hcju/proofwikiqa huggingface dataset:
Let $m \ne n$. {{begin-eqn}} {{eqn | l = \int \sin m x \sin n x \rd x       | r = \frac {\sin \paren {m - n} x} {2 \paren {m - n} } - \frac {\sin \paren {m + n} x} {2 \paren {m + n} } + C       | c = Primitive of $\sin m x \sin n x$ }} {{eqn | ll= \leadsto       | l = \int_\alpha^{\alpha + 2 \pi} \sin m x \sin n x \rd x       | r = \intlimits {\frac {\sin \paren {m - n} x} {2 \paren {m - n} } - \frac {\sin \paren {m + n} x} {2 \paren {m + n} } } \alpha {\alpha + 2 \pi}       | c =  }} {{eqn | r = \paren {\frac {\sin \paren {\paren {m - n} \paren {\alpha + 2 \pi} } } {2 \paren {m - n} } - \frac {\sin \paren {\paren {m + n} \paren {\alpha + 2 \pi} } } {2 \paren {m + n} } }       | c =  }} {{eqn | o =        | ro= -       | r = \paren {\frac {\sin \paren {m - n} \alpha} {2 \paren {m - n} } - \frac {\sin \paren {m + n} \alpha} {2 \paren {m + n} } }       | c =  }} {{eqn | r = \paren {\frac {\sin \paren {m - n} \alpha} {2 \paren {m - n} } - \frac {\sin \paren {m + n} \alpha} {2 \paren {m + n} } }       | c = Corollary to Sine of Angle plus Full Angle }} {{eqn | o =        | ro= -       | r = \paren {\frac {\sin \paren {m - n} \alpha} {2 \paren {m - n} } - \frac {\sin \paren {m + n} \alpha} {2 \paren {m + n} } }       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{qed|lemma}} When $m = n$ we have: {{begin-eqn}} {{eqn | l = \int \sin m x \sin m x \rd x       | r = \int \sin^2 m x \rd x       | c =  }} {{eqn | r = \frac x 2 - \frac {\sin 2 m x} {4 m} + C       | c = Primitive of $\sin^2 m x$ }} {{eqn | ll= \leadsto       | l = \int_\alpha^{\alpha + 2 \pi} \sin m x \sin m x \rd x       | r = \intlimits {\frac x 2 - \frac {\sin 2 m x} {4 m} } \alpha {\alpha + 2 \pi}       | c =  }} {{eqn | r = \paren {\frac {\alpha + 2 \pi} 2 - \frac {\sin \paren {2 m \paren {\alpha + 2 \pi} } } {4 m} } - \paren {\frac \alpha 2 - \frac {\sin 2 m \alpha} {4 m} }       | c =  }} {{eqn | r = \paren {\frac {\alpha + 2 \pi} 2 - \frac {\sin 2 m \alpha} {4 m} } - \paren {\frac \alpha 2 - \frac {\sin 2 m \alpha} {4 m} }       | c = Corollary to Sine of Angle plus Full Angle }} {{eqn | r = \frac {\alpha + 2 \pi - \alpha} 2       | c =  }} {{eqn | r = \pi       | c =  }} {{end-eqn}} {{qed}}
Let $m \ne n$. {{begin-eqn}} {{eqn | l = \int \cos m x \cos n x \rd x       | r = \frac {\sin \paren {m - n} x} {2 \paren {m - n} } + \frac {\sin \paren {m + n} x} {2 \paren {m + n} } + C       | c = Primitive of $\cos m x \cos n x$ }} {{eqn | ll= \leadsto       | l = \int_\alpha^{\alpha + 2 \pi} \cos m x \cos n x \rd x       | r = \intlimits {\frac {\sin \paren {m - n} x} {2 \paren {m - n} } + \frac {\sin \paren {m + n} x} {2 \paren {m + n} } } \alpha {\alpha + 2 \pi}       | c =  }} {{eqn | r = \paren {\frac {\sin \paren {\paren {m - n} \paren {\alpha + 2 \pi} } } {2 \paren {m - n} } + \frac {\sin \paren {\paren {m + n} \paren {\alpha + 2 \pi} } } {2 \paren {m + n} } }       | c =  }} {{eqn | o =        | ro= -       | r = \paren {\frac {\sin \paren {m - n} \alpha} {2 \paren {m - n} } + \frac {\sin \paren {m + n} \alpha} {2 \paren {m + n} } }       | c =  }} {{eqn | r = \paren {\frac {\sin \paren {m - n} \alpha} {2 \paren {m - n} } + \frac {\sin \paren {m + n} \alpha} {2 \paren {m + n} } }       | c = Corollary to Sine of Angle plus Full Angle }} {{eqn | o =        | ro= -       | r = \paren {\frac {\sin \paren {m - n} \alpha} {2 \paren {m - n} } + \frac {\sin \paren {m + n} \alpha} {2 \paren {m + n} } }       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{qed|lemma}} When $m = n$ we have: {{begin-eqn}} {{eqn | l = \int \cos m x \cos m x \rd x       | r = \int \cos^2 m x \rd x       | c =  }} {{eqn | r = \frac x 2 + \frac {\sin 2 m x} {4 m} + C       | c = Primitive of $\cos^2 m x$ }} {{eqn | ll= \leadsto       | l = \int_\alpha^{\alpha + 2 \pi} \cos m x \cos m x \rd x       | r = \intlimits {\frac x 2 + \frac {\sin 2 m x} {4 m} } \alpha {\alpha + 2 \pi}       | c =  }} {{eqn | r = \paren {\frac {\alpha + 2 \pi} 2 + \frac {\map \sin {2 m \paren {\alpha + 2 \pi} } } {4 m} } - \paren {\frac \alpha 2 + \frac {\sin 2 m \alpha} {4 m} }       | c =  }} {{eqn | r = \paren {\frac {\alpha + 2 \pi} 2 + \frac {\sin 2 m \alpha} {4 m} } - \paren {\frac \alpha 2 + \frac {\sin 2 m \alpha} {4 m} }       | c = Corollary to Sine of Angle plus Full Angle }} {{eqn | r = \frac {\alpha + 2 \pi - \alpha} 2       | c =  }} {{eqn | r = \pi       | c =  }} {{end-eqn}} {{qed}}
Let $m \ne n$. {{begin-eqn}} {{eqn | l = \int \sin m x \cos n x \rd x       | r = \frac {-\map \cos {m - n} x} {2 \paren {m - n} } - \frac {\map \cos {m + n} x} {2 \paren {m + n} } + C       | c = Primitive of $\sin m x \cos n x$ }} {{eqn | ll= \leadsto       | l = \int_\alpha^{\alpha + 2 \pi} \sin m x \cos n x \rd x       | r = \intlimits {\frac {-\map \cos {m - n} x} {2 \paren {m - n} } - \frac {\map \cos {m + n} x} {2 \paren {m + n} } } \alpha {\alpha + 2 \pi}       | c =  }} {{eqn | r = \paren {\frac {-\map \cos {\paren {m - n} \paren {\alpha + 2 \pi} } } {2 \paren {m - n} } - \frac {\map \cos {\paren {m + n} \paren {\alpha + 2 \pi} } } {2 \paren {m + n} } }       | c =  }} {{eqn | o =        | ro= -       | r = \paren {\frac {-\map \cos {m - n} \alpha} {2 \paren {m - n} } - \frac {\map \cos {m + n} \alpha} {2 \paren {m + n} } }       | c =  }} {{eqn | r = \paren {\frac {-\map \cos {m - n} \alpha} {2 \paren {m - n} } - \frac {\map \cos {m + n} \alpha} {2 \paren {m + n} } }       | c = Corollary to Cosine of Angle plus Full Angle }} {{eqn | o =        | ro= -       | r = \paren {\frac {-\map \cos {m - n} \alpha} {2 \paren {m - n} } - \frac {\map \cos {m + n} \alpha} {2 \paren {m + n} } }       | c =  }} {{eqn | r = 0       | c = after simplification }} {{end-eqn}} When $m = n$ we have: {{begin-eqn}} {{eqn | l = \int \sin m x \cos m x \rd x       | r = \frac {\sin^2 m x} {2 m} + C       | c = Primitive of $\sin m x \cos m x$ }} {{eqn | ll= \leadsto       | l = \int_\alpha^{\alpha + 2 \pi} \sin m x \cos m x \rd x       | r = \intlimits {\frac {\sin^2 m x} {2 m} } \alpha {\alpha + 2 \pi}       | c =  }} {{eqn | r = \frac {\map {\sin^2} {m \paren {\alpha + 2 \pi} } } {2 m} - \frac {\sin^2 m \alpha} {2 m}       | c =  }} {{eqn | r = \frac {\sin^2 m \alpha} {2 m} - \frac {\sin^2 m \alpha} {2 m}       | c = Corollary to Cosine of Angle plus Full Angle }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{qed}}
Let $m \ne n$. {{begin-eqn}} {{eqn | l = \int \sin m x \rd x       | r = -\frac {\cos m x} m + C       | c = Primitive of $\sin m x$ }} {{eqn | ll= \leadsto       | l = \int_\alpha^{\alpha + 2 \pi} \sin m x \rd x       | r = \intlimits {-\frac {\cos m x} m} \alpha {\alpha + 2 \pi}       | c =  }} {{eqn | r = \paren {-\frac {\map \cos {m \paren {\alpha + 2 \pi} } } m} - \paren {-\frac {\cos m \alpha} a}       | c =  }} {{eqn | r = \paren {-\frac {\cos m \alpha} m} - \paren {-\frac {\cos m \alpha} a}       | c = Corollary to Cosine of Angle plus Full Angle }} {{eqn | r = 0       | c = after simplification }} {{end-eqn}} {{qed|lemma}} Let $m = 0$. {{begin-eqn}} {{eqn | l = \int \sin 0 x \rd x       | r = \int 0 \rd x       | c = Sine of Zero is Zero }} {{eqn | ll= \leadsto       | l = \int_\alpha^{\alpha + 2 \pi} \sin 0 x \rd x       | r = \bigintlimits 0 \alpha {\alpha + 2 \pi}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{qed}}
Let $m \ne 0$. {{begin-eqn}} {{eqn | l = \int \cos m x \rd x       | r = \frac {\sin m x} m + C       | c = Primitive of $\cos m x$ }} {{eqn | ll= \leadsto       | l = \int_\alpha^{\alpha + 2 \pi} \cos m x \rd x       | r = \intlimits {\frac {\sin m x} m} \alpha {\alpha + 2 \pi}       | c =  }} {{eqn | r = \paren {\frac {\map \sin {m \paren {\alpha + 2 \pi} } } m} - \paren {\frac {\sin m \alpha} m}       | c =  }} {{eqn | r = \paren {\frac {\sin m \alpha} m} - \paren {\frac {\sin m \alpha} m}       | c = Corollary to Sine of Angle plus Full Angle }} {{eqn | r = 0 - 0       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{qed|lemma}} Let $m = 0$. {{begin-eqn}} {{eqn | l = \int \cos 0 x \rd x       | r = \int 1 \rd x       | c = Cosine of Zero is One }} {{eqn | r = x + C       | c = Primitive of Constant }} {{eqn | ll= \leadsto       | l = \int_\alpha^{\alpha + 2 \pi} \cos 0 x \rd x       | r = \bigintlimits x \alpha {\alpha + 2 \pi}       | c =  }} {{eqn | r = \alpha + 2 \pi - \alpha       | c =  }} {{eqn | r = 2 \pi       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map f x       | r = \dfrac {a_0} 2 + \sum_{m \mathop = 1}^\infty \paren {a_m \cos m x + b_m \sin m x}       | c =  }} {{eqn | ll= \leadsto       | l = \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x       | r = \int_\alpha^{\alpha + 2 \pi} \paren {\dfrac {a_0} 2 + \sum_{m \mathop = 1}^\infty \paren {a_m \cos m x + b_m \sin m x} } \cos n x \rd x       | c =  }} {{eqn | r = \int_\alpha^{\alpha + 2 \pi} \dfrac {a_0} 2 \cos n x \rd x       | c =  }} {{eqn | o =        | ro= +       | r = \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} \paren {a_m \cos m x + b_m \sin m x} \cos n x \rd x}       | c =  }} {{eqn | r = \frac {a_0} 2 2 \pi \delta_{n 0} + \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} \paren {a_m \cos m x + b_m \sin m x} \cos n x \rd x}       | c = $\displaystyle \int_\alpha^{\alpha + 2 \pi} \cos n x \rd x = \begin {cases} 0 & : n \ne 0 \\ 2 \pi & : n = 0 \end {cases}$ }} {{eqn | r = a_0 \pi \delta_{n 0} + \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} a_m \cos m x \cos n x \rd x}       | c =  }} {{eqn | o =        | ro= +       | r = \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} b_m \sin m x \cos n x \rd x}       | c =  }} {{eqn | r = a_0 \pi \delta_{n 0} + \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} a_m \cos m x \cos n x \rd x}       | c = $\displaystyle \int_\alpha^{\alpha + 2 \pi} \sin m x \cos n x \rd x = 0$ }} {{eqn | r = a_0 \pi \delta_{n 0} + \sum_{m \mathop = 1}^\infty a_m \pi \delta_{m n}       | c = $\displaystyle \int_\alpha^{\alpha + 2 \pi} \cos m x \cos n x \rd x = \begin {cases} 0 & : m \ne n \\ \pi & : m = n \end {cases}$ }} {{end-eqn}} Thus when $n = 0$ we have: :$\displaystyle \int_\alpha^{\alpha + 2 \pi} \map f x \rd x = \int_\alpha^{\alpha + 2 \pi} \map f x \cos 0 x \rd x = a_0 \pi$ and when $n \ne 0$ we have: :$\displaystyle \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x = a_n \pi$ Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \map f x       | r = \dfrac {a_0} 2 + \sum_{m \mathop = 1}^\infty \paren {a_m \cos m x + b_m \sin m x}       | c =  }} {{eqn | ll= \leadsto       | l = \int_\alpha^{\alpha + 2 \pi} \map f x \sin n x \rd x       | r = \int_\alpha^{\alpha + 2 \pi} \paren {\dfrac {a_0} 2 + \sum_{m \mathop = 1}^\infty \paren {a_m \cos m x + b_m \sin m x} } \sin n x \rd x       | c =  }} {{eqn | r = \int_\alpha^{\alpha + 2 \pi} \dfrac {a_0} 2 \sin n x \rd x + \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} \paren {a_m \cos m x + b_m \sin m x} \sin n x \rd x}       | c =  }} {{eqn | r = 0 + \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} \paren {a_m \cos m x + b_m \sin m x} \sin n x \rd x}       | c = $\displaystyle \int_\alpha^{\alpha + 2 \pi} \sin n x \rd x = 0$ }} {{eqn | r = \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} a_m \cos m x \sin n x \rd x} + \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} b_m \sin m x \sin n x \rd x}       | c =  }} {{eqn | r = \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} b_m \sin m x \sin n x \rd x}       | c = $\displaystyle \int_\alpha^{\alpha + 2 \pi} \sin m x \cos n x \rd x = 0$ }} {{eqn | r = \sum_{m \mathop = 1}^\infty b_m \pi \delta_{m n}       | c = $\displaystyle \int_\alpha^{\alpha + 2 \pi} \sin m x \sin n x \rd x = \begin{cases} 0 & : m \ne n \\ \pi & : m = n \end{cases}$ }} {{eqn | r = b_n \pi        | c = {{Defof|Kronecker Delta}} }} {{end-eqn}} {{qed}}
From Sine of Angle plus Full Angle: :$\map \sin {x + 2 \pi} = \sin x$ The result follows from the General Periodicity Property: If: :$\forall x \in X: \map f x = \map f {x + L}$ then: :$\forall n \in \Z: \forall x \in X: \map f x = \map f {x + n L}$ {{qed}} Category:Sine Function gusvgiqkof0mwgv40vcs6ccv9omg34v
From Cosine of Angle plus Full Angle: :$\cos \left({x + 2 \pi}\right) = \cos x$ The result follows from the General Periodicity Property: If: : $\forall x \in X: f \left({x}\right) = f \left({x + L}\right)$ then: :$\forall n \in \Z: \forall x \in X: f \left({x}\right) = f \left({x + n L}\right)$ {{qed}} Category:Cosine Function g7c826pvzjeuni1efuplz3n4fklnr87
=== Lemma 1 === {{:Fourier's Theorem/Lemma 1}} === Lemma 2 === {{:Fourier's Theorem/Lemma 2}} === Lemma 3 === {{:Fourier's Theorem/Lemma 3}} === Main Theorem === Let $\map {S_N} x$ denote the first $N$ terms of the Fourier series: :$(2): \quad \map {S_N} x = \displaystyle \frac {a_0} 2 + \sum_{n \mathop = 1}^N \paren {a_n \cos n x + b_n \sin n x}$ where: :$(3): \quad \displaystyle a_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x$ :$(4): \quad \displaystyle b_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \sin n x \rd x$ Substituting from $(3)$ and $(4)$ into $(2)$ and rearranging: :$\map {S_N} x = \displaystyle \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f u \paren {\frac 1 2 + \sum_{n \mathop = 1}^N \paren {\cos n x \cos n u + \sin n x \sin n u} } \rd u$ Now we have: :$\dfrac 1 2 + \displaystyle\sum_{n \mathop = 1}^N \paren {\cos n x \cos n u + \sin n x \sin n u} = \frac {\map \sin {\paren {N + \frac 1 2} \paren {u - x} } } {2 \, \map \sin {\frac 1 2 \paren {u - x} } }$ {{explain|There should be a link somewhere to give the above}} Hence: :$\displaystyle \map {S_N} x = \int_\alpha^{\alpha + 2 \pi} \map \psi u \frac {\map \sin {\paren {N + \frac 1 2} \paren {u - x} } } {2 \, \map \sin {\frac 1 2 \paren {u - x} } } \rd u$ where: :$\map \psi u = \dfrac 1 \pi \map f u \dfrac {\frac 1 2 \paren {u - x} } {\sin \frac 1 2 \paren {u - x} }$ We have that $\map f u$ satisfies the Dirichlet conditions on $\openint \alpha {\alpha + 2 \pi}$. {{proofread|In the source book given, when initially stated, this theorem was as given here. When it is proved in the next chapter, it is given as being piecewise smooth. I believe one implies the other, but not sure. Initial thoughts are that piecewise smooth is stronger.}} Hence $f$ is piecewise smooth on $\openint \alpha {\alpha + 2 \pi}$. That is, $f$ has right-hand derivative and left-hand derivative at all $x$ in $\openint \alpha {\alpha + 2 \pi}$. Thus at the point $u = x$, $f$ has right-hand derivative and left-hand derivative, and so does $\map \psi u$. So by Fourier's Theorem: Lemma 3: :$\displaystyle \lim_{n \mathop \to N} \map {S_N} x = \frac \pi 2 \paren {\map \psi {x^+} + \map \psi {x^-} }$ Now: :$\map \psi {x^+} = \displaystyle \frac 1 \pi \map f {x^+} \lim_{u \mathop \to x} \dfrac {\frac 1 2 \paren {u - x} } {\sin \frac 1 2 \paren {u - x} } = \frac 1 \pi \map f {x^+}$ and: :$\map \psi {x^-} = \displaystyle \frac 1 \pi \map f {x^-} \lim_{u \mathop \to x} \dfrac {\frac 1 2 \paren {u - x} } {\sin \frac 1 2 \paren {u - x} } = \frac 1 \pi \map f {x^-}$ and so: :$\displaystyle \lim_{n \mathop \to N} \map {S_N} x = \frac 1 2 \paren {\lim_{x \mathop \to a^+} \map f x + \lim_{x \mathop \to a^-} \map f x}$ {{qed}}
Each of the intervals $\mathbb I_k$ is such that $f \left[{\mathbb I_k}\right]$ is a constant function: :$\forall x \in \mathbb I_k: f \left({x}\right) = \lambda_k$ Thus: {{begin-eqn}} {{eqn | l = \int_{\mathbb I_k} f \left({x}\right) \, \mathrm d x       | r = \int_{\alpha_k}^{\beta_k} \lambda_k \, \mathrm d x       | c =  }} {{eqn | r = \lambda_k \left({\beta_k - \alpha_k}\right)       | c = Definite Integral of Constant }} {{end-eqn}} From the corollary to Sum of Integrals on Adjacent Intervals for Integrable Functions: {{begin-eqn}} {{eqn | l = \int_\alpha^\beta f \left({x}\right) \, \mathrm d x       | r = \sum_{k \mathop = 1}^n \int_{\mathbb I_k} f \left({x}\right) \, \mathrm d x       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n \lambda_k \left({\beta_k - \alpha_k}\right)       | c =  }} {{end-eqn}} {{qed}} Category:Step Functions Category:Definite Integrals 44ardltc52sdkvrwqx5lmkm9g1o583q
Recall the definition of step function: :{{Definition:Step Function}} Recall the Dirichlet conditions: {{:Definition:Dirichlet Conditions}} We inspect the Dirichlet conditions in turn. === $(\text D 1)$: Absolute Integrability === From Definite Integral of Step Function: :$\displaystyle \int_\alpha^\beta \map f x \rd x = \sum_{k \mathop = 1}^n \lambda_k \paren {\beta_k - \alpha_k}$ where $\alpha_k, \beta_k$ are the endpoints of $\mathbb I_k$ for $1 \le k \le n$. When $\lambda_k < 1$, we have that: :$\size {\lambda_k} = -\lambda_k$ and so: :$\displaystyle \int_{\alpha_k}^{\beta_k} \lambda_k \rd x = -\lambda_k \paren {\beta_k - \alpha_k}$ Absolute integrability follows. {{qed|lemma}} === $(\text D 2)$: Finite Number of Local Maxima and Minima === By nature of $f$ being a step function, the image set of $f$ is finite: :$f \openint \alpha \beta = \set {\lambda_1, \lambda_2, \ldots, \lambda n}$ The set of local maxima and the set of local minima must be subsets of $f \openint \alpha \beta$. Therefore there must be a finite number of each. {{qed|lemma}} === $(\text D 3)$: Discontinuities are Finite in Number and Nature === From Constant Real Function is Continuous, there are no discontinuities except perhaps at the endpoints of $\mathbb I_k$ for $1 \le k \le n$. Let us inspect those endpoints. Let $\alpha_k, \beta_k$ be the endpoints of $\mathbb I_k$ for $1 \le k \le n$, where $\alpha_k < \beta_k$. Let $\mathbb I_1, \mathbb I_2, \ldots, \mathbb I_n$ be ordered such that: :$\forall k: \alpha_k < \alpha_{k + 1}$ By definition, $\beta_{k - 1} = \alpha_k$. Thus we have that: {{begin-eqn}} {{eqn | l = \lim_{x \mathop \to {\beta_{k - 1} }^-}       | r = \lim_{x \mathop \to {\alpha_k}^-}       | c =  }} {{eqn | r = \lambda_{k - 1}       | c =  }} {{eqn | l = \lim_{x \mathop \to {\alpha_k}^+}       | r = \lambda_k       | c =  }} {{end-eqn}} and if $\lambda_k \ne \lambda_{k - 1}$ then this is indeed a finite discontinuity. By the nature of the step function, there are a finite number of them. {{qed|lemma}} Thus all Dirichlet conditions are satisfied by $f$. {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {2 \pi} \int_{-\pi}^\pi \size {\map f x}^2 \rd x       | r = \frac 1 {2 \pi} \int_{-\pi}^\pi \map f x \overline {\map f x} \rd x       | c = Modulus in Terms of Conjugate }} {{eqn | r = \frac 1 {2 \pi} \int_{-\pi}^\pi \sum_{n \mathop = -\infty}^\infty c_n e^{i n x} \overline {\sum_{m \mathop = -\infty}^\infty c_m e^{i m x} } \rd x }} {{eqn | r = \frac 1 {2 \pi} \int_{-\pi}^\pi \sum_{n \mathop = -\infty}^\infty c_n e^{i n x} \sum_{m \mathop = -\infty}^\infty \overline {c_m} e^{-i m x} \rd x       | c = Sum of Complex Conjugates }} {{eqn | r = \frac 1 {2 \pi} \int_{-\pi}^\pi \sum_{n \mathop = -\infty}^\infty \sum_{m \mathop = -\infty}^\infty c_n \overline {c_m} e^{i x \left({n - m}\right)} \rd x }} {{eqn | r = \frac 1 {2 \pi} \sum_{n \mathop = -\infty}^\infty \sum_{m \mathop = -\infty}^\infty c_n \overline {c_m} \int_{-\pi}^\pi e^{i x \left({n - m}\right)} \rd x       | c = Fubini's Theorem }} {{eqn | r = \frac 1 {2 \pi} \sum_{n \mathop = -\infty}^\infty \sum_{m \mathop = -\infty}^\infty c_n \overline {c_m} 2 \pi \delta_{n m}       | c = Integral over 2 pi of Exponential of i by n x }} {{eqn | r = \frac {2 \pi} {2 \pi} \sum_{n \mathop = -\infty}^\infty c_n \overline {c_n} }} {{eqn | r = \sum_{n \mathop = -\infty}^\infty \size {c_n}^2  }} {{end-eqn}} {{qed}} {{Namedfor|Marc-Antoine Parseval}} Category:Parseval's Theorem n7lwtqwrxeozi2tv31js1j5lz1lxomz
{{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^{n + 1} } {n^2}       | r = \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^2} - \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n}\right)^2}       | c = separating odd positive terms from even negative terms }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^2} - \frac 1 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} }} {{eqn | r = \frac {\pi^2} 8 - \frac 1 4 \times \frac {\pi^2} 6       | c = Sum of Reciprocals of Squares of Odd Integers, Basel Problem }} {{eqn | r = \frac {\pi^2 \left({3 - 1}\right)} {24} }} {{eqn | r = \frac {\pi^2} {12} }} {{end-eqn}} {{qed}}
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as: :$\map f x = \begin {cases} \paren {x - \pi}^2 & : 0 < x \le \pi \\ \pi^2 & : \pi < x < 2 \pi \end {cases}$ From Fourier Series: Square of x minus pi, Square of pi, its Fourier series can be expressed as: :$(1): \quad \map f x \sim \displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos n x} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin n x}$ We have that: {{begin-eqn}} {{eqn | l = \map f {\pi - 0}       | r = \paren {\pi - \pi}^2       | c =  }} {{eqn | r = 0       | c =  }} {{eqn | l = \map f {\pi + 0}       | r = \pi^2       | c =  }} {{end-eqn}} where $\map f {\pi - 0}$ and $\map f {\pi + 0}$ denote the limit from the left and limit from the right respectively of $\map f \pi$. It is apparent that $\map f x$ satisfies the Dirichlet conditions: :$(\text D 1): \quad f$ is bounded on $\openint 0 {2 \pi}$ :$(\text D 2): \quad f$ has a finite number of local maxima and local minima. :$(\text D 3): \quad f$ has $1$ of discontinuity, which is finite. Hence from Fourier's Theorem: {{begin-eqn}} {{eqn | l = \map f \pi       | r = \frac {\map f {\pi - 0} + \map f {\pi + 0} } 2       | c =  }} {{eqn | r = \frac {0 + \pi^2} 2       | c =  }} {{eqn | r = \frac {\pi^2} 2       | c =  }} {{end-eqn}} Thus setting $x = \pi$ in $(1)$: {{begin-eqn}} {{eqn | l = \map f \pi       | r = \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos n \pi} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin n \pi}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\pi^2} 2       | r = \frac {2 \pi^2} 3 + 2 \sum_{n \mathop = 1}^\infty \frac {\cos n \pi} {n^2}       | c = Sine of Multiple of Pi }} {{eqn | ll= \leadsto       | l = \frac {\pi^2} 4       | r = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {n^2}       | c = Cosine of Multiple of Pi and simplification }} {{eqn | ll= \leadsto       | l = -\frac {\pi^2} {12}       | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {n^2}        | c =  }} {{eqn | ll= \leadsto       | l = \frac {\pi^2} {12}       | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } {n^2}       | c = changing sign and subsuming into powers of $-1$ }} {{end-eqn}} {{qed}}
Let $\map f x$ be the real function defined on $\openint {-\pi} \pi$ as: :$\map f x = \pi^2 - x^2$ By Fourier Series: $\pi^2 - x^2$ over $\openint {-\pi} \pi$: :$\displaystyle \pi^2 - x^2 \sim \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\cos n x} {n^2}$ for $x \in \openint {-\pi} \pi$. Setting $x = 0$: {{begin-eqn}} {{eqn | l = \pi^2 - 0^2       | r = \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\cos 0} {n^2} }} {{eqn | r = \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac 1 {n^2}       | c = Cosine of Zero is One }} {{eqn | ll= \leadsto       | l = \frac {\pi^2} {12}       | r = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac 1 {n^2}       | c = simplification }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } {n^2}       | c = rearrangement }} {{end-eqn}} {{qed}}