Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 158687 by mahdipoor last updated on 07/Nov/21

∀x∈R  f(x)=f ′ (x)      and   f(0)=1  prove  f(a+b)=f(a)×f(b)

$$\forall{x}\in{R}\:\:{f}\left({x}\right)={f}\:'\:\left({x}\right)\:\:\:\:\:\:{and}\:\:\:{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${prove}\:\:{f}\left({a}+{b}\right)={f}\left({a}\right)×{f}\left({b}\right) \\ $$

Answered by mr W last updated on 07/Nov/21

y′=(dy/dx)=y  (dy/y)=dx  ∫(dy/y)=∫dx  ln y−ln C=x  ln (y/C)=x  ⇒y=Ce^x   y(0)=C=1  ⇒y=f(x)=e^x   f(a+b)=e^(a+b) =e^a e^b =f(a)f(b)

$${y}'=\frac{{dy}}{{dx}}={y} \\ $$$$\frac{{dy}}{{y}}={dx} \\ $$$$\int\frac{{dy}}{{y}}=\int{dx} \\ $$$$\mathrm{ln}\:{y}−\mathrm{ln}\:{C}={x} \\ $$$$\mathrm{ln}\:\frac{{y}}{{C}}={x} \\ $$$$\Rightarrow{y}={Ce}^{{x}} \\ $$$${y}\left(\mathrm{0}\right)={C}=\mathrm{1} \\ $$$$\Rightarrow{y}={f}\left({x}\right)={e}^{{x}} \\ $$$${f}\left({a}+{b}\right)={e}^{{a}+{b}} ={e}^{{a}} {e}^{{b}} ={f}\left({a}\right){f}\left({b}\right) \\ $$

Commented by mahdipoor last updated on 07/Nov/21

can you prove without use e^x  ?

$${can}\:{you}\:{prove}\:{without}\:{use}\:{e}^{{x}} \:? \\ $$

Commented by mr W last updated on 07/Nov/21

f(x)=e^x  is the only solution satifying  f′(x)=f(x) and f(0)=1.

$${f}\left({x}\right)={e}^{{x}} \:{is}\:{the}\:{only}\:{solution}\:{satifying} \\ $$$${f}'\left({x}\right)={f}\left({x}\right)\:{and}\:{f}\left(\mathrm{0}\right)=\mathrm{1}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com