Menu Close

Question-66498




Question Number 66498 by miracle wokama last updated on 16/Aug/19
Commented by MJS last updated on 16/Aug/19
too complicated  it′s easy to see that f′(0)=1 and f′(1)=100
$$\mathrm{too}\:\mathrm{complicated} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{that}\:{f}'\left(\mathrm{0}\right)=\mathrm{1}\:\mathrm{and}\:{f}'\left(\mathrm{1}\right)=\mathrm{100} \\ $$
Commented by mathmax by abdo last updated on 16/Aug/19
we have f(x)=1+x +(x^2 /2) +(x^3 /3)+....+(x^(100) /(100)) ⇒  f^′ (x)=1+x +x^2  +....+x^(99)  ⇒ f^′ (1) =1+1+....+1(100×)=100  but f^′ (0) =1 ⇒f^′ (1)=100f^′ (0)
$${we}\:{have}\:{f}\left({x}\right)=\mathrm{1}+{x}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+….+\frac{{x}^{\mathrm{100}} }{\mathrm{100}}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\mathrm{1}+{x}\:+{x}^{\mathrm{2}} \:+….+{x}^{\mathrm{99}} \:\Rightarrow\:{f}^{'} \left(\mathrm{1}\right)\:=\mathrm{1}+\mathrm{1}+….+\mathrm{1}\left(\mathrm{100}×\right)=\mathrm{100} \\ $$$${but}\:{f}^{'} \left(\mathrm{0}\right)\:=\mathrm{1}\:\Rightarrow{f}^{'} \left(\mathrm{1}\right)=\mathrm{100}{f}^{'} \left(\mathrm{0}\right) \\ $$
Commented by mathmax by abdo last updated on 16/Aug/19
no need to use all those relations....the Q is simple and eazy....
$${no}\:{need}\:{to}\:{use}\:{all}\:{those}\:{relations}….{the}\:{Q}\:{is}\:{simple}\:{and}\:{eazy}…. \\ $$
Answered by Cmr 237 last updated on 17/Aug/19
f(x)=Σ_(k=1) ^(100) −_k ^x^k  +1  f^((1)) (x)=Σ_(k=1) ^(100) x^(k−1)               =((1−x^(100) )/(1−x))  f^((1)) (0)=1  f^((1)) (1)=100  d.ou f^((1)) (1)=100f^((1)) (0)
$${f}\left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\underset{{k}} {\overset{{x}^{{k}} } {−}}+\mathrm{1} \\ $$$$\mathrm{f}^{\left(\mathrm{1}\right)} \left(\mathrm{x}\right)=\underset{{k}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}{x}^{{k}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}−{x}^{\mathrm{100}} }{\mathrm{1}−{x}} \\ $$$${f}^{\left(\mathrm{1}\right)} \left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}^{\left(\mathrm{1}\right)} \left(\mathrm{1}\right)=\mathrm{100} \\ $$$${d}.{ou}\:{f}^{\left(\mathrm{1}\right)} \left(\mathrm{1}\right)=\mathrm{100}{f}^{\left(\mathrm{1}\right)} \left(\mathrm{0}\right) \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published.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