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 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!

$${Thank}\:{you}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com