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Question Number 93225 by Ajao yinka last updated on 11/May/20

Commented by prakash jain last updated on 12/May/20

$$1+x+...+x^n\mathrm{has}\mathrm{no}\mathrm{root}\mathrm{in}\mathbb{R}\mathrm{for}n\mathrm{even}$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^n\delta (1+x+...+x^n)dx=0$$$$\mathrm{for}\mathrm{even}n.\mathrm{So}\mathrm{inequality}\mathrm{does}\mathrm{nold}$$$$\mathrm{for}\mathrm{even}n.$$

Commented by prakash jain last updated on 12/May/20

$$\mathrm{for}n\mathrm{odd}$$$$\delta (1+x+...+x^n)=\frac{\delta (x+1)}{\mid 1+2x+3x^2+..+{nx}^{n-1}{\mid }_{x=-1}}$$$$\mathrm{S}=1+2x+3x^2+...+{nx}^{n-1}$$$$xS=x+2x^2+....+(n-1)x^{n-1}+{nx}^n$$$$(1-x)S=(1+x+x^2+..+x^{n-1})-{nx}^n$$$$x=-1$$$$2S=n+1\left(nodd\right)$$$$S=\frac{n+1}{2}$$$$\delta (1+x+x^2+..+x^n)=\frac{2\delta (x+1)}{n+1}\left(nodd\right)$$$$\mathrm{So}\mathrm{for}\mathrm{odd}n$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{x}^{n+1}\delta (1+x+x^2+...+x^n)dx=\frac{2}{n+1}$$

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