find-n-1-u-n-n-if-u-n-u-n-1-u-n-1-

Question Number 119754 by Bird last updated on 26/Oct/20

f i n d \sum_{n = 1}^{\infty} \frac{u_{n}}{n!} i f u_{n} = u_{n + 1} + u_{n - 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 + \frac{1}{r} \Rightarrow r^{2} - r + 1 = 0 \Rightarrow r = \frac{1 \pm i \sqrt{3}}{2} = e^{\pm \frac{i π}{3}}

u_{n} = Υ e^{\frac{i π n}{3}} + Λ e^{- \frac{i π n}{3}}

\underset{n = 1}{\sum^{\infty}} \frac{u_{n}}{n!} = Υ \underset{n = 1}{\sum^{\infty}} \frac{{(e^{\frac{i π}{3}})}^{n}}{n!} + Λ \underset{n = 1}{\sum^{\infty}} \frac{{(e^{\frac{- i π}{3}})}^{n}}{n!}

= Υ (e^{e^{\frac{i π}{3}}} - 1) + Λ (e^{e^{\frac{- i π}{3}}} - 1)

Commented by talminator2856791 last updated on 27/Oct/20

WOW that is a very beautiful answer

where do you learn that ?

Commented by Dwaipayan Shikari last updated on 27/Oct/20

\underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n!} = e^{x} - 1

H e r e x = e^{\frac{i π}{3}} a n d e^{\frac{- i π}{3}}

:)

:)

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 = \frac{1 \pm \sqrt{3} i}{2} = e^{\pm i \frac{π}{3}}

u_{n} = λ e^{+ i n \frac{π}{3}} + μ e^{- i n \frac{π}{3}}

(λ and μ depend on u_{0} and u_{1})

S = \underset{n = 1}{\sum^{\infty}} \frac{u_{n}}{n!} = λ \underset{n = 0}{\sum^{\infty}} \frac{{(e^{i \frac{π}{3}})}^{n}}{n!} + μ \underset{n = 0}{\sum^{\infty}} \frac{{(e^{- i \frac{π}{3}})}^{n}}{n!} - u_{0}

S = λ e^{\frac{1}{2} + i \frac{\sqrt{3}}{2}} + μ e^{\frac{1}{2} - i \frac{\sqrt{3}}{2}} - u_{0}

S = \sqrt{e} (λ e^{i \frac{\sqrt{3}}{2}} + μ e^{- i \frac{\sqrt{3}}{2}}) - u_{0}

with :

u_{0} = λ + μ

u_{1} = λ e^{i \frac{π}{3}} + μ e^{- i \frac{π}{3}}

{\begin{cases}  \end{cases}

Answered by mathmax by abdo last updated on 26/Oct/20

u_{n} = u_{n + 1} + u_{n - 1} \Rightarrow u_{n + 1} - u_{n} + u_{n - 1} = 0 \Rightarrow u_{n + 2} - u_{n + 1} + u_{n} = 0

\to r^{2} - r + 1 = 0 \to Δ = 1 - 4 = - 3 \Rightarrow r_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}}

r_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{3}} \Rightarrow u_{n} = {ar}_{1}^{n} + {br}_{1}^{n} = {ae}^{\frac{in π}{3}} + {be}^{- i \frac{n π}{3}}

\Rightarrow \sum_{n = 1}^{\infty} \frac{u_{n}}{n} = a \sum_{n = 1}^{\infty} \frac{e^{\frac{in π}{3}}}{n} + b \sum_{n = 1}^{\infty} \frac{e^{- \frac{in π}{3}}}{n}

let determine f (z) = \sum_{n = 1}^{\infty} \frac{z^{n}}{n} we have for ∣ z ∣< 1

f^{^{'}} (z) = \sum_{n = 1}^{\infty} z^{n - 1} = \sum_{n = 0}^{\infty} z^{n} = \frac{1}{1 - z} \Rightarrow f (z) = - \ln (1 - z) + c

c = f (0) = 0 \Rightarrow \sum_{n = 1}^{\infty} \frac{z^{n}}{n} = - \ln (1 - z) \Rightarrow

\sum_{n = 1}^{\infty} \frac{u_{n}}{n} = - aln (1 - e^{\frac{i π}{3}}) - bln (1 - e^{- \frac{i π}{3}}) we have

\ln (1 - e^{\frac{i π}{3}}) = \ln (1 - \cos (\frac{π}{3}) - isin (\frac{π}{3}))

= \ln ({2 \sin}^{2} (\frac{π}{6}) - 2 isin (\frac{π}{6}) \cos (\frac{π}{6}))

= \ln (- 2 isin (\frac{π}{6}) e^{\frac{i π}{6}}) = \ln (2) + \ln (- i) + \ln (\sin \frac{π}{6}) + \frac{i π}{6}

= \ln (2) - \frac{i π}{2} + \ln (\sin (\frac{π}{6})) + \frac{i π}{6}

\ln (1 - e^{- \frac{i π}{3}}) = \ln (1 - \cos (\frac{π}{3}) + isin (\frac{π}{3}))

= \ln ({2 \sin}^{2} (\frac{π}{6}) + 2 isin (\frac{π}{6}) \cos (\frac{π}{6}))

= \ln (2 isin (\frac{π}{6}) e^{- \frac{i π}{6}}) = \ln (2) + \ln (i) + \ln (\sin (\frac{π}{6})) - \frac{i π}{6}

= \ln (2) + \frac{i π}{2} + \ln (\sin (\frac{π}{2})) - \frac{i π}{6} \Rightarrow

\sum_{n = 1}^{\infty} \frac{u_{n}}{n} = - a {\ln (2) - \frac{i π}{2} + \ln (\frac{1}{2}) + \frac{i π}{6}} - b {\ln (2) + \frac{i π}{2} + \ln (\frac{1}{2}) - \frac{i π}{6}}

= - ai (\frac{π}{6} - \frac{π}{2}) - bi (\frac{π}{2} - \frac{π}{6})

= - ai (- \frac{π}{3}) - bi (\frac{π}{3}) = (a - b) i \frac{π}{3}

a and b can be found by initial conditions

Commented by mathmax by abdo last updated on 27/Oct/20

sorry i have solved Σ \frac{u_{n}}{n} not Σ \frac{u_{n}}{n!} \dots!

Commented by mathmax by abdo last updated on 27/Oct/20

we have u_{n} = {ae}^{\frac{in π}{3}} + {be}^{- \frac{in π}{3}} \Rightarrow

\sum_{n = 1}^{\infty} \frac{u_{n}}{n!} = a \sum_{n = 1}^{\infty} \frac{{(e^{\frac{i π}{3}})}^{n}}{n!} + b \sum_{n = 1}^{\infty} \frac{{(e^{- \frac{i π}{3}})}^{n}}{n!}

= a (e^{e^{\frac{i π}{3}}} - 1) + b (e^{e^{- \frac{i π}{3}}} - 1)

= a {e^{\frac{1}{2} + \frac{i \sqrt{3}}{2}} - 1) + b (e^{\frac{1}{2} - \frac{i \sqrt{3}}{2}} - 1)

= a e^{\frac{1}{2}} (\cos (\frac{\sqrt{3}}{2}) + isin (\frac{\sqrt{3}}{2})) + {be}^{\frac{1}{2}} (\cos (\frac{\sqrt{3}}{2}) - isin (\frac{\sqrt{3}}{2})) - (a + b)

= (a + b) \sqrt{e} \cos (\frac{\sqrt{3}}{2}) + i (a - b) \sqrt{e} \sin (\frac{\sqrt{3}}{2}) - (a + b)