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Question Number 89913 by M±th+et£s last updated on 20/Apr/20

Commented by M±th+et£s last updated on 20/Apr/20

$$provethat$$

Answered by maths mind last updated on 20/Apr/20

$$1+i=\sqrt{2}{e}^{i\frac{\pi }{4}}\Rightarrow$$$$=\sum _{k⩾1}\frac{{\left(\sqrt{2}\right)}^{k^2}{e}^{i\frac{k^2\pi }{4}}}{{\left(\sqrt{2}\right)}^{k^2}{k}^2}$$$$=\sum _{k⩾1}(cos\left(\frac{k^2\pi }{4}\right)+isin\left(\frac{k^2\pi }{4}\right)).\frac{1}{k^2}$$$$=\sum _{k⩾1}\frac{cos\left(\frac{k^2\pi }{4}\right)}{k^2}+i\sum _{k⩾1}\frac{sin\left(\frac{k^2\pi }{4}\right)}{k^2}$$$$=\underset{k=0}{\sum ^{+\mathrm{\infty }}}\frac{cos\left((k^2\pi +k\pi +\frac{\pi }{4})\right)}{{(2k+1)}^2}+\sum _{k⩾1}\frac{cos\left(k^2\pi \right)}{4k^2}+i\sum _{k=0}\frac{sin(k^2\pi +k\pi +\frac{\pi }{4})}{{(2k+1)}^2}$$$$=\sum _{k⩾0}\frac{{(-1)}^{k(k+1)}}{{(2k+1)}^2}.\frac{\sqrt{2}}{2}+\sum _{k⩾1}\frac{{(-1)}^k}{4k^2}+i\sum _{k⩾0}\frac{\sqrt{2}}{2}.\frac{1}{{(2k+1)}^2}$$$$=\sum _{k⩾0}\frac{1}{{(2k+1)}^2}.\frac{\sqrt{2}}{2}+\sum _{k⩾1}\frac{{(-1)}^k}{4k^2}+\frac{i}{\sqrt{2}}\sum _{k⩾0}\frac{1}{{(2k+1)}^2}$$$$=\frac{\sqrt{2}}{2}\left(\frac{{\pi }^2}{8}\right)+\frac{1}{16}.\frac{{\pi }^2}{6}-\frac{{\pi }^2}{32}+\frac{i}{\sqrt{2}}.\frac{{\pi }^2}{8}$$$$=\frac{{\pi }^2\sqrt{2}}{16}-\frac{{\pi }^2}{48}+\frac{i{\pi }^2}{8\sqrt{2}}=\frac{(3\sqrt{2}-1){\pi }^2}{48}+\frac{i{\pi }^2}{8\sqrt{2}}$$

Commented by M±th+et£s last updated on 20/Apr/20

$$godblessyousir.$$$$arethereanybooksyourecommend$$$$reading?$$

Commented by maths mind last updated on 20/Apr/20

$$AlmostimpposibleSeriesandintegral$$$$iwillchektherighttitelandcomback$$

Commented by M±th+et£s last updated on 20/Apr/20

$$thankyouverrymuchsir$$

Commented by M±th+et£s last updated on 20/Apr/20

$$youmean(PaulJ.Nahin)book$$

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