Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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) sinhz = Σ_(n=0) ^∞ (z^(2n+1) /((2n+1)!)) sinhz^2 = Σ_(n=0) ^∞ (z^(4n+2) /((2n+1)!)) zsinhz^2 = Σ_(n=0) ^∞ (z^(4n+3) /((2n+1)!)) (b) ∀n, (d^n /dz^n )e^z = e^z ⇒ (d^n /dz^n )e^z ∣_(z=2) = e^2 e^z = Σ_(n=0) ^∞ (e^2 /(n!))(z−2)^n = e^2 Σ_(n=0) ^∞ (((z−2)^n )/(n!)) (c) f(z) = ((z(z+1))/((1−z)^2 )) f(z) = 1+(3/(z−1))+(2/((z−1)^2 )) • f(−1) = 0 • f^((n)) (z) = ((3(−1)^n n!)/((z−1)^(n+1) ))+((2(−1)^n (n+1)!)/((z−1)^(n+2) )) ⇒ f^((n)) (−1) = ((3(−1)^n n!)/((−2)^(n+1) ))+((2(−1)^n (n+1)!)/((−2)^(n+2) )) ⇒ f^((n)) (−1) = −((3n!)/2^(n+1) )+(((n+1)!)/2^(n+1) ) f^((n)) (−1) = ((n!)/2^(n+1) )(n+1−3) = (((n−2)n!)/2^(n+1) ) f(z) = f(−1)+Σ_(n=1) ^∞ ((f^((n)) (−1))/(n!))(z−(−1))^n f(z) = Σ_(n=1) ^∞ ((n−2)/2^(n+1) )(z+1)^n

$$\left(\mathrm{a}\right) \\ $$$$\mathrm{sinh}{z}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{z}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$\mathrm{sinh}{z}^{\mathrm{2}} \:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{z}^{\mathrm{4}{n}+\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$${z}\mathrm{sinh}{z}^{\mathrm{2}} \:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{z}^{\mathrm{4}{n}+\mathrm{3}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$\left(\mathrm{b}\right)\: \\ $$$$\forall{n},\:\frac{{d}^{{n}} }{{dz}^{{n}} }{e}^{{z}} \:=\:{e}^{{z}} \:\Rightarrow\:\frac{{d}^{{n}} }{{dz}^{{n}} }{e}^{{z}} \mid_{{z}=\mathrm{2}} \:=\:{e}^{\mathrm{2}} \\ $$$${e}^{\mathrm{z}} \:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{e}^{\mathrm{2}} }{{n}!}\left({z}−\mathrm{2}\right)^{{n}} \:=\:{e}^{\mathrm{2}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({z}−\mathrm{2}\right)^{{n}} }{{n}!} \\ $$$$\left(\mathrm{c}\right) \\ $$$${f}\left({z}\right)\:=\:\frac{{z}\left({z}+\mathrm{1}\right)}{\left(\mathrm{1}−{z}\right)^{\mathrm{2}} } \\ $$$${f}\left({z}\right)\:=\:\mathrm{1}+\frac{\mathrm{3}}{{z}−\mathrm{1}}+\frac{\mathrm{2}}{\left({z}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\bullet\:{f}\left(−\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\bullet\:{f}^{\left({n}\right)} \left({z}\right)\:=\:\frac{\mathrm{3}\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({z}−\mathrm{1}\right)^{{n}+\mathrm{1}} }+\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)!}{\left({z}−\mathrm{1}\right)^{{n}+\mathrm{2}} } \\ $$$$\Rightarrow\:{f}^{\left({n}\right)} \left(−\mathrm{1}\right)\:=\:\frac{\mathrm{3}\left(−\mathrm{1}\right)^{{n}} {n}!}{\left(−\mathrm{2}\right)^{{n}+\mathrm{1}} }+\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)!}{\left(−\mathrm{2}\right)^{{n}+\mathrm{2}} } \\ $$$$\Rightarrow\:{f}^{\left({n}\right)} \left(−\mathrm{1}\right)\:=\:−\frac{\mathrm{3}{n}!}{\mathrm{2}^{{n}+\mathrm{1}} }+\frac{\left({n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$$$\:{f}^{\left({n}\right)} \left(−\mathrm{1}\right)\:=\:\frac{{n}!}{\mathrm{2}^{{n}+\mathrm{1}} }\left({n}+\mathrm{1}−\mathrm{3}\right)\:=\:\frac{\left({n}−\mathrm{2}\right){n}!}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$$${f}\left({z}\right)\:=\:{f}\left(−\mathrm{1}\right)+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{f}^{\left({n}\right)} \left(−\mathrm{1}\right)}{{n}!}\left({z}−\left(−\mathrm{1}\right)\right)^{{n}} \\ $$$${f}\left({z}\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}−\mathrm{2}}{\mathrm{2}^{{n}+\mathrm{1}} }\left({z}+\mathrm{1}\right)^{{n}} \\ $$

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

$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} \:+\mathrm{z}}{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} }\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{z}+\mathrm{1}=\mathrm{y}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\varphi\left(\mathrm{y}\right)=\frac{\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{y}−\mathrm{1}}{\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} }=\frac{\mathrm{y}^{\mathrm{2}} −\mathrm{2y}+\mathrm{1}+\mathrm{y}−\mathrm{1}}{\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{y}^{\mathrm{2}} −\mathrm{y}}{\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\varphi\left(\mathrm{y}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\varphi^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)}{\mathrm{n}!}\mathrm{y}^{\mathrm{n}} \\ $$$$\varphi^{\left(\mathrm{n}\right)} \left(\mathrm{y}\right)=\left\{\left(\mathrm{y}^{\mathrm{2}} −\mathrm{y}\right)×\frac{\mathrm{1}}{\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{n}\right)} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(\mathrm{y}^{\mathrm{2}} −\mathrm{y}\right)^{\left(\mathrm{k}\right)} \left(\frac{\mathrm{1}}{\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} }\right)^{\left(\mathrm{n}−\mathrm{k}\right)} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \left(\mathrm{y}^{\mathrm{2}} −\mathrm{y}\right)^{\left(\mathrm{k}\right)} ×\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{k}} \left(\mathrm{n}−\mathrm{k}\right)!}{\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{n}−\mathrm{k}+\mathrm{1}} } \\ $$$$=\left(\mathrm{y}^{\mathrm{2}} −\mathrm{y}\right)\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{n}!}{\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{n}+\mathrm{1}} }\:+\mathrm{C}_{\mathrm{n}} ^{\mathrm{1}} \left(\mathrm{2y}−\mathrm{1}\right)\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{n}} } \\ $$$$+\:\mathrm{2C}_{\mathrm{n}} ^{\mathrm{2}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{2}} \left(\mathrm{n}−\mathrm{2}\right)!}{\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{n}−\mathrm{1}} }\:\Rightarrow \\ $$$$\varphi^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)=−\mathrm{C}_{\mathrm{n}} ^{\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(−\mathrm{2}\right)^{\mathrm{n}} }\:+\mathrm{2}\:\mathrm{C}_{\mathrm{n}} ^{\mathrm{2}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{2}} \left(\mathrm{n}−\mathrm{2}\right)!}{\left(−\mathrm{2}\right)^{\mathrm{n}−\mathrm{1}} } \\ $$$$=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)!}{\mathrm{2}^{\mathrm{n}} }\:−\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\frac{\left(\mathrm{n}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }=\frac{\mathrm{n}!}{\mathrm{2}^{\mathrm{n}} }−\frac{\mathrm{n}!}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }\:\Rightarrow \\ $$$$\varphi\left(\mathrm{y}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}!}\left(\frac{\mathrm{n}!}{\mathrm{2}^{\mathrm{n}} }−\frac{\mathrm{n}!}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }\right)\mathrm{y}^{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{z}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }\right)\left(\mathrm{z}+\mathrm{1}\right)^{\mathrm{n}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com