Midterm

Midterm

  • Next week is Reading week, and the midterm is on the Friday after that, March 3 7pm-9:30pm
    • It will cover the material up to 11.8: Power series.
    • I strongly suggest reviewing all the material, doing the practice problems, and starting to time yourself with previous midterms during Reading week. The midterm will have approximately 14 multiple choice and 6 written answer questions.
    • On the week of the midterm, you should already be familiar with all the concepts covered, and you should only have to solve more previous midterms.

Strategy for testing series

Steps for series

  • Just as calculating integrals, deciding if a given series is convergent, and maybe calculating the sum if it exists is a non-algorithmic process.
    • That is, there's no recipe for doing it, you have to keep trying. Still, there are a few steps you can take which might work.
    • If it's a \(p\)-series, that is of the form \(\sum\frac{1}{n^p}\), then we know it's convergent precisely when \(p>1\).
    • If it's a geometric series, that is of the form \(\sum_{n=0}^\infty ar^n\), then we know it's convergent precisely when \(|r|<1\), and in that case, the sum is \(\frac{a}{1-r}\).
    • If it resembles a \(p\)-series or a geometric series, then you can try using the Comparison test or the Limit comparison test. Note that the Comparison test requires the terms of the series to be nonnegative, which you can achieve by taking absolute values.
    • If you can see that \(\lim a_n\ne0\), then by the Divergence test, the series \(\sum a_n\) is divergent.
    • If it's an alternating series, then you can try the Alternating series test.
    • If the series involves factorials or other products, you can try the Ratio test. Note that this is going to give a limit of 1 for \(p\)-series, rational functions, and algebraic functions (what you get from rational functions by taking radicals), so for those, you should use another method.
    • If the series involves an \(n\)-th power, you can try the Root test.
    • If there exists a positive, decreasing, continuous function \(f(x)\) so that \(f(n)=a_n\), then you can try the Integral test.
  • Exercises. 11.7: 1, 3, 5, 9, 13, 18, 25, 27, 33

Power series

Introducing power series

  • Finally, we've made it to the new notion of functions we learned all this stuff for: power series.
    • A power series is a series of the form \[ \sum_{n=0}^\infty c_nx^n=c_0+c_1x+c_2x^2+\dotsb \]
    • Here, \(x\) is a variable, and the \(c_n\)'s are constants which are called the coefficients of the series.
    • Note that this is like a polynomial, but it has infinitely many terms.
    • The sum of the series is the function \[ f(x)=c_0+c_1x+c_2x^2+\dotsb \] with domain the collection of the \(x\) for which the series converges.
  • Example. Consider the power series with \(c_n=1\) for all \(n\), that is \[ \sum_{n=0}^\infty x^n=1+x+x^2+\dotsb \]
    • Note that it is a geometric series with common factor \(x\). Therefore, the sum has domain \(|x|<1\), and value \[ f(x)=\sum_{n=0}^\infty x^n=\frac{1}{1-x}. \]

Power series centered at \(a\)

  • Note that any power series is convergent at \(x=0\): \[ \sum_{n=0}^\infty c_n0^n=c_0. \]
    • For reasons which will become apparent later, we would also like to have series around numbers other than 0.
    • For any number \(a\), the series \[ \sum_{n=0}^\infty c_n(x-a)^n=c_0+c_1(x-a)+c_2(x-a)^2+\dotsb \] is called a power series centered at \(a\) (or in \((x-a)\) or about \(a\)).
    • Note that a series centerted at \(a\) is always convergent at \(x=a\).

Radius of convergence.

  • Example. Consider the series \(\sum_{n=1}^\infty\frac{(x+2)^n}{n}\).
    • Let us try to find its domain using the Ratio test: \[ \left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(x+2)^{n+1}}{(x+2)^n}\right|\cdot\frac{n+1}{n}=|x+2|\frac{n+1}{n}\to|x+2|. \]
    • This shows us that if \(|x+2|<1\) then the series is absolute convergent, and if \(|x+2|>1\), then the series is divergent.
    • The ratio test doesn't tell us anything about the cases \(x=-3\) and \(x=-1\), so we need to deal with those separately.
    • For \(x=-3\), we get the alternating harmonic series \(\sum_{n=1}^\infty(-1)^n\frac{1}{n}\), which we know is convergent.
    • For \(x=-1\), we get the harmonic series \(\sum_{n=1}^\infty\frac{1}{n}\), which we know is divergent.
    • In summary: the domain is \([-3,-1)\). Note that the interval is centered around \(-2\).
  • Example. Consider the series \(\sum_{n=1}^\infty n^nx^n\).
    • Let us try to find its domain using the Root test: \[ \sqrt[n]|a_n|=n|x|\to\begin{cases} \infty, & x\ne0,\\ 0, & x=0. \end{cases} \]
    • Therefore, in this case, the domain is \(0\) only.

Radius of convergence.

  • Example. The Bessel function of order \(0\) is the sum \(J_0(x)=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{2^{2n}(n!)^2}\). It's an important function in science.
    • Let's try to find its domain using the Ratio test: \[ \left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{x^{2(n+1)}}{x^{2n}}\cdot\frac{2^{2n}}{2^{2(n+1)}}\cdot\frac{(n!)^2}{((n+1)!)^2}\right|=\frac{x^2}{4(n+1)^2}\to0. \]
    • Therefore, the domain of \(J_0(x)\) is the entire real line \(\mathbf R\).
  • Theorem. Let \(\sum_{n=0}^\infty c_n(x-a)^n\) be a power series. Then there's only 3 possibilities.
    • There exists \(R>0\) such that the sum is convergent if \(|x-a|<R\) and divergent if \(|x-a|>R\). In this case, we say that the radius of convergence is \(R\).
    • The domain of the sum is only \(a\). In this case, we say that the radius of convergence is 0.
    • The domain of the sum is the entire real line \(\mathbf R\). In this case, we say that the radius of convergence is \(\infty\).
  • Exercises. 11.8: 3, 5, 9, 15, 21, 25, 27, 35