Assignment 14

Write down the expression for a(n+1) / a(n) and find its limit as n -> infinity. The radius of convergence is the reciprocal of this limit.

No reall tips to offer, but I'll be glad to answer questions on specific problems.

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Physics II

07-23-2008

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15:51:58

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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15:54:11

what are the degree 2 and degree 3 Taylor polynomials?

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Degree 2 polynomial: 3 + (-2)(x - 5) + (x - 5)^2 / 2

Degree 3 polynomial: 3 + (-2)(x - 5) + (x - 5)^2 / 2 + (-1)(x - 5)^3 / 2

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15:57:39

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

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I plugged in 4.9 into the polynomials as x, which I am not sure about. For the second degree polynomial I got 3.205 and for the third degree polynomial I got 3.2055

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16:09:52

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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I do not know exactly what to do here. But since we are using the graph point (5, 3) with the slope of -2 I turned it into the line y = 3 - 2(x - 5). Plugging 4.9 in as x, y = g(4.9) = 3.2

Oncea again, not sure what I was aiming for here.

You're right. See below to put your results in perspective:

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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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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Just to confirm your solution:

sin(t) = t - t^3 / 6 + t^5 / 120, approx.

So sin(t) / t = 1 - t^2 / 6 + t^4 / 120, approx.

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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

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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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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

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

yielding the Taylor series about x = 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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&#Good responses. See my notes and let me know if you have questions. &#