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Question Number 119754 by Bird last updated on 26/Oct/20
find Σ_(n=1) ^∞ (u_n /(n!)) if u_n =u_(n+1) +u_(n−1)
$${find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{u}_{{n}} }{{n}!}\:{if}\:\:{u}_{{n}} \:={u}_{{n}+\mathrm{1}} +{u}_{{n}−\mathrm{1}} \\ $$
Commented by Dwaipayan Shikari last updated on 26/Oct/20
u_n =u_(n+1) +u_(n−1) r^n =r^(n+1) +r^(n−1) 1=r+(1/r)⇒r^2 −r+1=0⇒r=((1±i(√3))/2)=e^(±((iπ)/3)) u_n =Υe^((iπn)/3) +Λe^(−((iπn)/3)) Σ_(n=1) ^∞ (u_n /(n!))=ΥΣ_(n=1) ^∞ (((e^((iπ)/3) )^n )/(n!))+ΛΣ_(n=1) ^∞ (((e^((−iπ)/3) )^n )/(n!)) =Υ(e^e^((iπ)/3) −1)+Λ(e^e^((−iπ)/3) −1)
$${u}_{{n}} ={u}_{{n}+\mathrm{1}} +{u}_{{n}−\mathrm{1}} \\ $$$${r}^{{n}} ={r}^{{n}+\mathrm{1}} +{r}^{{n}−\mathrm{1}} \\ $$$$\mathrm{1}={r}+\frac{\mathrm{1}}{{r}}\Rightarrow{r}^{\mathrm{2}} −{r}+\mathrm{1}=\mathrm{0}\Rightarrow{r}=\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}={e}^{\pm\frac{{i}\pi}{\mathrm{3}}} \\ $$$${u}_{{n}} =\Upsilon{e}^{\frac{{i}\pi{n}}{\mathrm{3}}} +\Lambda{e}^{−\frac{{i}\pi{n}}{\mathrm{3}}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{u}_{{n}} }{{n}!}=\Upsilon\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({e}^{\frac{{i}\pi}{\mathrm{3}}} \right)^{{n}} }{{n}!}+\Lambda\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({e}^{\frac{−{i}\pi}{\mathrm{3}}} \right)^{{n}} }{{n}!} \\ $$$$=\Upsilon\left({e}^{{e}^{\frac{{i}\pi}{\mathrm{3}}} } −\mathrm{1}\right)+\Lambda\left({e}^{{e}^{\frac{−{i}\pi}{\mathrm{3}}} } −\mathrm{1}\right) \\ $$
Commented by talminator2856791 last updated on 27/Oct/20
WOW that is a very beautiful answer where do you learn that?
$$\:\mathrm{WOW}\:\mathrm{that}\:\mathrm{is}\:\mathrm{a}\:\mathrm{very}\:\mathrm{beautiful}\:\mathrm{answer} \\ $$$$\mathrm{where}\:\mathrm{do}\:\mathrm{you}\:\mathrm{learn}\:\mathrm{that}? \\ $$
Commented by Dwaipayan Shikari last updated on 27/Oct/20
Σ_(n=1) ^∞ (x^n /(n!)) =e^x −1 Here x=e^((iπ)/3) and e^((−iπ)/3) :) :)
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!}\:={e}^{{x}} −\mathrm{1} \\ $$$${Here}\:{x}={e}^{\frac{{i}\pi}{\mathrm{3}}} \:\:{and}\:\:{e}^{\frac{−{i}\pi}{\mathrm{3}}} \\ $$$$\left.\:\:\:\:\:\:\:\:\::\right) \\ $$$$\left.\:\:\:\:\:\:\:\:\::\right) \\ $$
Answered by Olaf last updated on 26/Oct/20
u_(n+1) −u_n +u_(n−1) = 0 r^2 −r+1 = 0 Δ = (−1)^2 −4(1)(1) = −3 r = ((1±(√3)i)/2) = e^(±i(π/3)) u_n = λe^(+in(π/3)) +μe^(−in(π/3)) (λ and μ depend on u_0 and u_1 ) S = Σ_(n=1) ^∞ (u_n /(n!)) = λΣ_(n=0) ^∞ (((e^(i(π/3)) )^n )/(n!))+μΣ_(n=0) ^∞ (((e^(−i(π/3)) )^n )/(n!))−u_0 S = λe^((1/2)+i((√3)/2)) +μe^((1/2)−i((√3)/2)) −u_0 S = (√e)(λe^(i((√3)/2)) +μe^(−i((√3)/2)) )−u_0 with : u_0 = λ+μ u_1 = λe^(i(π/3)) +μe^(−i(π/3)) { (),() :}
$${u}_{{n}+\mathrm{1}} −{u}_{{n}} +{u}_{{n}−\mathrm{1}} \:=\:\mathrm{0} \\ $$$${r}^{\mathrm{2}} −{r}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\Delta\:=\:\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{1}\right)\:=\:−\mathrm{3} \\ $$$${r}\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}{i}}{\mathrm{2}}\:=\:{e}^{\pm{i}\frac{\pi}{\mathrm{3}}} \\ $$$${u}_{{n}} \:=\:\lambda{e}^{+{in}\frac{\pi}{\mathrm{3}}} +\mu{e}^{−{in}\frac{\pi}{\mathrm{3}}} \\ $$$$\left(\lambda\:\mathrm{and}\:\mu\:\mathrm{depend}\:\mathrm{on}\:{u}_{\mathrm{0}} \:\mathrm{and}\:{u}_{\mathrm{1}} \right) \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{u}_{{n}} }{{n}!}\:=\:\lambda\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({e}^{{i}\frac{\pi}{\mathrm{3}}} \right)^{{n}} }{{n}!}+\mu\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({e}^{−{i}\frac{\pi}{\mathrm{3}}} \right)^{{n}} }{{n}!}−{u}_{\mathrm{0}} \\ $$$$\mathrm{S}\:=\:\lambda{e}^{\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} +\mu{e}^{\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} −{u}_{\mathrm{0}} \\ $$$$\mathrm{S}\:=\:\sqrt{{e}}\left(\lambda{e}^{{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} +\mu{e}^{−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \right)−{u}_{\mathrm{0}} \\ $$$$\mathrm{with}\:: \\ $$$${u}_{\mathrm{0}} \:=\:\lambda+\mu \\ $$$${u}_{\mathrm{1}} \:=\:\lambda{e}^{{i}\frac{\pi}{\mathrm{3}}} +\mu{e}^{−{i}\frac{\pi}{\mathrm{3}}} \\ $$$$\begin{cases}\\\end{cases} \\ $$
Answered by mathmax by abdo last updated on 26/Oct/20
u_n =u_(n+1) +u_(n−1) ⇒u_(n+1) −u_n +u_(n−1) =0 ⇒u_(n+2) −u_(n+1) +u_n =0 →r^2 −r +1 =0 →Δ=1−4 =−3 ⇒r_1 =((1+i(√3))/2) =e^((iπ)/3) r_2 =((1−i(√3))/2)=e^(−((iπ)/3)) ⇒u_n =ar_1 ^n +br_1 ^n =ae^((inπ)/3) +be^(−i((nπ)/3)) ⇒Σ_(n=1) ^∞ (u_n /n) =a Σ_(n=1) ^∞ (e^((inπ)/3) /n) +bΣ_(n=1) ^∞ (e^(−((inπ)/3)) /n) let determine f(z) =Σ_(n=1) ^∞ (z^n /n) we have for ∣z∣<1 f^′ (z)=Σ_(n=1) ^∞ z^(n−1) =Σ_(n=0) ^∞ z^n =(1/(1−z)) ⇒f(z) =−ln(1−z)+c c=f(0)=0 ⇒Σ_(n=1) ^∞ (z^n /n)=−ln(1−z) ⇒ Σ_(n=1) ^∞ (u_n /n) =−aln(1−e^((iπ)/3) )−bln(1−e^(−((iπ)/3)) ) we have ln(1−e^((iπ)/3) )=ln(1−cos((π/3))−isin((π/3))) =ln(2sin^2 ((π/6))−2isin((π/6))cos((π/6))) =ln(−2isin((π/6))e^((iπ)/6) ) =ln(2)+ln(−i)+ln(sin(π/6))+((iπ)/6) =ln(2)−((iπ)/2) +ln(sin((π/6)))+((iπ)/6) ln(1−e^(−((iπ)/3)) ) =ln(1−cos((π/3))+isin((π/3))) =ln(2sin^2 ((π/6))+2isin((π/6))cos((π/6))) =ln(2isin((π/6))e^(−((iπ)/6)) )=ln(2)+ln(i)+ln(sin((π/6)))−((iπ)/6) =ln(2)+((iπ)/2) +ln(sin((π/2)))−((iπ)/6) ⇒ Σ_(n=1) ^∞ (u_n /n) =−a{ln(2)−((iπ)/2)+ln((1/2))+((iπ)/6)}−b{ln(2)+((iπ)/2)+ln((1/2))−((iπ)/6)} =−ai((π/6)−(π/2))−bi((π/2)−(π/6)) =−ai(−(π/3))−bi((π/3)) =(a−b)i(π/3) a and b can be found by initial conditions
$$\mathrm{u}_{\mathrm{n}} =\mathrm{u}_{\mathrm{n}+\mathrm{1}} +\mathrm{u}_{\mathrm{n}−\mathrm{1}} \:\Rightarrow\mathrm{u}_{\mathrm{n}+\mathrm{1}} −\mathrm{u}_{\mathrm{n}} +\mathrm{u}_{\mathrm{n}−\mathrm{1}} =\mathrm{0}\:\Rightarrow\mathrm{u}_{\mathrm{n}+\mathrm{2}} −\mathrm{u}_{\mathrm{n}+\mathrm{1}} +\mathrm{u}_{\mathrm{n}} =\mathrm{0} \\ $$$$\rightarrow\mathrm{r}^{\mathrm{2}} −\mathrm{r}\:+\mathrm{1}\:=\mathrm{0}\:\rightarrow\Delta=\mathrm{1}−\mathrm{4}\:=−\mathrm{3}\:\Rightarrow\mathrm{r}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \\ $$$$\mathrm{r}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\Rightarrow\mathrm{u}_{\mathrm{n}} =\mathrm{ar}_{\mathrm{1}} ^{\mathrm{n}} \:+\mathrm{br}_{\mathrm{1}} ^{\mathrm{n}} \:=\mathrm{ae}^{\frac{\mathrm{in}\pi}{\mathrm{3}}} \:+\mathrm{be}^{−\mathrm{i}\frac{\mathrm{n}\pi}{\mathrm{3}}} \\ $$$$\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{u}_{\mathrm{n}} }{\mathrm{n}}\:=\mathrm{a}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{e}^{\frac{\mathrm{in}\pi}{\mathrm{3}}} }{\mathrm{n}}\:+\mathrm{b}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{e}^{−\frac{\mathrm{in}\pi}{\mathrm{3}}} }{\mathrm{n}} \\ $$$$\mathrm{let}\:\mathrm{determine}\:\mathrm{f}\left(\mathrm{z}\right)\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{z}^{\mathrm{n}} }{\mathrm{n}}\:\:\mathrm{we}\:\mathrm{have}\:\:\mathrm{for}\:\mid\mathrm{z}\mid<\mathrm{1} \\ $$$$\mathrm{f}^{'} \left(\mathrm{z}\right)=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{z}^{\mathrm{n}−\mathrm{1}} \:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\:\mathrm{z}^{\mathrm{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{z}}\:\Rightarrow\mathrm{f}\left(\mathrm{z}\right)\:=−\mathrm{ln}\left(\mathrm{1}−\mathrm{z}\right)+\mathrm{c} \\ $$$$\mathrm{c}=\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{z}^{\mathrm{n}} }{\mathrm{n}}=−\mathrm{ln}\left(\mathrm{1}−\mathrm{z}\right)\:\Rightarrow \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{u}_{\mathrm{n}} }{\mathrm{n}}\:=−\mathrm{aln}\left(\mathrm{1}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)−\mathrm{bln}\left(\mathrm{1}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{ln}\left(\mathrm{1}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)=\mathrm{ln}\left(\mathrm{1}−\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}\right)−\mathrm{isin}\left(\frac{\pi}{\mathrm{3}}\right)\right) \\ $$$$=\mathrm{ln}\left(\mathrm{2sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{6}}\right)−\mathrm{2isin}\left(\frac{\pi}{\mathrm{6}}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{6}}\right)\right) \\ $$$$=\mathrm{ln}\left(−\mathrm{2isin}\left(\frac{\pi}{\mathrm{6}}\right)\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{6}}} \right)\:=\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{ln}\left(−\mathrm{i}\right)+\mathrm{ln}\left(\mathrm{sin}\frac{\pi}{\mathrm{6}}\right)+\frac{\mathrm{i}\pi}{\mathrm{6}} \\ $$$$=\mathrm{ln}\left(\mathrm{2}\right)−\frac{\mathrm{i}\pi}{\mathrm{2}}\:+\mathrm{ln}\left(\mathrm{sin}\left(\frac{\pi}{\mathrm{6}}\right)\right)+\frac{\mathrm{i}\pi}{\mathrm{6}} \\ $$$$\mathrm{ln}\left(\mathrm{1}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:=\mathrm{ln}\left(\mathrm{1}−\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}\right)+\mathrm{isin}\left(\frac{\pi}{\mathrm{3}}\right)\right) \\ $$$$=\mathrm{ln}\left(\mathrm{2sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{6}}\right)+\mathrm{2isin}\left(\frac{\pi}{\mathrm{6}}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{6}}\right)\right) \\ $$$$=\mathrm{ln}\left(\mathrm{2isin}\left(\frac{\pi}{\mathrm{6}}\right)\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{6}}} \right)=\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{ln}\left(\mathrm{i}\right)+\mathrm{ln}\left(\mathrm{sin}\left(\frac{\pi}{\mathrm{6}}\right)\right)−\frac{\mathrm{i}\pi}{\mathrm{6}} \\ $$$$=\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{i}\pi}{\mathrm{2}}\:+\mathrm{ln}\left(\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\right)\right)−\frac{\mathrm{i}\pi}{\mathrm{6}}\:\Rightarrow \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{u}_{\mathrm{n}} }{\mathrm{n}}\:=−\mathrm{a}\left\{\mathrm{ln}\left(\mathrm{2}\right)−\frac{\mathrm{i}\pi}{\mathrm{2}}+\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{i}\pi}{\mathrm{6}}\right\}−\mathrm{b}\left\{\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{i}\pi}{\mathrm{2}}+\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{i}\pi}{\mathrm{6}}\right\} \\ $$$$=−\mathrm{ai}\left(\frac{\pi}{\mathrm{6}}−\frac{\pi}{\mathrm{2}}\right)−\mathrm{bi}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{6}}\right) \\ $$$$=−\mathrm{ai}\left(−\frac{\pi}{\mathrm{3}}\right)−\mathrm{bi}\left(\frac{\pi}{\mathrm{3}}\right)\:=\left(\mathrm{a}−\mathrm{b}\right)\mathrm{i}\frac{\pi}{\mathrm{3}}\:\: \\ $$$$\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{can}\:\mathrm{be}\:\mathrm{found}\:\mathrm{by}\:\mathrm{initial}\:\mathrm{conditions} \\ $$
Commented by mathmax by abdo last updated on 27/Oct/20
sorry i have solved Σ (u_n /n) not Σ (u_n /(n!))...!
$$\mathrm{sorry}\:\mathrm{i}\:\mathrm{have}\:\mathrm{solved}\:\Sigma\:\frac{\mathrm{u}_{\mathrm{n}} }{\mathrm{n}}\:\:\mathrm{not}\:\Sigma\:\frac{\mathrm{u}_{\mathrm{n}} }{\mathrm{n}!}…! \\ $$
Commented by mathmax by abdo last updated on 27/Oct/20
we have u_n =ae^((inπ)/3) +be^(−((inπ)/3)) ⇒ Σ_(n=1) ^∞ (u_n /(n!)) =a Σ_(n=1) ^∞ (((e^((iπ)/3) )^n )/(n!)) +b Σ_(n=1) ^∞ (((e^(−((iπ)/3)) )^n )/(n!)) =a( e^e^((iπ)/3) −1) +b(e^e^(−((iπ)/3)) −1) =a{e^((1/2)+((i(√3))/2)) −1)+b(e^((1/2)−((i(√3))/2)) −1) =a e^(1/2) (cos(((√3)/2))+isin(((√3)/(2 ))))+be^(1/2) (cos(((√3)/2))−isin(((√3)/2)))−(a+b) =(a+b)(√e)cos(((√3)/2))+i(a−b)(√e)sin(((√3)/2))−(a+b)
$$\mathrm{we}\:\mathrm{have}\:\mathrm{u}_{\mathrm{n}} =\mathrm{ae}^{\frac{\mathrm{in}\pi}{\mathrm{3}}} \:+\mathrm{be}^{−\frac{\mathrm{in}\pi}{\mathrm{3}}} \:\Rightarrow \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{u}_{\mathrm{n}} }{\mathrm{n}!}\:=\mathrm{a}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \frac{\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{n}} }{\mathrm{n}!}\:+\mathrm{b}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{n}} }{\mathrm{n}!} \\ $$$$=\mathrm{a}\left(\:\mathrm{e}^{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} } −\mathrm{1}\right)\:+\mathrm{b}\left(\mathrm{e}^{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} } −\mathrm{1}\right) \\ $$$$=\mathrm{a}\left\{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}} −\mathrm{1}\right)+\mathrm{b}\left(\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}} −\mathrm{1}\right) \\ $$$$=\mathrm{a}\:\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+\mathrm{isin}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\:}\right)\right)+\mathrm{be}^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\mathrm{isin}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right)−\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$=\left(\mathrm{a}+\mathrm{b}\right)\sqrt{\mathrm{e}}\mathrm{cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+\mathrm{i}\left(\mathrm{a}−\mathrm{b}\right)\sqrt{\mathrm{e}}\mathrm{sin}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\left(\mathrm{a}+\mathrm{b}\right) \\ $$

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