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Question Number 148071 by tabata last updated on 25/Jul/21

Commented by tabata last updated on 25/Jul/21

$? ? ? ? ? ? ?$
	Answered by Olaf_Thorendsen last updated on 25/Jul/21

	$(a)$ $\sinh z = \underset{n = 0}{\sum^{\infty}} \frac{z^{2 n + 1}}{(2 n + 1)!}$ $\sinh z^{2} = \underset{n = 0}{\sum^{\infty}} \frac{z^{4 n + 2}}{(2 n + 1)!}$ $z \sinh z^{2} = \underset{n = 0}{\sum^{\infty}} \frac{z^{4 n + 3}}{(2 n + 1)!}$ $(b)$ $\forall n, \frac{d^{n}}{{d z}^{n}} e^{z} = e^{z} \Rightarrow \frac{d^{n}}{{d z}^{n}} e^{z} ∣_{z = 2} = e^{2}$ $e^{z} = \underset{n = 0}{\sum^{\infty}} \frac{e^{2}}{n!} {(z - 2)}^{n} = e^{2} \underset{n = 0}{\sum^{\infty}} \frac{{(z - 2)}^{n}}{n!}$ $(c)$ $f (z) = \frac{z (z + 1)}{{(1 - z)}^{2}}$ $f (z) = 1 + \frac{3}{z - 1} + \frac{2}{{(z - 1)}^{2}}$ $∙ f (- 1) = 0$ $∙ f^{(n)} (z) = \frac{3 {(- 1)}^{n} n!}{{(z - 1)}^{n + 1}} + \frac{2 {(- 1)}^{n} (n + 1)!}{{(z - 1)}^{n + 2}}$ $\Rightarrow f^{(n)} (- 1) = \frac{3 {(- 1)}^{n} n!}{{(- 2)}^{n + 1}} + \frac{2 {(- 1)}^{n} (n + 1)!}{{(- 2)}^{n + 2}}$ $\Rightarrow f^{(n)} (- 1) = - \frac{3 n!}{2^{n + 1}} + \frac{(n + 1)!}{2^{n + 1}}$ $f^{(n)} (- 1) = \frac{n!}{2^{n + 1}} (n + 1 - 3) = \frac{(n - 2) n!}{2^{n + 1}}$ $f (z) = f (- 1) + \underset{n = 1}{\sum^{\infty}} \frac{f^{(n)} (- 1)}{n!} {(z - (- 1))}^{n}$ $f (z) = \underset{n = 1}{\sum^{\infty}} \frac{n - 2}{2^{n + 1}} {(z + 1)}^{n}$
	Answered by mathmax by abdo last updated on 25/Jul/21
	$f(z)=((z^2 +z)/((z−1)^2 )) we do the changement z+1=y ⇒ f(z)=ϕ(y)=(((y−1)^2 +y−1)/((y−2)^2 ))=((y^2 −2y+1+y−1)/((y−2)^2 )) =((y^2 −y)/((y−2)^2 )) ϕ(y)=Σ_(n=0) ^∞ ((ϕ^((n)) (0))/(n!))y^n ϕ^((n)) (y)={(y^2 −y)×(1/((y−2)^2 ))}^((n)) =Σ_(k=0) ^n C_n ^k (y^2 −y)^((k)) ((1/((y−2)^2 )))^((n−k)) =Σ_(k=0) ^n C_n ^k (y^2 −y)^((k)) ×(((−1)^(n−k) (n−k)!)/((y−2)^(n−k+1) )) =(y^2 −y)(((−1)^n n!)/((y−2)^(n+1) )) +C_n ^1 (2y−1)(((−1)^(n−1) (n−1)!)/((y−2)^n )) + 2C_n ^2 (((−1)^(n−2) (n−2)!)/((y−2)^(n−1) )) ⇒ ϕ^((n)) (0)=−C_n ^1 (((−1)^(n−1) (n−1)!)/((−2)^n )) +2 C_n ^2 (((−1)^(n−2) (n−2)!)/((−2)^(n−1) )) =((n(n−1)!)/2^n ) −n(n−1)(((n−2)!)/2^(n−1) )=((n!)/2^n )−((n!)/2^(n−1) ) ⇒ ϕ(y)=Σ_(n=0) ^∞ (1/(n!))(((n!)/2^n )−((n!)/2^(n−1) ))y^n ⇒f(z)=Σ_(n=0) ^∞ ((1/2^n )−(1/2^(n−1) ))(z+1)^n$
	$f (z) = \frac{z^{2} + z}{{(z - 1)}^{2}} we do the changement z + 1 = y \Rightarrow$ $f (z) = φ (y) = \frac{{(y - 1)}^{2} + y - 1}{{(y - 2)}^{2}} = \frac{y^{2} - 2 y + 1 + y - 1}{{(y - 2)}^{2}}$ $= \frac{y^{2} - y}{{(y - 2)}^{2}}$ $φ (y) = \sum_{n = 0}^{\infty} \frac{φ^{(n)} (0)}{n!} y^{n}$ $φ^{(n)} (y) = {(y^{2} - y) \times \frac{1}{{(y - 2)}^{2}}}^{(n)}$ $= \sum_{k = 0}^{n} C_{n}^{k} {(y^{2} - y)}^{(k)} {(\frac{1}{{(y - 2)}^{2}})}^{(n - k)}$ $= \sum_{k = 0}^{n} C_{n}^{k} {(y^{2} - y)}^{(k)} \times \frac{{(- 1)}^{n - k} (n - k)!}{{(y - 2)}^{n - k + 1}}$ $= (y^{2} - y) \frac{{(- 1)}^{n} n!}{{(y - 2)}^{n + 1}} + C_{n}^{1} (2 y - 1) \frac{{(- 1)}^{n - 1} (n - 1)!}{{(y - 2)}^{n}}$ $+ {2 C}_{n}^{2} \frac{{(- 1)}^{n - 2} (n - 2)!}{{(y - 2)}^{n - 1}} \Rightarrow$ $φ^{(n)} (0) = - C_{n}^{1} \frac{{(- 1)}^{n - 1} (n - 1)!}{{(- 2)}^{n}} + 2 C_{n}^{2} \frac{{(- 1)}^{n - 2} (n - 2)!}{{(- 2)}^{n - 1}}$ $= \frac{n (n - 1)!}{2^{n}} - n (n - 1) \frac{(n - 2)!}{2^{n - 1}} = \frac{n!}{2^{n}} - \frac{n!}{2^{n - 1}} \Rightarrow$ $φ (y) = \sum_{n = 0}^{\infty} \frac{1}{n!} (\frac{n!}{2^{n}} - \frac{n!}{2^{n - 1}}) y^{n}$ $\Rightarrow f (z) = \sum_{n = 0}^{\infty} (\frac{1}{2^{n}} - \frac{1}{2^{n - 1}}) {(z + 1)}^{n}$
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