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Question Number 25374 by christoph last updated on 09/Dec/17

Commented by prakash jain last updated on 09/Dec/17

F(w)=F{e^(−πx^2 ) } =∫_(−∞) ^( ∞) e^(−πx^2 ) e^(−ixw) dx f(x)=e^(−πx^2 ) f ′(x)=−2πxe^(−πx^2 ) =−2πxf(x) (i) F(f ′(x))=iwF(f(x)) F(xf(x))=−iF′(f (x)) from (i) taking fourier transform iwF(f(x))=−2πiF′(f(x)) we need to solve the above differential equation to to get F(f(x)). F(w)=c_1 e^(−(w^2 /(4π))) for two variable function i think f(x,y)=e^(−π(x^2 +y^2 )) F(f(x,y))=∫∫e^(−π(x^2 +y^2 )) e^(−i(w_1 x+w_2 y)) dxdy F(f(x,y))=c_1 e^(−(w_1 ^2 /(4π))) ∙c_2 e^(−(w_2 ^2 /(4π))) i am not 100% sure about the answer. pls recheck.

$$\mathscr{F}\left({w}\right)=\mathscr{F}\left\{{e}^{−\pi{x}^{\mathrm{2}} } \right\} \\ $$$$=\int_{−\infty} ^{\:\infty} {e}^{−\pi{x}^{\mathrm{2}} } {e}^{−{ixw}} {dx} \\ $$$${f}\left({x}\right)={e}^{−\pi{x}^{\mathrm{2}} } \\ $$$${f}\:'\left({x}\right)=−\mathrm{2}\pi{xe}^{−\pi{x}^{\mathrm{2}} } =−\mathrm{2}\pi{xf}\left({x}\right)\:\left({i}\right) \\ $$$$\mathscr{F}\left({f}\:'\left({x}\right)\right)={iw}\mathscr{F}\left({f}\left({x}\right)\right) \\ $$$$\mathscr{F}\left({xf}\left({x}\right)\right)=−{i}\mathscr{F}'\left({f}\:\left({x}\right)\right) \\ $$$${from}\:\left({i}\right)\:{taking}\:{fourier}\:{transform} \\ $$$${iw}\mathscr{F}\left({f}\left({x}\right)\right)=−\mathrm{2}\pi{i}\mathscr{F}'\left({f}\left({x}\right)\right) \\ $$$${we}\:{need}\:{to}\:{solve}\:{the}\:{above}\:{differential} \\ $$$${equation}\:{to}\:{to}\:{get}\:\mathscr{F}\left({f}\left({x}\right)\right). \\ $$$$\mathscr{F}\left({w}\right)={c}_{\mathrm{1}} {e}^{−\frac{{w}^{\mathrm{2}} }{\mathrm{4}\pi}} \\ $$$${for}\:{two}\:{variable}\:{function}\:{i}\:{think} \\ $$$${f}\left({x},{y}\right)={e}^{−\pi\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} \\ $$$$\mathscr{F}\left({f}\left({x},{y}\right)\right)=\int\int{e}^{−\pi\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {e}^{−{i}\left({w}_{\mathrm{1}} {x}+{w}_{\mathrm{2}} {y}\right)} {dxdy} \\ $$$$\mathscr{F}\left({f}\left({x},{y}\right)\right)={c}_{\mathrm{1}} {e}^{−\frac{{w}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{4}\pi}} \centerdot{c}_{\mathrm{2}} {e}^{−\frac{{w}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{4}\pi}} \\ $$$$\mathrm{i}\:\mathrm{am}\:\mathrm{not}\:\mathrm{100\%}\:\mathrm{sure}\:\mathrm{about}\:\mathrm{the}\:\mathrm{answer}. \\ $$$$\mathrm{pls}\:\mathrm{recheck}. \\ $$

Commented by christoph last updated on 10/Dec/17

Thank you sir!

$${Thank}\:{you}\:{sir}! \\ $$

Commented by christoph last updated on 10/Dec/17

You are right! Thank so much:)

$$\left.{You}\:{are}\:{right}!\:{Thank}\:{so}\:{much}:\right) \\ $$

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