1) calculate U_n =∫_0 ^∞ e^(−n[x]) sin(((πx)/n))dx nnatural and n≥1 2)determine nature of Σ U


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Question Number 87530 by mathmax by abdo last updated on 04/Apr/20

$1) c a l c u l a t e U_{n} = \int_{0}^{\infty} e^{- n [x]} s i n (\frac{π x}{n}) d x n n a t u r a l a n d n ⩾ 1$ $2) d e t e r m i n e n a t u r e o f Σ U_{n}$
Commented by mathmax by abdo last updated on 05/Apr/20
$1) U_n =Σ_(p=0) ^∞ ∫_p ^(p+1) e^(−np) sin(((πx)/n))dx =Σ_(p=0) ^∞ e^(−np) ∫_p ^(p+1) sin(((πx)/n))dx =Σ_(p=0) ^∞ e^(−np) [−(n/π)cos(((πx)/n))]_p ^(p+1) =−(n/π)Σ_(p=0) ^∞ e^(−np) { cos((π/n)(p+1)}−cos(((pπ)/n))} =−(n/π)Σ_(p=0) ^∞ e^(−np) cos(((π(p+1))/n))+(n/π)Σ_(p=0) ^∞ e^(−np) cos(((pπ)/n)) we have Σ_(p=0) ^∞ e^(−np) cos(((pπ)/n)) =Re(Σ_(p=0) ^∞ e^(−np+i((pπ)/n)) )=Re(A_n ) and A_n =Σ_(p=0) ^∞ (e^(−n+((iπ)/n)) )^p =(1/(1−e^(−n+((iπ)/n)) )) =(1/(1−e^(−n) {cos((π/n))+isin((π/n))})) =(1/(1−e^(−n) cos((π/n))−ie^(−n) sin((π/n)))) =((1−e^(−n) cos((π/n))+ie^(−n) sin((π/n)))/((1−e^(−n) cos((π/n)))^2 +e^(−2n) sin^2 ((π/n)))) ⇒ A_n =((1−e^(−n) cos((π/n)))/((1−e^(−n) cos((π/n)))^2 +e^(−2n) sin^2 ((π/n)))) Σ_(p=0) ^∞ e^(−np) cos(((π(p+1))/n)) =Σ_(p=1) ^∞ e^(−n(p−1)) cos(((πp)/n)) =e^n Σ_(p=1) ^∞ e^(−np) cos(((πp)/n)) =e^n (Σ_(p=1) ^∞ e^(−np) cos(((πp)/n))−1) =e^n (A_n −1) ⇒ U_n =−(n/π)e^n (A_n −1) +(n/π) A_n ⇒ U_n =(n/π)(1−e^n )A_n +((ne^n )/π)$
$1) U_{n} = \sum_{p = 0}^{\infty} \int_{p}^{p + 1} e^{- n p} s i n (\frac{π x}{n}) d x$ $= \sum_{p = 0}^{\infty} e^{- n p} \int_{p}^{p + 1} s i n (\frac{π x}{n}) d x$ $= \sum_{p = 0}^{\infty} e^{- n p} {[- \frac{n}{π} c o s (\frac{π x}{n})]}_{p}^{p + 1}$ $= - \frac{n}{π} \sum_{p = 0}^{\infty} e^{- n p} {c o s (\frac{π}{n} (p + 1)} - c o s (\frac{p π}{n})}$ $= - \frac{n}{π} \sum_{p = 0}^{\infty} e^{- n p} c o s (\frac{π (p + 1)}{n}) + \frac{n}{π} \sum_{p = 0}^{\infty} e^{- n p} c o s (\frac{p π}{n})$ $w e h a v e \sum_{p = 0}^{\infty} e^{- n p} c o s (\frac{p π}{n}) = R e (\sum_{p = 0}^{\infty} e^{- n p + i \frac{p π}{n}}) = R e (A_{n})$ $a n d A_{n} = \sum_{p = 0}^{\infty} {(e^{- n + \frac{i π}{n}})}^{p} = \frac{1}{1 - e^{- n + \frac{i π}{n}}}$ $= \frac{1}{1 - e^{- n} {c o s (\frac{π}{n}) + i s i n (\frac{π}{n})}} = \frac{1}{1 - e^{- n} c o s (\frac{π}{n}) - {i e}^{- n} s i n (\frac{π}{n})}$ $= \frac{1 - e^{- n} c o s (\frac{π}{n}) + {i e}^{- n} s i n (\frac{π}{n})}{{(1 - e^{- n} c o s (\frac{π}{n}))}^{2} + e^{- 2 n} {s i n}^{2} (\frac{π}{n})} \Rightarrow$ $A_{n} = \frac{1 - e^{- n} c o s (\frac{π}{n})}{{(1 - e^{- n} c o s (\frac{π}{n}))}^{2} + e^{- 2 n} {s i n}^{2} (\frac{π}{n})}$ $\sum_{p = 0}^{\infty} e^{- n p} c o s (\frac{π (p + 1)}{n}) = \sum_{p = 1}^{\infty} e^{- n (p - 1)} c o s (\frac{π p}{n})$ $= e^{n} \sum_{p = 1}^{\infty} e^{- n p} c o s (\frac{π p}{n}) = e^{n} (\sum_{p = 1}^{\infty} e^{- n p} c o s (\frac{π p}{n}) - 1)$ $= e^{n} (A_{n} - 1) \Rightarrow U_{n} = - \frac{n}{π} e^{n} (A_{n} - 1) + \frac{n}{π} A_{n}$ $\Rightarrow U_{n} = \frac{n}{π} (1 - e^{n}) A_{n} + \frac{{n e}^{n}}{π}$
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