Assignment 15 Query

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course Mth 174

query 14

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Question:

**** query (problem has been omitted from 5th edition but should be worked and self-critiqued) 4th edition 10.1.23 (3d edition 10.1.20 formerly problem 9.1.12)

If g is a functin with continuous derivatives, and the following are known:

Х g(5) = 3,

Х g'(5) = -2,

Х g''(5) = 1,

Х g'''(5) = -3

Then what are the corresponding degree 2 and degree 3 Taylor polynomials?

What are the values of these polynomials at x = 4.9?

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Your solution:

The general polynomial for this would be:

G(x)=g(a)+gТ(a)(x-a) + gТТ(a)(x-a)^2/2! + g(n)(a)(x-a)^n/n!

The 2nd degree would be:

G(5)+gТ(5)(x-5)+gТТ(5)(x-5)^2/2!

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

X=3.205

The 3rd degree would be:

G(5) +gТ(5)(x-5)+gТТ(5)(x-5)^2/2! + gТТТ(5)(x-5)^3/3!

=3-2(x-5) + (x-5)^2/2 Ц 3(x-5)^3/6

X=3.2055

confidence rating #$&*: very

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Given Solution:

The degree-n Taylor polynomial about a is

g(x) = g(a) + g С (a) ( x Ц a ) + g С С (a) (x Ц a)^2 / 2! + Е + g [n] (a) ( x Ц a)^n / n!.

The degree-2 polynomial is

g(5) + g С (5) ( x Ц 5 ) + g С С (5) (x Ц 5)^2 / 2! =
3 Ц 2 ( x Ц 5) + 1 ( x Ц 5)^2 / 2!
= 3 Ц 2 ( x Ц 5) + 1 ( x Ц 5)^2 / 2

The degree-3 polynomial is

g(5) + g С (5) ( x Ц 5 ) + g С С (5) (x Ц 5)^2 / 2! + g С С С (5) (x Ц 5)^3 / 3!
= 3 Ц 2 ( x Ц 5) + 1 ( x Ц 5)^2 / 2! Ц 3 ( x Ц 5)^3 / 3!
= 3 Ц 2 ( x Ц 5) + 1 ( x Ц 5)^2 / 2 Ц 3 ( x Ц 5)^3 / 6

for degree 2, we obtain g(4.9)= 3.205

for degree 3, we obtain g(4.9)=3.2055

**** 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?

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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Question:

**** query problem 10.1.32 (4th edition problem 10.1.35 3d edition 9.1.31) estimate the integral of sin(t) / t from t=0 to t=1

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Your solution:

3rd degree:

Sin(x)/x = x/x Цx^3/3!/x=1-x^2/3!,x,0,1 = .944444444444444

5th degree:

Sin(x)/x = x/x Цx^3/3!/x + x^5/5!/x = 1- x^2/6 + x^4/120, x, 0, 1 = .94611111

So, it converges to the above integral

confidence rating #$&*: very

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Given Solution:

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. **

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

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STUDENT ANSWER: I think that a trapezoid approximation won`t work well because of the shape of the graph of the sin t. Also, it definently won`t work for our initial expression from the book, because the fn is undefined at t=0. The altitudes of each line go from being too small to too large and so on. Perhaps that is the reason, but I am not completely certain.

I'm not positive about this but I think it has something to do with the fact that the integral is undefined at t = 0.

INSTRUCTOR RESPONSE: ** The biggest problem is the undefined value at t = 0. **

The first trapezoid will always begin with an undefined side. So the trapezoidal approximation on any interval including t = 0 cannot be defined.

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Question:

**** Query omitted from 5th edition but should be worked and self-critiqued (4th edition problem 10.2.25 3d edition problem 10.2.21 formerly 9.2.12) Taylor series for ln(1+2x) **** show how you obtained the series by taking derivatives

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Your solution:

F(y)=ln(y) fТ(y)=1/y

G(x)=1+2x GТ(x)=2

F(x)=ln(1+2x) fТ(x)= 2/1+2x

FТТ(x)= -4(1+2x)^2

FТТТ(x)=16/(1+2x)^3

At x=0, f(x)=0 , fТ(x)=2, fТТ(x)= -4 fТТТ(x)=16

confidence rating #$&*: very

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Given Solution:

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 + ...

**** 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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16:59:42

The derivatives of g(x) = ln(x) are

g'(x) = 1/x
g''(x) = -1/x^2
g'''(x) = -2 / x^3
g''''(x) = 6 / x^4

...

g[n](x) = (-1)^n * (n-1)! / x^n

yielding the Taylor series about n = 1:

ln(x) = (x - 1) - (x - 1)^2 / 2 + (x-1)^3 / 3 - (x - 1)^4 / 4 ... + (-1)^(n-1) (x-1)^n / n + ...

To get ln(1 + x), we can just substitute 1 + 2x for x in the above. It follows that

ln(1 + 2x) = (1 + x - 1) - (1 + x - 1)^2 / 2 + (1 + x - 1)^3 / 3 - (1 + x - 1)^4 / 4 ... + (-1)^(n-1)(1 + x-1)^n / n + ...
= x Ц x^2 / 2 + x^3 / 3 Ц x^4 / 4 + Е + (-1)^(n-1) x^n / n.

The function ln(1 + 2x) is obtained by just substituting 2x into the previous:

ln(1 + 2x) = 2x - (2x)^2 / 2 + (2x)^3 / 3 - (2x)^4 / 4 ... + (-1)^(n-1) ( 2x )^n / n + ...

or writing out the terms more explicitly

ln(1 + 2x) = 2 x Ц 2^2 x^2 / 2 + 2^3 x^3 / 3 Ц 2^4 x^4 / 4 + Е + (-1)^(n-1) x^n / n + Е

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**** What is your expected interval of convergence?

For your function we have | a(n) | = 2 for all n, so | a(n+1) / a(n) | = 2 / 2 = 1 for all n. By the ratio test the interval of convergence is therefore 1 / 1 = 1.

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19:04:22

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Self-critique (if necessary):

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self-critique rating #$&*:

This assignment has helped me to get a better understanding of Taylor approximations and series. It is good to see another way in which you can evaluate the int2egral for an expression that cannot normally be reasonably integrated.

I do have a question, though:

When something is a factorial, such as 3!, does that mean 3*2*1?

** That's right. You appear to have been using the factorial correctly. **

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

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Assignment 15 Query

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course Mth 174

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:

FТ(x)=1/3x^(-2/3)

FТТ(x)=(-2/9)x^(-5/3)

FТТТ(x)=(10/27)x^(-8/3)

FТТТТ(x)=(-80/81)x^(-11/3)

All the derivatives are undefined at x=0

Abs.fТТТТ(x)=80/81(.5)^(-11/3)=12.54

Abs (12.54/4! * (.5-1)^4) =.0327

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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

Х f ' (x) = 1/3 x^(-2/3),

Х f '' (x) = -2/9 x^(-5/3),

Х f ''' (x) = 10/27 x^(-8/3),

Х 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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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:

When we derive an odd number of times about x=0, we get zero.

When we derive an even number of times about x=0 we get 1/n! *x^n

Lim n->inf. X^(n+1)/(n+1)!/x^n/n! = lim x->inf. x/n = 0

En(x)=M/(n+1)! * x^(n+1)

Since M=1, we have 1/(n+1)! * x^(n+1)

As n-infinity, this also approaches zero, proving convergence.

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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:

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

Х 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.

Х En(x) = M / (n+1)! * x^(n+1)

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

Х As n -> infinity this approaches zero.

Х 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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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:

First we will derive y=sqrt.x 4 times about x=1

Y=sqrt.x

YТ=1/2x^(-1/2)

YТТ= -1/4x^(-3/2)

YТТТ= 3/8x^(-5/2)

YТТТТ = -15/16x^(-7/2)

Plugging in 1 we will get 1, ╜, -1/4, 3/8, -15/16

Sqrt.x = 1+ ╜(x-1) -1/4(x-1)^2/2! +3/8(x-1)^3/3! -15/16 (x-1)^4/4!

Sqrt.x+1 = 1+ ╜(x) -1/4(x)^2/2! +3/8(x)^3/3! -15/16 (x)^4/4!

Sinx=x-x^3/3!

Sin^2x=x^2-2x^4/3!

Sin^3x=x^3

Sin^4x=x^4

Sin(x+1)= 1+1/2(sinx) =1/4(sin^2x)/2! +3/8(sin^3x)/3! -15/16(sin^4x)/4!

The polynomial representation of sqrt.(1+sinx):

1+1/2(x-x^3/3!) Ц ╝(x^2 -2x^4/3!)/2! +3/8x^3/3! Ц 15/16x^4/4!

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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)?

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! + ,,,

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

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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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