Review Quiz 1

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Math 2300 Section 005 – Calculus II – Fall 2010

Quiz – Friday, November 12, 2010

We will be having a review in class every day until the next midterm exam on Wednesday of next 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. If f(2)=g(2)=h(2)=0, f^{{\prime}}(2)=h^{{\prime}}(2)=0, g^{{\prime}}(2)=22, f^{{\prime\prime}}(2)=3, g^{{\prime\prime}}(2)=5 and h^{{\prime\prime}}(2)=7, calculate

    \lim _{{x\rightarrow 2}}\frac{f(x)}{h(x)}.

    Solution: Using L'Hopital's rule we have

    \lim _{{x\rightarrow 2}}\frac{f(x)}{h(x)}=\lim _{{x\rightarrow 2}}\frac{f^{{\prime}}(x)}{h^{{\prime}}(x)}=\lim _{{x\rightarrow 2}}\frac{f^{{\prime\prime}}(x)}{h^{{\prime\prime}}(x)}=\frac{3}{7}.
  2. Find the first four nonzero terms of the Taylor series about 0 for the function

    \frac{t}{1+t}.

    Solution: Notice that

    \frac{d}{dt}\frac{t}{1+t}=\frac{1}{(1+t)^{2}}=(1+t)^{{-2}}.

    Since this is a binomial series with p=-2 we will expand this series and then integrate it term by term to obtain the series that we need. We have

    \displaystyle(1+t)^{{-2}} \displaystyle=1-2t+\frac{(-2)(-3)}{2!}t^{2}+\frac{(-2)(-3)(-4)}{3!}t^{3}+\cdots
    \displaystyle=1-2t+\frac{6}{2!}t^{2}-\frac{24}{3!}t^{3}+\cdots
    \displaystyle=1-2t+3t^{2}-4t^{3}+\cdots

    Now integrate term by term and find

    \displaystyle\int(1+t)^{{-2}}\, dt \displaystyle=\int 1-2t+3t^{2}-4t^{3}+\cdots\, dt
    \displaystyle\frac{t}{1+t} \displaystyle=t-t^{2}+t^{3}-t^{4}+\cdots

    Since we're only looking for the first four terms, we're done.

  3. Find the Taylor polynomials of degree n approximating \ln(1+x) for x near 0.

    Solution: This is actually one of the Taylor series that you should remember. In any case, we first find the derivatives using the chain rule. We'll denote f(x)=\ln(1+x) and then

    \displaystyle f^{{\prime}}(x) \displaystyle=\frac{1}{1+x}
    \displaystyle f^{{\prime\prime}}(x) \displaystyle=-\frac{1}{(1+x)^{2}}
    \displaystyle f^{{(3)}}(x) \displaystyle=\frac{2}{(1+x)^{3}}
    \displaystyle\vdots
    \displaystyle f^{{(n)}}(x) \displaystyle=\frac{(n-1)!(-1)^{{n+1}}}{(1+x)^{n}}

    But, notice that all of these derivatives evaluated at zero are f^{{(n)}}(0)=(n-1)!(-1)^{{n+1}}. Also, since \ln(1)=0, the constant term of P_{n}(x) is zero. (Thus, we write the Taylor polynomials as a sum starting at k=1 since the first degree term is the first that will appear in the expansion.) So, we form the nth degree Taylor polynomial as

    \displaystyle P_{n}(x) \displaystyle=\sum _{{k=1}}^{n}\frac{f^{{(k)}}(0)}{k!}x^{k}
    \displaystyle=\sum _{{k=1}}^{n}\frac{(k-1)!(-1)^{{k+1}}}{k!}x^{k}
    \displaystyle=\sum _{{k=1}}^{n}\frac{(-1)^{{k+1}}}{k}x^{k}
    \displaystyle=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots+\frac{(-1)^{{n+1}}x^{n}}{n}.
  4. Use a fourth degree Taylor approximation for e^{h} for h near 0 to evaluate the limit

    \lim _{{h\rightarrow 0}}\frac{e^{h}-1-h}{h^{2}}

    Solution: We know that the fourth degree Taylor polynomial for e^{h} is

    P_{4}(h)=1+h+\frac{h^{2}}{2!}+\frac{h^{3}}{3!}+\frac{h^{4}}{4!}.

    Using this in the limit we find

    \displaystyle\lim _{{h\rightarrow 0}}\frac{e^{h}-1-h}{h^{2}} \displaystyle\approx\lim _{{h\rightarrow 0}}\frac{\left(1+h+\frac{h^{2}}{2!}+\frac{h^{3}}{3!}+\frac{h^{4}}{4!}\right)-1-h}{h^{2}}
    \displaystyle=\lim _{{h\rightarrow 0}}\frac{\frac{h^{2}}{2!}+\frac{h^{3}}{3!}+\frac{h^{4}}{4!}}{h^{2}}
    \displaystyle=\lim _{{h\rightarrow 0}}\left(\frac{1}{2!}+\frac{h}{3!}+\frac{h^{2}}{4!}\right)
    \displaystyle=\frac{1}{2}.
  5. Find the first four nonzero terms of the Taylor series about 0 for the function e^{{-x}}.

    Solution: Just take the Taylor series for e^{x} and replace the x by -x.

    e^{{-x}}\approx 1-x+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\frac{x^{4}}{4!}
  6. Find the third-degree Taylor polynomial for f(x)=x^{3}+7x^{2}-5x+1 about x=0.

    Solution: Remember that the first four terms make up P_{4}(x). We discussed the Taylor polynomials of polynomials a long time ago. The result really shouldn't come as any surprise. We have

    \displaystyle f(0) \displaystyle=1
    \displaystyle f^{{\prime}}(0) \displaystyle=-5
    \displaystyle f^{{\prime\prime}}(0) \displaystyle=14
    \displaystyle f^{{(3)}}(0) \displaystyle=6.

    Then

    \displaystyle P_{4}(x) \displaystyle=1-\frac{5}{1!}x+\frac{14}{2!}x^{2}+\frac{6}{3!}x^{3}
    \displaystyle=1-5x+7x^{2}+x^{3}
    \displaystyle=f(x).

    That is, the nth degree Taylor polynomial (and all subsequent higher degree Taylor polynomials) of an n-degree polynomial is just the polynomial itself.

  7. Find the first four terms of the Taylor series for the given function about 0.

    f(x)=\frac{1}{\sqrt{1+x}}

    Solution: Notice that this is a binomial series

    f(x)=\frac{1}{\sqrt{1+x}}=(1+x)^{{-1/2}}.

    So, the first four terms are given as

    \displaystyle P_{4}(x) \displaystyle=1-\frac{1}{2}x+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}x^{2}+\cdots+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\left(-\frac{7}{2}\right)}{4!}x^{4}
    \displaystyle=1-\frac{1}{2}x+\frac{3}{8}x^{2}-\frac{5}{16}x^{3}+\frac{35}{128}x^{4}
  8. By recognizing the series as a Taylor series evaluated at a particular value of x, find the sum of the following convergent series.

    1+\frac{2}{1!}+\frac{4}{2!}+\frac{8}{3!}+\cdots+\frac{2^{n}}{n!}+\cdots

    Solution: Notice that this is just the Taylor series for e^{x} centered at zero with x=2.

    \displaystyle 1+\frac{2}{1!}+\frac{4}{2!}+\frac{8}{3!}+\cdots+\frac{2^{n}}{n!}+\cdots \displaystyle=1+\frac{1}{1!}2+\frac{1}{2!}2^{2}+\frac{1}{3!}2^{3}+\cdots+\frac{1}{n!}2^{n}+\cdots
    \displaystyle=\sum _{{k=0}}^{\infty}\frac{1}{k!}2^{k}
    \displaystyle=e^{2}
  9. Use the Taylor series expansion of \arctan x to approximate \pi.

    Solution: We know that

    \frac{d}{dx}\arctan x=\frac{1}{1+x^{2}}.

    Thus, if we can find the Taylor series for this function, we can integrate it and get back to \arctan x. We can perform a substitution and find

    \displaystyle\frac{1}{1+x} \displaystyle=1-x+x^{2}-x^{3}+\cdots=\sum _{{k=0}}^{\infty}(-1)^{k}x^{k}
    \displaystyle\frac{1}{1+(x^{2})} \displaystyle=1-(x^{2})+(x^{2})^{2}-(x^{3})^{2}+\cdots=\sum _{{k=0}}^{\infty}(-1)^{k}(x^{2})^{k}
    \displaystyle\frac{1}{1+x^{2}} \displaystyle=1-x^{2}+x^{4}-x^{6}+\cdots=\sum _{{k=0}}^{\infty}(-1)^{k}x^{{2k}}

    Then, we integrate and find

    \displaystyle\arctan x \displaystyle=\int\frac{1}{1+x^{2}}\, dx
    \displaystyle=\int 1-x^{2}+x^{4}-x^{6}+\cdots\, dx
    \displaystyle=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\cdots
    \displaystyle=\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{2k+1}x^{{2k+1}}.

    So, we now have a Taylor series for \arctan x around x=0. Approxmating \pi is then a simple trick. We recall that

    \frac{\pi}{4}=\arctan 1\qquad\text{and so}\qquad\pi=4\arctan 1.

    Using our Taylor series we then have

    \displaystyle\pi \displaystyle=4\arctan(1)
    \displaystyle=4\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots\right)
    \displaystyle=4\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{2k+1}
  10. Find the first four terms of the Taylor series for 1/x about x=2.

    Solution: This can be done directly. The derivatives are

    \displaystyle f(x) \displaystyle=\frac{1}{x}
    \displaystyle f^{{\prime}}(x) \displaystyle=-\frac{1}{x^{2}}
    \displaystyle f^{{\prime\prime}}(x) \displaystyle=\frac{2}{x^{3}}
    \displaystyle f^{{(3)}}(x) \displaystyle=-\frac{3!}{x^{4}}
    \displaystyle f^{{(4)}}(x) \displaystyle=\frac{4!}{x^{5}}.

    We evaluate each derivative at x=2 and find (some reducing happens in the first couple of terms)

    \displaystyle P_{4}(x) \displaystyle=\frac{1}{2}+\frac{-\frac{1}{4}}{1!}(x-2)+\frac{\frac{2}{8}}{2!}(x-2)^{2}+\frac{-\frac{6}{16}}{3!}(x-2)^{3}+\frac{\frac{24}{32}}{4!}(x-2)^{4}
    \displaystyle=1-\frac{1}{4}x+\frac{1}{8}(x-2)^{2}-\frac{1}{16}(x-2)^{3}+\frac{1}{32}(x-2)^{4}
  11. Find the radius and interval of convergence for the Taylor series of \ln(1+x) around x=0.

    Solution: Whenever you want to find the radius and interval of convergence of a Taylor series, you're really just using the fact that a Taylor series is a power series. So, we use the ratio test. We know that the Taylor series for \ln(1+x) around x=0 is given as

    \ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots=\sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}x^{k}}{k}.

    That is, a general term of the series is given as

    a_{k}=\frac{(-1)^{{k+1}}x^{k}}{k}.

    Thus, by the ratio test, the series will converge whenever

    \displaystyle\lim _{{k\rightarrow\infty}}\frac{\left|a_{{k+1}}\right|}{\left|a_{k}\right|} \displaystyle<1
    \displaystyle\lim _{{k\rightarrow\infty}}\frac{\displaystyle\left|\frac{(-1)^{{k+2}}x^{{k+1}}}{k+1}\right|}{\displaystyle\left|\frac{(-1)^{{k+1}}x^{k}}{k}\right|} \displaystyle<1
    \displaystyle\lim _{{k\rightarrow\infty}}\frac{\displaystyle\left|\frac{x^{{k+1}}}{k+1}\right|}{\displaystyle\left|\frac{x^{k}}{k}\right|} \displaystyle<1
    \displaystyle\lim _{{k\rightarrow\infty}}|x|\frac{k}{k+1} \displaystyle<1
    \displaystyle|x| \displaystyle<1
    \displaystyle|x-0| \displaystyle<1.

    That final step may seem a bit confusing. Why in the world would we rewrite |x| as |x-0|? The reason is, as I've mentioned before, that |a-b| provides a convenient way to say “the distance between a and b.” So |x-0| represents the distance between zero and x. Saying |x-0|<1 just says “the distance between x and 0 is less than one.” Now, by the ratio test, we have that the given Taylor series converges when |x-0|<1, and so the radius of convergence is 1.

    We now find the interval of convergence. It will be one of the following intervals:

    (-1,1),\quad(-1,1],\quad[-1,1),\quad[-1,1].

    We need to plug in -1 and 1 to the series and determine convergence individually at these points. At x=-1 we have

    \sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}(-1)^{k}}{k}=\sum _{{k=1}}^{\infty}\frac{(-1)^{{2k+1}}}{k}=-\sum _{{k=1}}^{\infty}\frac{1}{k}

    does not converge. So, we do not include -1 in the interval of convergence. At x=1 we have

    \sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}(1)^{k}}{k}=\sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}}{k}

    converges by the alternating series test. Thus, the interval of convergence is (-1,1].

© 2011 Jason B. Hill. All Rights Reserved.