#$&*
course Mth 174
query 14
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Question:
**** query (problem has been omitted from 5th edition but should be worked and self-critiqued) 4th edition 10.1.23 (3d edition 10.1.20 formerly problem 9.1.12)
If g is a functin with continuous derivatives, and the following are known:
• g(5) = 3,
• g'(5) = -2,
• g''(5) = 1,
• g'''(5) = -3
Then what are the corresponding degree 2 and degree 3 Taylor polynomials?
What are the values of these polynomials at x = 4.9?
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Your solution:
g(x) = g(a) + g`(a)(x - a) + f``(a)(x - a)^2/2! + f```(a)(x -a)^3/3! + …
degree 2: g(x) = g(a) + (x-a)g`(a) + (x-a)^2f``(a) /2! = 3 - 2(x-5) + (x -a)^2/2
degree 3: g(x) = g(a) + (x-a)g`(a) + (x-a)^2f``(a) /2! + (x-a)^3f```(a)/3! = 3 - 2(x-5) + (x -a)^2/2 - 3(x-a)^3/6 = 3 - 2(x-5) + (x -a)^2/2 - (x-a)^3/2
For degree 2, g(4.9) = 3 - 2(4.9 - 5) + (4.9 - 5)^2/2 = 3.205
For degree 3, g(4.9) = 3 - 2(4.9 - 5) + (4.9 - 5)^2/2 - (4.9-5)^3/2 = 3.2055
confidence rating #$&*: 3
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Given Solution:
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
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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Self-critique (if necessary):
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Self-critique Rating: ok
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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:
f(0) = sin t = 0
f`(0) = cos t = 1
f``(0) = -sin t = 0
f```(0) = -cos t = -1
f(4)(0) = sin t = 0
f(5)(0) = cos t = 1
degree 3 for sin t = 0 + t
sin t / t = 1, int (sin t / t dt, 0, 1) = int( 1, dt, 0, 1) = 1
degree 5 for sin t = 0 + t - t^3/3! + t^5/5!
sint / t = 1 - t^2/6 + t^4/120, int (sin t / t dt, 0, 1) = t - t^3/18 + t^5/600] 0 to 1 = 0.9461
confidence rating #$&*: 3
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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. **
STUDENT QUESTION
Still a little unsure of how antiderivatives and definite integrals were set up??????
INSTRUCTOR RESPONSE
For example the degree 3 approximation of sin(t) / t is P3(t) = 1 - t^2 / 6.
Integrating that from t = 0 to 1 we get antiderivative t - t^3 / 18. Between t = 0 and t = 1 this expression changes from 0 to 1 - 1/18 = 17/18. So the degree 3 approximation results in an integral of 17/18, which is about .945.
**** 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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Self-critique (if necessary):
why you use the degree 4 approximation to represents degree 3?
@& Using the degree-4 approximation of the sine function gives us a degree-3 approximation of sin(t) / t.*@
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Self-critique Rating: why you use the degree 4 approximation to represents degree 3?
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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(0) = ln(1 + 2x) = 0
f`(0) = 2/(1+2x) = 2
f``(0) = -2^2/(1+2x)^2 = 2^2
f```(0) = 2^3*2!/(1+2x)^3 = -2^3 * 2!
f(4)(0) = -2^4*3!/ (1+2x)^4 = 2^4 * 3!
f(5)(0) = 2^5*4!/ (1+2x)^5 = -2^5 * 4!
f(a) = 0 + 2a -2^2a^2/2 - 2^3*2! a^3/3! + 2^4 a^4 *3!/4! - 2^5 * 4! a^5/5!
= 2a - 2a^2 - 8a^3/3 + 16a^4/4 - 32a^5/5 + … + (-1)^n 2^n a^n/n
confidence rating #$&*: 3
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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 + ...
STUDENT QUESTION
I am still a little confused about the general term n and wanted to ask where I can find more info on the factorial of general term or just info on (2n+1)! how that is calculated????????
INSTRUCTOR RESPONSE
Here is a key explanation in the given solution:
'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.'
Let's deconstruct it a bit. You have taken the first four derivatives correctly, and you should first refresh yourself on how you got them.
Do you see how every derivative gets a factor of 2 (resulting from application of the chain rule to the (1 + 2x) in the denominator? This is what leads to the 2^n term; with every derivative we pack on another multiple of 2.
I'm sure you see how the exponent in the denominator keeps increasing by 1, so that the exponent on the nth derivative is n. It is this that leads with every step to a new factor of n - 1 (which comes from then n - 1 power in the denominator of the previous term).
The (-1)^(n-1) comes from the fact that the power of (1 + 2x) is always in our denominator. So with every new derivative we get another - sign. On the nth term we will have taken n - 1 derivatives of negative powers, so we have multiplied by -1 a total of n-1 times. This leads to the factor (-1)^(n-1).
**** 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 - x^2 / 2 + x^3 / 3 - x^4 / 4 + … + (-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 - 2^2 x^2 / 2 + 2^3 x^3 / 3 - 2^4 x^4 / 4 + … + (-1)^(n-1) x^n / n + …
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16:59:42
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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: ok
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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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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"
Self-critique (if necessary):
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Self-critique rating:
#*&!
&#Good responses. See my notes and let me know if you have questions. &#