Science:Math Exam Resources/Courses/MATH307/December 2010/Question 06 (a)

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MATH307 December 2010

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q2 (e)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q3 (d)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q6 (d)  •  Q6 (e)  •  Q6 (f)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  

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Question 06 (a)
Determine the coefficients cn in the expansion $f(x)=\sum _{n=-\infty }^{\infty }c_{n}e^{2\pi inx}$ where ƒ(x) = x and $0\leq x\leq 1$.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that the functions $\displaystyle {}e^{2\pi {}inx/L}$ are an orthogonal basis for square integrable functions, i.e. $\int _{0}^{L}e^{2\pi {}inx/L}e^{-2\pi {}imx/L}{\textrm {d}}x={\begin{cases}L,\qquad {}n=m\\0,\qquad {}n\neq {}m\end{cases}}$ How can we use this orthogonality to get the coefficients in the series?

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Solution

Let $\displaystyle {}e_{n}(x)=e^{2\pi {}inx}$ so that we want $c_{i}$ such that $x=\sum _{i=-\infty }^{\infty }{}c_{i}e_{i}(x).$ From Hint 1 (and setting L=1) we have that $e_{n}$ are orthogonal i.e., $\int _{0}^{1}e_{n}(x){\overline {e_{m}}}(x){\textrm {d}}x={\begin{cases}1,\qquad {}n=m\\0,\qquad {}n\neq {}m\end{cases}}$ where the overbar indicates the complex conjugate $\displaystyle {}{\overline {e_{n}}}(x)=e^{-2\pi {}inx}.$ We can get the coefficients of $x$ using this orthogonality. Take the inner product with each $e_{n}(x)$ $\langle e_{n},x\rangle =\sum _{i=-\infty }^{\infty }c_{i}\langle e_{n},e_{i}\rangle =c_{n}\langle e_{n},e_{n}\rangle =c_{n}$ $c_{n}=\int _{0}^{1}e^{-2\pi inx}xdx$. If $n\neq 0,$ we use integration by parts: $\int udv=uv-\int vdu$ with $u=x$, $dv=e^{-2\pi inx}dx$ to obtain ${\begin{aligned}c_{n}&=\left[{\frac {x}{-2\pi in}}e^{-2\pi inx}\right]_{0}^{1}+\int _{0}^{1}{\frac {1}{2\pi in}}e^{-2\pi inx}dx\\&=\left[-{\frac {x}{2\pi in}}e^{-2\pi inx}-{\frac {1}{4\pi ^{2}i^{2}n^{2}}}e^{-2\pi inx}\right]_{0}^{1}\\&=\left[-{\frac {x}{2\pi in}}e^{-2\pi inx}+{\frac {1}{4\pi ^{2}n^{2}}}e^{-2\pi inx}\right]_{0}^{1}\\&=\left[\left({\frac {1}{4\pi ^{2}n^{2}}}-{\frac {x}{2\pi in}}\right)e^{-2\pi inx}\right]_{0}^{1}\\&=\left[\left({\frac {1}{4\pi ^{2}n^{2}}}-{\frac {1}{2\pi in}}\right)e^{-2\pi in(1)}\right]-\left[\left({\frac {1}{4\pi ^{2}n^{2}}}-0\right)e^{-2\pi in(0)}\right]\\&=\left({\frac {1}{4\pi ^{2}n^{2}}}-{\frac {1}{2\pi in}}\right)e^{-2\pi in}-{\frac {1}{4\pi ^{2}n^{2}}}\\&=\left({\frac {1}{4\pi ^{2}n^{2}}}-{\frac {1}{2\pi in}}\right)(1)-{\frac {1}{4\pi ^{2}n^{2}}}\\&=-{\frac {1}{2\pi in}}\\&={\frac {i}{2\pi {}n}}\end{aligned}}$ Here we see the importance of having $n\neq 0$, otherwise the coefficient diverges using this formula. If n = 0, $c_{0}=\int _{0}^{1}f(x)dx=\left[{\frac {1}{2}}x^{2}\right]_{0}^{1}={\frac {1}{2}}$ Therefore, we have all the coefficients for the series.

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