620-295 Real Analysis with Applications, Lecture 29

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 4 February 2009

Taylor Series

If $f : [a, b] \to ℝ$ and $f^{(N + 1)} : [a, b] \to ℝ$ is continuous then $\begin{array}{rcl} f (x) & = & a_{0} + a_{1} (x - a) + a_{2} {(x - a)}^{2} + \dots a_{N} {(x - a)}^{N} + Error (N), \end{array}$ where $a_{l} = \frac{1}{l!} f^{(l)} (a)$ and there exists $c \in (a, x)$ such that $Error (N) = \frac{1}{(N + 1)!} f^{(N + 1)} (c) {(x - a)}^{N + 1} .$

Stone-Weierstrass

If $f : [a, b] \to ℂ$ is a continuous function then there exists a sequence of polynomials $(p_{1} p_{2} \dots)$ such that $(p_{1} p_{2} \dots)$ converges uniformly to $f .$

So essentially any continuous function is close to a power series $f (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots$

Triginometric Series

$f (x) = c_{0} + c_{1} e^{i x} + c_{- 1} e^{- i x} + c_{2} e^{2 i x} + c_{- 2} e^{- 2 i x} + \dots$

If $k \in ℤ$ then $\sin (k x) = \frac{e^{k i x} - e^{- k i x}}{2 i} and \cos (k x) = \frac{e^{k i x} + e^{- k i x}}{2}$ and $e^{i k x} = \cos (k x) + i \sin (k x) and e^{- i k x} = \cos (k x) - i \sin (k x) .$

$\begin{array}{rcl} f (x) & = & c_{0} + c_{1} (\cos (x) + i \sin (x)) + c_{2} (\cos (2 x) + i \sin (2 x)) + \dots \\ + c_{- 1} (\cos (x) - i \sin (x)) + c_{- 2} (\cos (2 x) - i \sin (2 x)) + \dots \\ = & c_{0} + (c_{1} + c_{- 1}) \cos (x) + i (c_{1} - c_{- 1}) \sin (x) + (c_{2} + c_{- 2}) \cos (2 x) + i (c_{2} + c_{- 2}) + \dots \sin (2 x) \\ = & a_{0} + a_{1} \cos (x) + b_{1} \sin (x) + a_{2} \cos (2 x) + b_{2} \sin (2 x) + \dots \end{array}$ where $a_{0} = c_{0}, a_{k} = c_{k} + c_{- k} and b_{k} = i (c_{k} - c_{- k}) .$

Let $k \in ℤ .$

\begin{array}{rcl} (a) & \frac{1}{2 π} \int_{0}^{2 π} e^{i k x} dx & = & \{\begin{array}{rcl} 0, if k \neq 0 \\ 1, if k = 0 \end{array} \\ (b) & \frac{1}{2 π} \int_{0}^{2 π} e^{i k x} e^{- i l x} dx & = & δ_{k l}, where δ_{k l} = \{\begin{array}{rcl} 0, if k \neq l \\ 1, if k = l \end{array} \end{array}

f (x) = c_{0} + c_{1} e^{i x} + c_{- 1} e^{- i x} + c_{2} e^{i 2 x} + c_{- 2} e^{i (- 2 x)} + \dots

then

c_{k} = \frac{1}{2 π} \int_{0}^{2 π} f (x) e^{i (- k x)} dx

Proof (a)

If $k \neq 0$ then $\begin{array}{rcl} \frac{1}{2 π} \int_{0}^{2 π} e^{i k x} dx & = & \frac{- i}{2 π k} e^{i k x} |_{x = 0}^{x = 2 π} \\ = \frac{- i}{2 π k} (e^{2 π i} - e^{0}) = & \frac{- i}{2 π k} (1 - 1) = \frac{- i}{2 π k} . 0 = 0 . \end{array}$

If $k = 0$ then $\begin{array}{rcl} \frac{1}{2 π} \int_{0}^{2 π} e^{i k x} dx & = & \frac{1}{2 π} \int_{0}^{2 π} e^{0} dx = \frac{1}{2 π} \int_{0}^{2 π} dx = \frac{1}{2 π} (x, |_{x = 0}^{x = 2 π}) = \frac{1}{2 π} (2 π - 0) = 1 . \end{array}$

Proof (b)

$\frac{1}{2 π} \int_{0}^{2 π} e^{i k x} e^{- i l x} dx = \frac{1}{2 π} \int_{0}^{2 π} e^{i (k - l) x} dx$ $= \{\begin{array}{rcl} 0, if k \neq l \\ 1, if k = l, \end{array}$

by part (a).

		Proof (c)
	$\frac{1}{2 π} \int_{0}^{2 π} f (x) e^{- i k x} dx = \frac{1}{2 π} \int_{0}^{2 π} (\sum_{l \in ℤ} c_{l} e^{i l x}) e^{i (- k) x} dx$ $= \frac{1}{2 π} \sum_{l \in ℤ} c_{l} \int_{0}^{2 π} e^{i (l - k) x} dx = c_{k} .$

Example: Consider $f : [0, 2 π] \to ℝ$ given by $f (x) = x^{2} .$ Write $f (x) = c_{0} + c_{1} e^{i x} + c_{- 1} e^{- i x} + c_{2} e^{i 2 x} + c_{- 2} e^{i (- 2 x)} + \dots$

Then $c_{0} = \frac{1}{2 π} \int_{0}^{2 π} f (x) dx = \frac{1}{2 π} \int_{0}^{2 π} x^{2} dx = \frac{1}{2 π} (\frac{x^{3}}{3} |_{0}^{2 π}) = \frac{x^{3}}{3} \frac{{(2 π)}^{3}}{3} = \frac{4 π^{2}}{3}$

and $c_{k} = \frac{1}{2 π} \int_{0}^{2 π} x^{2} e^{- k x} dx .$

$\begin{array}{rcl} \int x^{2} e^{- i k x} dx & = & \frac{1}{- i k} x^{2} e^{- i k x} - \int \frac{2 x e^{- i k x}}{- i k} dx \\ = & \frac{i x^{2}}{k} e^{- i k x} - (\frac{2 x e^{- i k x}}{{(- i k)}^{2}} - \int \frac{2 x e^{- i k x}}{{(- i k)}^{2}} dx) \\ = & \frac{i x^{2}}{k} e^{- i k x} + \frac{2 x e^{- i k x}}{k^{2}} + \frac{2 e^{- i k x}}{{(- i k)}^{3}} \end{array}$

So $\begin{array}{rcl} c_{k} & = & \frac{1}{2 π} \int_{0}^{2 π} x^{2} e^{- i k x} dx & = & \frac{i {(2 π)}^{2}}{2 π k} e^{- i k 2 π} + \frac{2 (2 π) e^{- i k 2 π}}{2 π k^{2}} \\ = & \frac{2 π i}{k} + \frac{2}{k^{2}} . \end{array}$

So $a_{k} = c_{k} + c_{- k} = \frac{2 π i}{k} + \frac{2}{k^{2}} + \frac{2 π i}{- k} + \frac{2}{k^{2}} = \frac{4}{k^{2}}$

and $b_{k} = i (c_{k} - c_{- k}) = i ((\frac{2 π i}{k} + \frac{2}{k^{2}}) - (\frac{2 π i}{- k} + \frac{2}{k^{2}})) = i . 2 . \frac{2 π i}{k} = - \frac{4 π}{k} .$

So $\begin{array}{rcl} x^{2} & = & \frac{4 π^{2}}{3} + 4 \cos (x) - 4 π \sin (x) + \frac{4}{4} \cos (2 x) - \frac{4 π}{2} \sin (2 x) + \dots \\ = & \frac{4 π^{2}}{3} + \sum_{k = 1}^{\infty} (\frac{4}{k^{2}} \cos (k x) - \frac{4 π}{k} \sin (k x)) . \end{array}$

If $x = π$ then

\begin{array}{rcl} π^{2} & = & \frac{4 π^{2}}{3} + \sum_{k = 1}^{\infty} \frac{4}{k^{2}} \cos (k π) - \frac{4 π}{k} \sin (k π) \\ = & \frac{4 π^{2}}{3} + \sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{4}{k^{2}} . \end{array}

So $- \frac{π^{2}}{3} = \sum_{k = 1}^{\infty} \frac{4}{k^{2}} and \frac{π^{2}}{12} = \sum_{k = 1}^{\infty} {(- 1)}^{k - 1} \frac{1}{k^{2}}$

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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