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Find-0-pi-f-x-cosxdx-given-that-f-x-1-2-k-1-n-coskx-

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Question Number 1780 by 112358 last updated on 23/Sep/15
Find ∫_0 ^( π) f(x)cosxdx given that f(x)=1+2Σ_(k=1) ^n coskx.
$${Find}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\pi} {f}\left({x}\right){cosxdx} \\ $$$${given}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=\mathrm{1}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{coskx}.\: \\ $$
Commented by Rasheed Soomro last updated on 23/Sep/15
∫_0 ^( π) f(x)cosxdx f(x)=1+2Σ_(k=1) ^n coskx. −−−−−−−−−−−−−−− ∫_0 ^( π) (1+2Σ_(k=1) ^n coskx)cosxdx ∫_0 ^( π) cosxdx+2Σ_(k=1) ^n ∫_0 ^( π) (coskx)(cosx)dx
$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\pi} {f}\left({x}\right){cosxdx} \\ $$$${f}\left({x}\right)=\mathrm{1}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{coskx}.\: \\ $$$$−−−−−−−−−−−−−−− \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \left(\mathrm{1}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{coskx}\right){cosxdx} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} {cosxdx}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\int_{\mathrm{0}} ^{\:\pi} \left({coskx}\right)\left({cosx}\right){dx} \\ $$
Answered by 123456 last updated on 24/Sep/15
f(x)=1+2Σ_(k=1) ^n cos kx f(x)cos x=cos x+2Σ_(k=1) ^n cos kx cos x ∫_0 ^π f(x)cos x dx=∫_0 ^π cos x+2Σ_(k=1) ^n ∫_0 ^π cos kx cos xdx ∫_0 ^π cos xdx=sin π−sin 0=0 ∫_0 ^π cos^2 xdx=(1/2)∫_0 ^π dx+(1/2)∫_0 ^π cos 2xdx=(1/2)π continue
$${f}\left({x}\right)=\mathrm{1}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{cos}\:{kx} \\ $$$${f}\left({x}\right)\mathrm{cos}\:{x}=\mathrm{cos}\:{x}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{cos}\:{kx}\:\mathrm{cos}\:{x} \\ $$$$\underset{\mathrm{0}} {\overset{\pi} {\int}}{f}\left({x}\right)\mathrm{cos}\:{x}\:{dx}=\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{cos}\:{x}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{cos}\:{kx}\:\mathrm{cos}\:{xdx} \\ $$$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{cos}\:{xdx}=\mathrm{sin}\:\pi−\mathrm{sin}\:\mathrm{0}=\mathrm{0}\: \\ $$$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{cos}^{\mathrm{2}} {xdx}=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi} {\int}}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{cos}\:\mathrm{2}{xdx}=\frac{\mathrm{1}}{\mathrm{2}}\pi \\ $$$$\mathrm{continue} \\ $$
Commented by 112358 last updated on 29/Sep/15
Graphically I found the value of the integral to be approximately π. Consider the identity coskxcosx=(1/2)(cos(k+1)x+cos(k−1)x) Then the integral under the summation becomes ∫_0 ^π (cos(k+1)x+cos(k−1)x)dx =(1/(k+1))sin(k+1)x+(1/(k−1))sin(k−1)x∣_0 ^π =((sin(k+1)π)/(k+1))+((sin(k−1)π)/(k−1)) With k∈Z^+ and k≥1 , sin(k+1)π=0 and sin(k−1)π=0. But ((sin(k−1)π)/(k−1)) is undefined at k=1. ∴ ∫_0 ^( π) f(x)cosx dx=πΣ_(k=1) ^n ((sin(k−1)π)/((k−1)π)) If k were a continuous variable we get that lim_(k→1) ((sin(k−1)π)/((k−1)π))=1 so that ∫_0 ^π f(x)cosxdx=π but I don′t feel comfortable with this approach since k is a discrete quantity.
$${Graphically}\:{I}\:{found}\:{the}\:{value} \\ $$$${of}\:{the}\:{integral}\:{to}\:{be}\:{approximately}\:\pi. \\ $$$${Consider}\:{the}\:{identity} \\ $$$${coskxcosx}=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left({k}+\mathrm{1}\right){x}+{cos}\left({k}−\mathrm{1}\right){x}\right) \\ $$$${Then}\:{the}\:{integral}\:{under}\:{the}\: \\ $$$${summation}\:{becomes} \\ $$$$\int_{\mathrm{0}} ^{\pi} \left({cos}\left({k}+\mathrm{1}\right){x}+{cos}\left({k}−\mathrm{1}\right){x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{{k}+\mathrm{1}}{sin}\left({k}+\mathrm{1}\right){x}+\frac{\mathrm{1}}{{k}−\mathrm{1}}{sin}\left({k}−\mathrm{1}\right){x}\mid_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{{sin}\left({k}+\mathrm{1}\right)\pi}{{k}+\mathrm{1}}+\frac{{sin}\left({k}−\mathrm{1}\right)\pi}{{k}−\mathrm{1}} \\ $$$${With}\:{k}\in\mathbb{Z}^{+} \:{and}\:{k}\geqslant\mathrm{1}\:,\:{sin}\left({k}+\mathrm{1}\right)\pi=\mathrm{0} \\ $$$${and}\:{sin}\left({k}−\mathrm{1}\right)\pi=\mathrm{0}.\:{But}\:\:\frac{{sin}\left({k}−\mathrm{1}\right)\pi}{{k}−\mathrm{1}} \\ $$$${is}\:{undefined}\:{at}\:{k}=\mathrm{1}. \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\:\pi} {f}\left({x}\right){cosx}\:{dx}=\pi\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{sin}\left({k}−\mathrm{1}\right)\pi}{\left({k}−\mathrm{1}\right)\pi} \\ $$$${If}\:{k}\:{were}\:{a}\:{continuous}\:{variable} \\ $$$${we}\:{get}\:{that}\:\underset{{k}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{sin}\left({k}−\mathrm{1}\right)\pi}{\left({k}−\mathrm{1}\right)\pi}=\mathrm{1} \\ $$$${so}\:{that}\:\int_{\mathrm{0}} ^{\pi} {f}\left({x}\right){cosxdx}=\pi \\ $$$${but}\:{I}\:{don}'{t}\:{feel}\:{comfortable}\:{with} \\ $$$${this}\:{approach}\:{since}\:{k}\:{is}\:{a}\:{discrete} \\ $$$${quantity}. \\ $$$$ \\ $$

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