critique Rating:3

@&

For x > 0, it is clear that x^(-11/3) is a decreasing power function.

So its maximum value on an interval will occur at the beginning of that interval.

M is the magnitude of the maximum error. So the - sign in front doesn't matter.

*@

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Question:

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Question: Query 10.4.20 (was 10.4.16 3d edition) (was p. 487 problem 12) Taylor series of sin(x) converges to sin(x) for all x (note change from cos(x) in previous edition)

explain how you proved the result.

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Your solution:

We begin with the sin(x) and are asked to show that the series converges for all x using taylor series, expanding about 0.

We first find the derivatives

f(a)= sinx f(0)=0

f’(a)= cosx f(0)=1

f’’(a)= -sinx f(0)=0

f’’’(a)=-cosx f(0)=-1

f’’’’(a)=sinx = f(0)=0

f(a) + f(a)(x-a)/1! + f’’(a)(x-a)^2/2! + f’’’(a)*(x-a)^3/3! + f’’’’(a)(x-a)^4/4! + f(n)(x-a)^n+1/n!

filling in for f(0) we get

0+(1*(x-0)^1/1!+ 0 + -1(x^3)/6 + 0 + x-(x^3)/6 . . . + x^(n+1)/(n+1)!

Now we can see above that all even derivatives are the sin function and will cancel out to zero. So we are left with only odd terms in the taylor expansion

We can also see that since the odd terms are cos terms the only values they can have is 1 or -1 so we see that M=1 for or maximum.

So, M/(m+1)! * x^(n+1)

1/(n+1)! * x^(n+1)

Now if we can prove the limit of the error term is 0 as n approaches infinity, we will know that the series coverges for all values of x.

So we have x^(n+1)/ (n+1!) we can’t really tell if this converges so we can do the ratio test

Finding a(n+1) and multiplying if by the reciprocal of a.

x^(n+1+1) / (n+2)! * (n+1)!/x^(n+1)

so cancelling we get x^(n+1)/(n+2) which as x approaches infinity, the limit is 0 so the series converges thus all values of x converge.

confidence rating #$&*: 2

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Given Solution:

** For even values of n the nth derivative is sin(x); when expanding about 0 this will result in terms of the form 0 * x^n / n!, or just 0.

If n is odd, the nth derivative is +- cos(x). Expanding about 0 this derivative has magnitude 1 for all n. So the nth term, for n odd, is just 1 / n! * x^n.

None of the Taylor coefficients exceeds the maximum magnitude M = 1.

For any x, lim(n -> infinity} (x^n / n!) = 0:

• lim { n -> infinity} ( [ x^(n+1) / (n+1)! ] / [ x^n / n! ] ) = lim (n -> infinity) ( x / n ) = 0

• the limit is zero since x is fixed and n increases without bound.

Putting this together formally in terms of the definition of the error term.

• En(x) = M / (n+1)! * x^(n+1)

• Since M = 1, En(x) becomes = 1 / (n+1)! * x^(n+1)

• As n -> infinity this approaches zero.

• If the error term approaches zero as n -> infinity, the series converges for all x.

It's not obvious that x^(n + 1)/ (n + 1)! approaches zero for any x. If x gets large, x^(n+1) gets very, very large.

However a ratio test will show that x^(n+1) / (n+1)! does approach zero for any value of x, giving us the stated result. The series does converge for all values of x.**

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Self-critique (if necessary):

### So we have x^(n+1)/ (n+1!) we can’t really tell if this converges so we can do the ratio test

Finding a(n+1) and multiplying if by the reciprocal of a.

x^(n+1+1) / (n+1+1)! * (n+1)!/x^(n+1)

x^(n+2)/(n+2)(n+1)(n-1) * (n+1)(n-1)/x^(n+1)

x^(n+1)/(n+2)

so cancelling we get x^(n+1)/(n+2) which approaches 0 as x approaches infinity, the limit is 0 so the series converges thus all values of x converge.

Am I cancelling the above out correctly??? I still feel like I am doing something wrong. Also do we plug in larger and larger values of x into x^(n+1) / (n+2) which we got from the ratio test to check for limit/divergence. If so I’m not sure I understand how we know that x^(n+1) / (n+2) would approach zero. I can see that if we use large values of n that the result would approach infinity but I don’t see that same result for x values approaching infinity. Please help me understand where I am confused????

@&

I seldom use the word 'cancel' . What we do is factor identical terms out of the numerator and denominator, which then divide one another with result 1.

The expresion

x^(n+1+1) / (n+2)! * (n+1)!/x^(n+1)

can be rearranged to

x^(n+1+1) / x^(n+1) * (n+1) ! / (n+2) !

The x terms have n + 1 + 1 factors, each equal to x, in the numerator, and n + 1 identical factors in the denominator. S0 n + 1 of these factors can be matched up with the n + 1 identical factors in the denominator, leaving only one x in the numerator.

(n+1)! = (n+1) * n * (n-1) * ... * 1

and

(n+2)! = (n + 2) * (n+1) * n * (n-1) * ... * 1

The entire numerator matches up with the (n+1) * n * (n-1) * ... * 1 in the denominator, leaving just (n+2) in the denominator.

Thus the expression simplifies to

x / (n + 2).

*@

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Self-critique Rating:3

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The payoff is that for any value of x, the expression x / n approaches zero as n approaches infinity.

We aren't interested in what happens as x approaches infinity. It's enough that there is no value of x for which x / n fails to approach zero as n approaches infinity.

*@

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Question:

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Question: Query problem 10.3.10 (3d edition 10.3.12) (was 9.3.12) series for `sqrt(1+sin(`theta))

what are the first four nonzero terms of the series?

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Your solution:

We are asked to find the first 4 non zero terms of the series squareroot of (1+sin(theta))

We begin by rewriting as (1+sin(theta))^1/2

We can easily perform the taylor expansion on (1+x)^1/2

f(a)= (1+x)^(1/2) f(0)= 1

f’(a)= ½(1+x)^-1/2 f’(0)=1/2

f’’(a)= -1/4 (1+x)^-(3/2) f’’(0)=-1/4

f’’’(a)= (-1/4)(1+x)^(-5/2) f’’’(0)=3/8

f’’’’(a)= (-15/16)(1+x)^(-7/2) f’’’’(0)= -15/16

so taylor expansion of (1+x)^(1/2) is

1+ (1/2)*x-0)/1 - (1/4) * (x-0)^2/2! + (3/8)(x-0)^3 / 3! - (15/16) * (x-0)^4 / 4!

1 + (1/2)x - (1/8) x^2 + (3/48)x^3 - (5/128)x^4

Now we can substitute in theta for x and get

1+ (1/2)theta - (1/8) theta^2 + (1/16) theat ^3 - (5/128) theta ^4

confidence rating #$&*: 3

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Given Solution:

We substitute the Taylor expansion of sin(x) into the Taylor expansion of 1 / sqrt(x) and ignore terms with powers exceeding 4:

Expanding y = sqrt(x) about x = 1 we get derivatives

y ' = 1/2 x^(-1/2)

y '' = -1/4 x^(-3/2)

y ''' = 3/8 x^(-5/2)

y '''' = -5/16 x^(-7/2).

Substituting 1 we get values 1, 1/2, -1/4, 3/8, -5/16.

The degree-4 Taylor polynomial is therefore

sqrt(x) = 1 + 1/2 (x-1) - 1/4 (x-1)^2 / 2 + 3/8 (x-1)^3/3! - 5/16 (x-1)^4 / 4!.

It follows that the polynomial for sqrt(1 + x) is

sqrt( 1 + x ) = 1 + 1/2 ( 1 + x - 1) - 1/4 (1 + x-1)^2 / 2 + 3/8 (1 + x-1)^3/3! - 5/16 (1 + x-1)^4 / 4!

= 1 + 1/2 (x) - 1/4 (x)^2 / 2 + 3/8 (x)^3/3! - 5/16 (x)^4 / 4!.

Now the first four nonzeo terms of the expansion of sin(theta) give us the polynomial

sin(theta) = theta-theta^3/3!+theta^5/5!-theta^7/7!.

We note that since sin(theta) contains theta as a term, the first four powers of sin(theta) will contain theta, theta^2, theta^3 and theta^4, so our expansion will ultimately contain these powers of theta.

We will expand the expressions for sin^2(theta), sin^3(theta) and sin^4(theta), but since we are looking for only the first four nonzero terms of the expansion we will not include powers of theta which exceed 4:

sin^2(theta) = (theta-theta^3/3!+theta^5/5!-theta^7/7!)^2 = theta^2 - 2 theta^4 / 3! + powers of theta exceeding 4

sin^3(theta) = (theta-theta^3/3!+theta^5/5!-theta^7/7!)^3 = theta^3 + powers of theta exceeding 4

sin^4(theta) = (theta-theta^3/3!+theta^5/5!-theta^7/7!)^4 = theta^4 + powers of theta exceeding 4

Substituting sin(theta) for x in the expansion of sqrt(1 + x) we obtain

sqrt(1 + sin(theta)) = 1 + 1/2 (sin(theta)) - 1/4 (sin(theta))^2 / 2 + 3/8 (sin(theta))^3/3! - 5/16 (sin(theta))^4 / 4!

= 1 + 1/2 (theta-theta^3/3! + powers of theta exceeding 4) - 1/4 (theta^2 - 2 theta^4 / 3! + powers of theta exceeding 4)/2 + 3/8 ( theta^3 + powers of theta exceeding 4) / 3! - 5/16 (theta^4 + powers of theta exceeding 4)/4!

= 1 + 1/2 theta - 1/4 theta^2/2 - 1/2 theta^3 / 3! + 3/8 theta^3 / 3! + 2 / ( 4 * 3! * 2) theta ^4 - 5/16 theta^4 / 4!

= 1 + 1/2 theta - 1/8 theta^2 + 1/16 theta^3 - 5/128 theta^4.

STUDENT SOLUTION:

Rewrite as sqrt (1+sin theta) = (1+sin theta)^(1/2)

Produce Taylor series for (1+x)^(1/2) using binomial expansion with p = 1/2:

(1+x)^(1/2) = 1 + x/2 + [(.5)(.5-1)x^2]/2! + [(.5)(.5-1)(.5-2)x^3]/3! ...

= 1 + x/2 + x^2/8 + x^3/16 ...

Produce Taylor series for sin theta

sin theta = theta - theta^3/3! + theta^5/5! - theta^7/7! ...

Substitute sin theta for x in series for (1+x)^(1/2):

sqrt (1+ sin theta) = 1 + (1/2)(theta - theta^3/3! + theta^5/5! - theta^7/7! ...) - (1/8)(theta - theta^3/3! + theta^5/5! - theta^7/7! ...)^2 + (1/16)(theta - theta^3/3! + theta^5/5! - theta^7/7! ...)^3 ...

sqrt (1+ sin theta) = 1 + theta/2 - (theta^2)/8 - (theta^3)/48 ...

INSTRUCTOR RESPONSE:

Your approach should work; you got -1/48 for the theta^3 term, whereas I believe the correct coefficient is -1/16. Otherwise your expansion agrees with that obtained below. However a different approach, which doesn't involve the binomial expansion, was used above.

What is the Taylor series for `sqrt(z)?

&&** we can let f(x) = squareroot of x+1 and do the series expansion because if we use the squareroot of z for the taylor expression about 0 all values are 0 so we know from the previous example that

f(a)= (1+x)^(1/2) f(0)= 1

f’(a)= ½(1+x)^-1/2 f’(0)=1/2

f’’(a)= -1/4 (1+x)^-(3/2) f’’(0)=-1/4

f’’’(a)= (-1/4)(1+x)^(-5/2) f’’’(0)=3/8

f’’’’(a)= (-15/16)(1+x)^(-7/2) f’’’’(0)= -15/16

so we have 1+ (1/2)x - x^2/8 + x^3/16 + 5(x^4)/128

now we have to substitute back in for 1+x we originally had the squareroot of z and we evaluated the squareroot of 1+x we we will need to plug in z-1 to the taylor expansion getting

1+ (1/2)(z-1) - (z-1)^2/8 + (z-1)^3/16 + 5*(z-1)^4)/128**********

RESPONSE -->

Using f(x) = sqrt (z) will not work because f(0) and all derivatives of f(x) evaluated at 0 = 0.

Let f(x) = sqrt (1+x)

Using binomial expansion:

sqrt (1+x) = 1 + x/2 - x^2/8 + x^3/16 + ...

Substituting z-1 for x (since 1 + (z-1) = z)

sqrt (z) = 1 + (z-1)/2 - (z-1)^2/8 + (z-1)^3/16 + ...

What is the Taylor series for 1+sin(`theta)?

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RESPONSE -->

Since 1 is a constant, its Taylor series is itself (if f(x) =1, f(0) =1 and all derivatives of f(x) = 0

Taylor series for sin (theta) = theta - theta^3/3! + theta^5/5! - theta^7/7! + ,,,

Adding the two series gives

1 + sin (theta) = 1 + (theta - theta^3/3! + theta^5/5! - theta^7/7! + ,,,

= 1 + theta - theta^3/3! + theta^5/5! - theta^7/7! + ,,,

&&***I can use taylor expansion for (1+cosx) getting

f(a)= 1+sinx f’(0)=1

f’(a)= cosx f’(0)=1

f’(a)= -sinx f’(0)=0

f’(a)= -cosx f’(0)=-1

f’(a)= sinx f’(0)=0

f’(a)= cosx f’(0)=1

so for taylor expansion we have 1+x-x^3/+ + x^5/120…..

so now we sub back in theta for x getting

1+theta - theta^3/6 + theta^5/120 …..****************

How are the two series combined to obtain the desired series?

&&*** a taylor series is a power series that follows power rules and can be added a term at a time, or term by term*******

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RESPONSE -->

See previous response. Since a Taylor series is a power series it follows the rules of a power series, including the rule that power series can be added term by term.

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STUDENT COMMENT:

The Taylor series for sqrt(z) provides an interesting way to manually calculate approximations of square roots.

Along with a table containing some reference values, your calculator uses Taylor expansions to evaluate functions.

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Self-critique (if necessary):

ok

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Self-critique Rating:3

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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