Menu Close

It-is-given-f-x-1-x-n-n-N-Find-f-0-f-0-f-0-2-f-0-3-f-n-0-n-

Tinku Tara 0 Comments
Pin




Question Number 183459 by Matica last updated on 26/Dec/22
It is given f(x)=(1+x)^n , n∈N. Find f(0)+f^′ (0)+((f^(′′) (0))/(2!))+((f′′′(0))/(3!))+...+((f^((n)) (0))/(n!)) .
$$\:\:{It}\:{is}\:{given}\:{f}\left({x}\right)=\left(\mathrm{1}+{x}\right)^{{n}} \:,\:{n}\in\mathbb{N}.\:{Find} \\ $$$$\:\:{f}\left(\mathrm{0}\right)+{f}^{'} \left(\mathrm{0}\right)+\frac{{f}^{''} \left(\mathrm{0}\right)}{\mathrm{2}!}+\frac{{f}'''\left(\mathrm{0}\right)}{\mathrm{3}!}+…+\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:\:. \\ $$
Answered by mahdipoor last updated on 26/Dec/22
=1+n+((n×(n−1))/2)+((n×(n−1)×(n−2))/(3!))+... ((n!)/(n!))=(_0 ^n )+(_1 ^n )+...(_n ^n )=2^n
$$=\mathrm{1}+{n}+\frac{{n}×\left({n}−\mathrm{1}\right)}{\mathrm{2}}+\frac{{n}×\left({n}−\mathrm{1}\right)×\left({n}−\mathrm{2}\right)}{\mathrm{3}!}+… \\ $$$$\frac{{n}!}{{n}!}=\left(_{\mathrm{0}} ^{{n}} \right)+\left(_{\mathrm{1}} ^{{n}} \right)+…\left(_{{n}} ^{{n}} \right)=\mathrm{2}^{{n}} \\ $$
Commented by Matica last updated on 26/Dec/22
please more detail
$${please}\:{more}\:{detail} \\ $$
Commented by mahdipoor last updated on 26/Dec/22
f(0)=(1+0)^n =1=(_0 ^n ) f^′ (0)=n(1+0)^(n−1) =n=(_1 ^n ) (1/(2!))f^(′′) (0)=((n(n−1))/(2!))(1+0)^(n−2) =(_2 ^n ) .... (1/(n!))f^n (0)=((n!)/(n!))(1+0)^(n−n) =(_n ^n ) ⇒A=(_0 ^n )+(_1 ^n )+...(_n ^n ) B=(a+b)^n =(_0 ^n )a^n b^0 +(_1 ^n )a^(n−1) b^1 +(_2 ^n )a^(n−2) b^2 +... +(_n ^n )a^(n−n) b^n ⇒a=b=1 ⇒A=B=(1+1)^n =2^n
$${f}\left(\mathrm{0}\right)=\left(\mathrm{1}+\mathrm{0}\right)^{{n}} =\mathrm{1}=\left(_{\mathrm{0}} ^{{n}} \right) \\ $$$${f}^{'} \left(\mathrm{0}\right)={n}\left(\mathrm{1}+\mathrm{0}\right)^{{n}−\mathrm{1}} ={n}=\left(_{\mathrm{1}} ^{{n}} \right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}!}{f}^{''} \left(\mathrm{0}\right)=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}\left(\mathrm{1}+\mathrm{0}\right)^{{n}−\mathrm{2}} =\left(_{\mathrm{2}} ^{{n}} \right) \\ $$$$…. \\ $$$$\frac{\mathrm{1}}{{n}!}{f}^{{n}} \left(\mathrm{0}\right)=\frac{{n}!}{{n}!}\left(\mathrm{1}+\mathrm{0}\right)^{{n}−{n}} =\left(_{{n}} ^{{n}} \right) \\ $$$$\Rightarrow{A}=\left(_{\mathrm{0}} ^{{n}} \right)+\left(_{\mathrm{1}} ^{{n}} \right)+…\left(_{{n}} ^{{n}} \right) \\ $$$${B}=\left({a}+{b}\right)^{{n}} =\left(_{\mathrm{0}} ^{{n}} \right){a}^{{n}} {b}^{\mathrm{0}} +\left(_{\mathrm{1}} ^{{n}} \right){a}^{{n}−\mathrm{1}} {b}^{\mathrm{1}} +\left(_{\mathrm{2}} ^{{n}} \right){a}^{{n}−\mathrm{2}} {b}^{\mathrm{2}} +… \\ $$$$+\left(_{{n}} ^{{n}} \right){a}^{{n}−{n}} {b}^{{n}} \: \\ $$$$\Rightarrow{a}={b}=\mathrm{1}\:\Rightarrow{A}={B}=\left(\mathrm{1}+\mathrm{1}\right)^{{n}} =\mathrm{2}^{{n}} \\ $$
Commented by Matica last updated on 26/Dec/22
Thank you !
$${Thank}\:{you}\:! \\ $$
Answered by mr W last updated on 26/Dec/22
f(x)=(1+x)^n =C_0 ^n +C_1 ^n x+C_2 ^n x^2 +...+C_n ^n x^n (i) on the other side acc. taylor series: f(x)=f(0)+f′(0)x+((f′′(0))/(2!))x^2 +((f′′′(0))/(3!))x^3 +...+((f^((n)) (0))/(n!))x^n +((f^((n+1)) (0))/((n+1)!))x^(n+1) +... (ii) compare (i) and (ii): ((f^((k)) (0))/(k!))=C_k ^n for 0≤k≤n and ((f^((k)) (0))/(k!))=0 for k≥n+1 ⇒f(0)+f′(0)x+((f′′(0))/(2!))x^2 +((f′′′(0))/(3!))x^3 +...+((f^((n)) (0))/(n!))x^n =(1+x)^n set x=1, ⇒f(0)+f′(0)+((f′′(0))/(2!))+((f′′′(0))/(3!))+...+((f^((n)) (0))/(n!))=(1+1)^n =2^n ✓
$${f}\left({x}\right)=\left(\mathrm{1}+{x}\right)^{{n}} ={C}_{\mathrm{0}} ^{{n}} +{C}_{\mathrm{1}} ^{{n}} {x}+{C}_{\mathrm{2}} ^{{n}} {x}^{\mathrm{2}} +…+{C}_{{n}} ^{{n}} {x}^{{n}} \:\:\:\:\left({i}\right) \\ $$$${on}\:{the}\:{other}\:{side}\:{acc}.\:{taylor}\:{series}: \\ $$$${f}\left({x}\right)={f}\left(\mathrm{0}\right)+{f}'\left(\mathrm{0}\right){x}+\frac{{f}''\left(\mathrm{0}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{{f}'''\left(\mathrm{0}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} +…+\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}{x}^{{n}} +\frac{{f}^{\left({n}+\mathrm{1}\right)} \left(\mathrm{0}\right)}{\left({n}+\mathrm{1}\right)!}{x}^{{n}+\mathrm{1}} +…\:\:\:\:\left({ii}\right) \\ $$$${compare}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\frac{{f}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}={C}_{{k}} ^{{n}} \:{for}\:\mathrm{0}\leqslant{k}\leqslant{n}\:{and} \\ $$$$\frac{{f}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}=\mathrm{0}\:{for}\:{k}\geqslant{n}+\mathrm{1} \\ $$$$\Rightarrow{f}\left(\mathrm{0}\right)+{f}'\left(\mathrm{0}\right){x}+\frac{{f}''\left(\mathrm{0}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{{f}'''\left(\mathrm{0}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} +…+\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}{x}^{{n}} =\left(\mathrm{1}+{x}\right)^{{n}} \\ $$$${set}\:{x}=\mathrm{1}, \\ $$$$\Rightarrow{f}\left(\mathrm{0}\right)+{f}'\left(\mathrm{0}\right)+\frac{{f}''\left(\mathrm{0}\right)}{\mathrm{2}!}+\frac{{f}'''\left(\mathrm{0}\right)}{\mathrm{3}!}+…+\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}=\left(\mathrm{1}+\mathrm{1}\right)^{{n}} =\mathrm{2}^{{n}} \:\checkmark \\ $$
Commented by Matica last updated on 26/Dec/22
Thank you!
$${Thank}\:{you}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *