Finding M in a Taylor's Inequality problem

[freudchicken06]

I was wondering if anybody could help me with a general rule for finding M in a Taylor's Inequality problem. I have my Calc II final exam tomorrow and I have no professor (only a grad student as a teacher, who hasn't been ANY help) so if anyone could help me out with how to solve the following kind of problem, that would be great. I'm not even sure where to start except with the formula for Taylor's Inequality:

"Use Taylor's Inequality to estimate ln(1.1) to within an error of 0.001. Write you answer as a single number."

I was also given the following information:

ln(1+X) = (sigma) (-1)^(n+1)(x^n)/(n)

 

[stapel]

How does "X" relate to "x"? How does "M" relate to the exercise? Are you referring to the inequality displayed here?

Thank you.

Eliz.

 

[freudchicken06]

Yes, I'm referring to the inequality:

abs(Rn(x)) (is less than or equal to) [M abs(x-a)^(n+1)]/(n+1)!

I'm having a ton of trouble understanding how to do these kinds of estimation problems and like I said, the grad student I have for a "professor" is of no help and can hardly figure them out himself.

 

[galactus]

Maybe I can explain a little.

The remainder is the 'meat' of the theorem because it tells you how close

you are when you replace the function with the polynomial.

Let's say the function is continuous on some interval we'll call I.

'a' is any point in the interval I, ususally its midpoint.

Let's use a polynomial of degree 3 with a=0 and x=1.

\(\displaystyle \L\\S_{n}(x)=f(a)+\frac{f'(a)(x-a)}{1!}+\frac{f''(a)(x-a)^{2}}{2!}+\frac{f'''(a)(x-a)^{3}}{3!}+.....+\frac{f^{n}(a)(x-a)^{n}}{n!}\)

\(\displaystyle \L\\S_{n}(x)\) is the sum of the polynomial terms up to term of degree n.

The remainder \(\displaystyle \L\\R_{n}(x)=f(x)-S_{n}(x)\) for all x in I.

Then there is a point let's call w in I, w is between a and x, s.t.

\(\displaystyle \L\\R_{n}(x)=\frac{f^{n+1}(w)(x-a)^{n+1}}{(n+1)!}\)


\(\displaystyle \L\\f(x)=ln(x+1)\\f'(x)=(x+1)^{-1}\\f''(x)=(-1)(x+1)^{-2}\\ f'''(x)=(-1)(-2)(x+1)^{-3}\\f''''(x)=(-1)(-2)(-3)(x+1)^{-4}\)

\(\displaystyle \L\\f(0)=0, \;\ f'(0)=1, \;\ f''(0)=-1, \;\ f'''(0)=2, \;\ f''''(w)=-6(w+1)^{-4}\)

\(\displaystyle \L\\f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^{2}}{2!}+\frac{f'''(a)(x-a)^{3}}{3!}+\frac{f''''(w)(x-a)^{4}}{4!}\)

\(\displaystyle \L\\ln(1+x)=0+x-\frac{x^{2}}{2}+\frac{2x^{3}}{6}-6(w+1)^{-4}\frac{x^{4}}{4!}\)

Therefore, \(\displaystyle \L\\S_{3}(x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}\) and \(\displaystyle \L\\ln(1.1)=0.1-\frac{(0.1)^{2}}{2}+\frac{(0.1)^{3}}{3}=.095333.....\)

\(\displaystyle \L\\R_{3}(x)=\frac{-x^{4}}{4}(w+1)^{4}.\)

Let's estimate:

\(\displaystyle \L\\0<w<0.1, \;\ so \;\ 1<w+1<1.1\)

\(\displaystyle \L\\\frac{1}{1.1}<\frac{1}{w+1}<1; \;\ |R_{3}(0.1)|=\frac{(-1)^{4}}{4}(w+1)^{4}<\frac{0.1}{4}=0.000025\)

That's closer than you need, but what the heck.