** For this 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. *&*&

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

&#See my notes and let me know if you have questions. &#

assignment 14

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

You appear to be treating g as a number rather than as a function with derivatives. You hve the right overall form so you should have a good basis for understanding the corrections neccesary:

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, the approximation is g(4.9)= 3.205

for degree 3, the approximation is g(4.9)=3.2055

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

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.

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

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. **

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

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 + ...

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................

.................................................

......!!!!!!!!...................................

......!!!!!!!!...................................