Sean Fitzpatrick |
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University of Lethbridge |
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Find the radius and interval of convergence:
\(\di \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}x^n\)
\(\di \sum_{n=0}^\infty \frac{n}{3^n}x^n\)
Identify the function whose Maclaurin series is given by:
\(\di \sum_{n=0}^\infty \frac{x^n}{n!}\)
\(\di \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!}x^{2n+1}\)
Determine radius and interval of convergence:
\(\di \sum_{n=0}^\infty (-2)^nx^n\)
\(\di \sum_{n=0}^\infty \frac{1}{n}(x-2)^n\)
\(\di \sum_{n=0}^\infty \frac{(-1)^n}{n^3}x^n\)
\(\di \sum_{n=0}^\infty \frac{(2x)^n}{n!}\)
One of the main reasons power series are useful is that derivatives and integrals work the same as they do for polynomials.
Let \(\di f(x)=\sum_{n=0}^\infty a_n(x-c)^n\) be a power series with radius of convergence \(R\text{.}\) Then:
\(\di f'(x) = \sum_{n=1}^\infty na_n(x-c)^{n-1}\text{,}\) with radius of convergence \(R\text{.}\)
\(\di \int f(x)\,dx = C+\sum_{n=0}^\infty \frac{a_n}{n+1}(x-c)^{n+1}\text{,}\) with radius of convergence \(R\text{.}\)
\(\di\sum_{n=0}^\infty x^n\)
\(\di\sum_{n=1}^\infty \frac{n}{2^n}x^n\)
\(\di\sum_{n=1}^\infty \frac{2^n}{n!}x^n\)
\(\di \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!}x^{2n+1}\)
\(\di \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n)!}x^{2n}\)
A common use for power series is finding solutions to differential equations.
We simply suppose that there is a solution \(f(x)=\sum_{n=0}^\infty a_nx^n\text{,}\) and plug it in.
Use power series to solve the following:
\(y'-2xy=0\text{,}\) with \(y(0)=5\)
\(y''+4y = 0\text{,}\) with \(y(0)=4\) and \(y'(0)=-2\)
Recall: given a function \(f\) differentiable at \(c\text{,}\) its linear approximation at \(c\) is given by
Higher-order approximations are given by Taylor polynomials:
If a Taylor polynomial is centred at \(0\text{,}\) we call it a Maclaurin polynomial.
The Taylor series of an infinitely differentiable function \(f(x)\) is given by
When \(c\) is zero, we get the Maclaurin series for \(f\text{.}\)
Recall the Lagrange formula for the remainder \(R_n(x)=f(x)-P_n(x)\) obtained when we approximate a function \(f\) by its degree \(n\) Taylor polynomial:
For many common functions, it's not hard to show that \(\lim\limits_{n\to \infty}R_n(x)=0\) for all \(x\) in the radius of convergence of the Taylor series.
Letting \(n\to\infty\) in the equality \(f(x)=P_n(x)+R_n(x)\text{,}\) we get
Give the Taylor series for:
\(f(x) = e^x\text{,}\) centred at \(0\text{.}\)
\(g(x) = \cos(x)\text{,}\) centred at \(0\text{.}\)
\(h(x)=\ln(x)\text{,}\) centred at \(1\text{.}\)
Give the interval of convergence for each one.
We can add, subtract, multiply and divide power series. Substitution works, too. (Sort of.)
In the case of Taylor series, this is equivalent to doing the same to the corresponding functions.
Determine Taylor series for:
\(\di f(x) = \frac{x^3}{1-x}\)
\(\di g(x) = \frac{\sin(x)}{e^x}\)
\(h(x) = \sec(x)\)
\(s(x) = \sin(x^3)\)
Many new functions (not expressible in terms of elementary functions) arise as power series, often as solutions to differential equations.
The Bessel functions are named after Friedrich Bessel, who found them as solutions to Kepler's equations. They also show up in problems involving vibrations.
The Bessel function of order zero is given by
The Bessel function of order one is given by
Evaluate the limit:
\(\di \lim_{x\to 0}\frac{\cos(x)-1-x^2}{x^4}\)
\(\di\lim_{x\to 0}\frac{\sqrt{1+x}-1-\frac12 x}{x^2}\)