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Question Number 185012 by aba last updated on 15/Jan/23

Σ_(k=0) ^n sin(kx)=Σ_(k=0) ^n Im(e^(ikx) ) =Im(Σ_(k=0) ^n (e^(ix) )^k ) =Im(((1−e^(i(n+1)x) )/(1−e^(ix) ))) ((1−e^(i(n+1)x) )/(1−e^(ix) ))=((e^(i(n+1)(x/2)) (e^(−i(n+1)(x/2)) −e^(i(n+1)(x/2)) ))/(e^(i(x/2)) (e^(−i(x/2)) −e^(i(x/2)) ))) =e^(in(x/2)) ×((−2sin((n+1)(x/2)))/(−2sin((x/2)))) =e^(in(x/2)) ((sin((n+1)(x/2)))/(sin((x/2)))) Σ_(k=0) ^n sin(kx)=((sin((n+1)(x/2) ))/(sin((x/2))))Im(e^(in(x/2)) ) =((sin((n+1)(x/2)))/(sin((x/2))))sin(((nx)/2))

$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{sin}\left(\mathrm{kx}\right)=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{Im}\left(\mathrm{e}^{\mathrm{ikx}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{Im}\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{e}^{\mathrm{ix}} \right)^{\mathrm{k}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{Im}\left(\frac{\mathrm{1}−\mathrm{e}^{\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}} }{\mathrm{1}−\mathrm{e}^{\mathrm{ix}} }\right) \\ $$$$\frac{\mathrm{1}−\mathrm{e}^{\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}} }{\mathrm{1}−\mathrm{e}^{\mathrm{ix}} }=\frac{\mathrm{e}^{\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}} \left(\mathrm{e}^{−\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}} −\mathrm{e}^{\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}} \right)}{\mathrm{e}^{\mathrm{i}\frac{\mathrm{x}}{\mathrm{2}}} \left(\mathrm{e}^{−\mathrm{i}\frac{\mathrm{x}}{\mathrm{2}}} −\mathrm{e}^{\mathrm{i}\frac{\mathrm{x}}{\mathrm{2}}} \right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{e}^{\mathrm{in}\frac{\mathrm{x}}{\mathrm{2}}} ×\frac{−\mathrm{2sin}\left(\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}\right)}{−\mathrm{2sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{e}^{\mathrm{in}\frac{\mathrm{x}}{\mathrm{2}}} \frac{\mathrm{sin}\left(\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} \\ $$$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{sin}\left(\mathrm{kx}\right)=\frac{\mathrm{sin}\left(\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}\:\right)}{\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\mathrm{Im}\left(\mathrm{e}^{\mathrm{in}\frac{\mathrm{x}}{\mathrm{2}}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{sin}\left(\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\mathrm{sin}\left(\frac{\mathrm{nx}}{\mathrm{2}}\right) \\ $$

Commented by Frix last updated on 15/Jan/23

!!!! Yes now it′s right.

$$!!!! \\ $$$$\mathrm{Yes}\:{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{right}. \\ $$

Commented by aba last updated on 15/Jan/23

thank

$$\mathrm{thank}\: \\ $$

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