Review Quiz 2

AttachmentSize
F2010M2300_quiz013.pdf42.62 KB
F2010M2300_quiz013_sol.pdf54.16 KB

Math 2300 Section 005 – Calculus II – Fall 2010

Quiz – Monday, November 15, 2010

We will be having a review in class every day until the next midterm exam on Wednesday of this week. For this quiz, you are given 10 minutes to decide which problems you think you can solve and which ones you want to see solved during our review session today.

  1. Use the Lagrange Error Bound Formula for P_{n}(x) to find a reasonable bound for the error in approximating the quantity e^{{0.40}} with a third-degree Taylor polynomial for the function f(x)=e^{x} about x=0. Choose the best error estimate.

    Solution: We use the formula

    |E_{3}(x)|\le\frac{M}{4!}|x|^{4}

    where M\ge|f^{{(4)}}(x)| on the interval [0,0.4] (i.e., the interval between where our Taylor polynomial is centered and where we're approximating the value). Since f^{{(4)}}(x)=e^{x} has a maximum value of e^{{0.4}} on the interval [0,0.4], we obtain

    |E_{3}(0.4)|\le\frac{e^{{0.4}}}{4!}0.4^{4}.
  2. Use the Lagrange Error Bound Formula for P_{n}(x) to find a reasonable bound for the error in approximating the quantity 17/\sqrt{3} with a third-degree Taylor polynomial for the function

    g(x)=\frac{17}{\sqrt{4-x}}

    about x=0. Choose the best error estimate.

    Solution: Here, we're approximating the given function g(x)=17/\sqrt{4-x} using a third degree Taylor polynomial P_{3}(x) about zero, and we're plugging in x=1 since the function evaluated at that point is the value we're interested in. (And so the Taylor polynomial P_{3}(x) approximates 17/\sqrt{3} when we plug in x=1 and find P_{3}(1). How badly does the Taylor polynomial evaluated at x=1 approximate the true value of 17/\sqrt{3}? That's what we're looking for.

    |E_{3}(1)|\le\frac{M}{4!}1^{4}=\frac{M}{4!}

    where M\ge|g^{{(4)}}(x)| on the interval [0,1]. We find the fourth derivative to be

    g^{{(4)}}(x)=\frac{1785}{16\,{\left(4-x\right)}^{{\left(\frac{9}{2}\right)}}}.

    If you graph this function on [0,1], you find that it is always increasing. So, the maximum value is found (as in the previous example) at the rightmost point, which in this case is x=1. So, we have M=0.7951. Thus,

    |E_{3}(1)|\le\frac{0.7951}{4!}=0.033133.
  3. For x in the interval [0,\infty) develop a formula (as a function of x) that yields the error bound for approximating the value of e^{x} using the nth Taylor series for f(x)=e^{x} centered about x=0.

    Solution: This is much like problem 1, but instead of plugging in a specific value of x and a specific degree for the Taylor polynomial, the result will be a function of x and n. (This is basically a multivariable function. If you want to know how well the nth degree Taylor polynomial of e^{x} approximates e^{x} at x, this function will tell you a bound for the error.) We covered some of the details in class.

    |E_{n}(x)|\le\frac{M}{(n+1)}|x|^{{n+1}}=\frac{e^{x}}{(n+1)!}x^{{n+1}}.
  4. Prove that the Taylor series for e^{x} centered at x=0 converges to e^{x} for all real numbers x.

    Solution: The Taylor series is the “infinite degree” Taylor polynomial. So, we consider the limit of the error bounds for P_{n}(x) as n\rightarrow\infty. That is, we're looking at

    \lim _{{n\rightarrow\infty}}|E_{n}(x)|\le\lim _{{n\rightarrow\infty}}\frac{M}{(n+1)!}x^{{n+1}}.

    Since the nth derivatives of e^{x} are always e^{x} this becomes

    \lim _{{n\rightarrow\infty}}\frac{e^{x}}{(n+1)!}x^{{n+1}}=e^{x}\lim _{{n\rightarrow\infty}}\frac{x^{{n+1}}}{(n+1)!}=e^{x}\lim _{{n\rightarrow\infty}}\frac{x^{n}}{n!}=0.

    We have noted several times in class that the limit on the right is zero. This shows that the difference between e^{x} and the Taylor polynomials of degree n as n\rightarrow\infty (the Taylor series) is zero. Thus, e^{x} is actually equal to its Taylor series for all values of x.

  5. Prove that the Taylor series for \sin x centered at x=0 converges to \sin x for all real numbers x.

    Solution: The Taylor series is the “infinite degree” Taylor polynomial. So, we consider the limit of the error bounds for P_{n}(x) as n\rightarrow\infty. That is, we're looking at

    \lim _{{n\rightarrow\infty}}|E_{n}(x)|\le\lim _{{n\rightarrow\infty}}\frac{M}{(n+1)!}x^{{n+1}}.

    Since all of the derivatives of \sin(x) satisfy -1\le\displaystyle\frac{d^{n}}{dx^{n}}\sin x\le 1, we know that M=1. Thus, we have

    \lim _{{n\rightarrow\infty}}|E_{n}(x)|\le\lim _{{n\rightarrow\infty}}\frac{x^{{n+1}}}{(n+1)!}=\lim _{{n\rightarrow\infty}}\frac{x^{n}}{n!}.

    We have noted several times in class that the limit on the right is zero. So, the proof is done. (The error always goes to zero, so the difference between the value of \sin x and the value of the Taylor series of \sin x is zero.)

© 2011 Jason B. Hill. All Rights Reserved.