Friday, April 27, 2012

Practice Exam 3

Practice Test 3
Administered on 20120427. Updated and made available online 20120430 6:15 PM.

1) Solve the initial value problem

\[
y^{\prime}+\frac{1}{x}y-\sin x=0
\]

\[
y(\pi)=0
\]

2) Integrate
\[
\int\frac{1}{x^{2}+2x+3}dx
\]

$2^{\prime}$) Integrate
\[
\int\frac{1}{x^{2}-2x+3}dx
\]

3) Convergent or divergent. State the test you use. You may not use
the same test twice. Extra: What about absolutely convergent?

a)
\[
\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{2n+1}
\]

b)
\[
\sum_{n=2}^{\infty}\frac{1}{n\ln n}
\]

c)
\[
\sum_{n=1}^{\infty}\frac{3^{n}}{(5+n)^{n}}
\]

d)
\[
\sum_{n=2}^{\infty}(1+e^{-n})
\]

e)
\[
\sum_{n=1}^{\infty}\frac{1}{2^{n}}
\]

4) State the tests you didn't use.

5) Determine the interval of converegence.

a)
\[
\sum_{n=1}^{\infty}\frac{(n-1)^{2}}{n!}x^{n}
\]

b)
\[
\sum_{n=1}^{\infty}\frac{1}{n}(x-2)^{n}
\]

6) Evaluate the indefinite integral as a power series. What is the
radius of convergence?

\[
\int x^{3}\ln(1+x)dx
\]

7) Conceptual Check

a) How many of the important Macluaurin Series and their radii of
convergence do you know? If necessary, which would you know how to
derive?

b) What is the definition of $\mathrm{binom}(k,n)$ given in your book? (page
760)

9 comments:

  1. If you have a question about a problem not already discussed, simply begin your post with the problem number followed by your question/answer/discussion.

    ReplyDelete
  2. For 3b I used the integral test and got the limit as t goes to infinity of (ln(lnn)) from 2 to t. It is therefore divergent.

    Is this correct?

    ReplyDelete
  3. For 3d I think the series converges because (1/e)^n converges by the geometric series test and (1/e)^n converges because (1/e)^n < (1/2)^n, so by the comparison test the series should converge.

    Is this correct?

    ReplyDelete
    Replies
    1. No, you have overlooked the 1. The terms are $$1+\frac{1}{e^n}$$.

      Delete
  4. doesn't the 1 just shift the series up one unit and not have any bearing on the convergence?

    ReplyDelete
    Replies
    1. There's a distinction between $$1+\sum _{n=2}^\infty e^{-n}$$ and $$\sum _{n=2}^\infty (1+e^{-n})$$.

      Delete
    2. Oh I didn't make that distinction. Thank you for the clarification.

      Delete