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Question Number 218769 by SdC355 last updated on 15/Apr/25
Fourier Series f(θ)=e^(izsin(θ))
$$\mathrm{Fourier}\:\mathrm{Series}\:{f}\left(\theta\right)={e}^{\boldsymbol{{i}}{z}\mathrm{sin}\left(\theta\right)} \\ $$
Answered by MrGaster last updated on 16/Apr/25
e^(iz sin(θ)) =Σ_(n=0) ^∞ (((iz sin(θ))^n )/(n!)) sin(θ)=^ sin(θ)=((e^(iθ) −e^(−iθ) )/(2i)) ∴iz sin(θ)=iz∙((e^(iθ) −e^(−iθ) )/(2i))=(z/2)(e^(iθ) −e^(iθ) ) e^(iz sin(θ)) =e^((x/2)(e^(iθ) −e^(iθ) )) =Σ_(n=0) ^∞ ((((z/2)(e^(iθ) −e^(−iθ) ))^n )/(n!)) (e^(iθ) −e^(iθ) )=Σ_(k=0) ^n ((n),(k) )i^(ikθ) (−1)^(n−k) e^(−i(n−k)θ) =Σ_(k=0) ^n ((n),(k) )(−1)^(n−k) e^(i(2k−n)θ) ∴e^(iz sin(θ)) =Σ_(n=0) ^∞ (z^n /(2^n n!))Σ_(k=0) ^n ((n),(k) )(−1)^(n−k) e^(i(2k−n)θ) e^(iz sin(θ)) =Σ_(k=0) ^∞ Σ_(n=k) ^∞ (z^n /(2^n n!)) ((n),(k) )(−1)^(n−k) e^(i(2k−n)θ) let m=2k−n,n=2k−m ⇒e^(z sin(θ)) =Σ_(m=−∞) ^∞ (Σ_(k=max(0,m)) ^∞ (z^(2k−m) /(2^(2k−m) (2k−m)!)) (((2k−m)),(k) )(−1)^((2k−m)−k) ) e^(iz sin(θ)) =Σ_(m=−∞) ^∞ c_m e^(imθ) Where c_m Fourier coefficient is: c_m =Σ_(k=max(0,m)) ^∞ (z^(2k−m) /(2^(2k−m) (2k−m)!)) (((2k−m)),(k) )(−1)^(k−m) Compared with Besselsr geneating function it can bex epressed as: e^(iz sin(θ)) =Σ_(k=−∞) ^∞ J_k (iz)e^(ikθ) =Σ_(k=−∞) ^∞ i^k I_k (z)e^(ikθ) Where I_k (z) is a modified Besselfunction.
$${e}^{{iz}\:\mathrm{sin}\left(\theta\right)} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({iz}\:\mathrm{sin}\left(\theta\right)\right)^{{n}} }{{n}!} \\ $$$$\mathrm{sin}\left(\theta\right)\overset{ } {=}\mathrm{sin}\left(\theta\right)=\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}{i}} \\ $$$$\therefore{iz}\:\mathrm{sin}\left(\theta\right)={iz}\centerdot\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}{i}}=\frac{{z}}{\mathrm{2}}\left({e}^{{i}\theta} −{e}^{{i}\theta} \right) \\ $$$${e}^{{iz}\:\mathrm{sin}\left(\theta\right)} ={e}^{\frac{{x}}{\mathrm{2}}\left({e}^{{i}\theta} −{e}^{{i}\theta} \right)} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{{z}}{\mathrm{2}}\left({e}^{{i}\theta} −{e}^{−{i}\theta} \right)\right)^{{n}} }{{n}!} \\ $$$$\left({e}^{{i}\theta} −{e}^{{i}\theta} \right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{i}^{{ik}\theta} \left(−\mathrm{1}\right)^{{n}−{k}} {e}^{−{i}\left({n}−{k}\right)\theta} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\left(−\mathrm{1}\right)^{{n}−{k}} {e}^{{i}\left(\mathrm{2}{k}−{n}\right)\theta} \\ $$$$\therefore{e}^{{iz}\:\mathrm{sin}\left(\theta\right)} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{z}^{{n}} }{\mathrm{2}^{{n}} {n}!}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\left(−\mathrm{1}\right)^{{n}−{k}} {e}^{{i}\left(\mathrm{2}{k}−{n}\right)\theta} \\ $$$${e}^{{iz}\:\mathrm{sin}\left(\theta\right)} =\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}={k}} {\overset{\infty} {\sum}}\frac{{z}^{{n}} }{\mathrm{2}^{{n}} {n}!}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\left(−\mathrm{1}\right)^{{n}−{k}} {e}^{{i}\left(\mathrm{2}{k}−{n}\right)\theta} \\ $$$$\mathrm{let}\:{m}=\mathrm{2}{k}−{n},{n}=\mathrm{2}{k}−{m} \\ $$$$\Rightarrow{e}^{{z}\:\mathrm{sin}\left(\theta\right)} =\underset{{m}=−\infty} {\overset{\infty} {\sum}}\left(\underset{{k}=\mathrm{max}\left(\mathrm{0},{m}\right)} {\overset{\infty} {\sum}}\frac{{z}^{\mathrm{2}{k}−{m}} }{\mathrm{2}^{\mathrm{2}{k}−{m}} \left(\mathrm{2}{k}−{m}\right)!}\begin{pmatrix}{\mathrm{2}{k}−{m}}\\{{k}}\end{pmatrix}\left(−\mathrm{1}\right)^{\left(\mathrm{2}{k}−{m}\right)−{k}} \right) \\ $$$${e}^{{iz}\:\mathrm{sin}\left(\theta\right)} =\underset{{m}=−\infty} {\overset{\infty} {\sum}}{c}_{{m}} {e}^{{im}\theta} \\ $$$$\mathrm{Where}\:{c}_{{m}} \:\mathrm{Fourier}\:\mathrm{coefficient}\:\mathrm{is}: \\ $$$${c}_{{m}} =\underset{{k}=\mathrm{max}\left(\mathrm{0},{m}\right)} {\overset{\infty} {\sum}}\frac{{z}^{\mathrm{2}{k}−{m}} }{\mathrm{2}^{\mathrm{2}{k}−{m}} \left(\mathrm{2}{k}−{m}\right)!}\begin{pmatrix}{\mathrm{2}{k}−{m}}\\{{k}}\end{pmatrix}\left(−\mathrm{1}\right)^{{k}−{m}} \\ $$$$\mathrm{Compared}\:\mathrm{with}\:\mathrm{Besselsr} \\ $$$$\mathrm{geneating}\:\mathrm{function}\:\mathrm{it}\:\mathrm{can}\:\mathrm{bex} \\ $$$$\mathrm{epressed}\:\mathrm{as}: \\ $$$${e}^{{iz}\:\mathrm{sin}\left(\theta\right)} =\underset{{k}=−\infty} {\overset{\infty} {\sum}}{J}_{{k}} \left({iz}\right){e}^{{ik}\theta} =\underset{{k}=−\infty} {\overset{\infty} {\sum}}{i}^{{k}} {I}_{{k}} \left({z}\right){e}^{{ik}\theta} \\ $$$$\mathrm{Where}\:{I}_{{k}} \left({z}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{modified}\: \\ $$$$\mathrm{Besselfunction}. \\ $$
Commented by SdC355 last updated on 16/Apr/25
ohhh....! Thx!
$$\mathrm{ohhh}….!\:\mathrm{Thx}!\: \\ $$

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