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Question Number 103205 by DGmichael last updated on 13/Jul/20

Answered by OlafThorendsen last updated on 13/Jul/20

$$f\left(\lambda \right)=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{\lambda }^k}{k!}$$$$\Rightarrow f^{\prime }\left(\lambda \right)=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{k{\lambda }^{k-1}}{k!}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{{\lambda }^{k-1}}{(k-1)!}$$$$\Rightarrow f^{\prime }\left(\lambda \right)=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{\lambda }^k}{k!}=f\left(\lambda \right)$$$$f^{\prime }\left(\lambda \right)=f\left(\lambda \right)\Rightarrow f\left(\lambda \right)=\mathrm{C}{e}^{\lambda }$$$$\mathrm{But}f\left(0\right)=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{0^k}{k!}=1=\mathrm{C}{e}^0\Rightarrow \mathrm{C}=1$$$$\mathrm{Then}f\left(\lambda \right)=1\times e^{\lambda }=e^{\lambda }$$$$$$

Answered by OlafThorendsen last updated on 13/Jul/20

$$f(x+h)=f\left(x\right)+\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{f^{\left(k\right)}\left(x\right)}{k!}{h}^n$$$$\mathrm{If}x=0,h=\lambda ,f=\exp$$$$e^{\lambda }=e^0+\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{e^0}{k!}{\lambda }^k=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{\lambda }^k}{k!}$$

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