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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 has no root in R for n even ∫_(−∞) ^∞ x^n δ(1+x+...+x^n )dx=0 for even n. So inequality does nold for even n.

$$\mathrm{1}+{x}+...+{x}^{{n}} \:\mathrm{has}\:\mathrm{no}\:\mathrm{root}\:\mathrm{in}\:\mathbb{R}\:\mathrm{for}\:{n}\:\mathrm{even} \\ $$$$\int_{−\infty} ^{\infty} {x}^{{n}} \delta\left(\mathrm{1}+{x}+...+{x}^{{n}} \right){dx}=\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{even}\:{n}.\:\mathrm{So}\:\mathrm{inequality}\:\mathrm{does}\:\mathrm{nold} \\ $$$$\mathrm{for}\:\mathrm{even}\:{n}. \\ $$

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

for n odd δ(1+x+...+x^n )=((δ(x+1))/(∣1+2x+3x^2 +..+nx^(n−1) ∣_(x=−1) )) 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 (n odd) S=((n+1)/2) δ(1+x+x^2 +..+x^n )=((2δ(x+1))/(n+1)) (n odd) So for odd n ∫_(−∞) ^(+∞) x^(n+1) δ(1+x+x^2 +...+x^n )dx=(2/(n+1))

$$\mathrm{for}\:{n}\:\mathrm{odd} \\ $$$$\delta\left(\mathrm{1}+{x}+...+{x}^{{n}} \right)=\frac{\delta\left({x}+\mathrm{1}\right)}{\mid\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +..+{nx}^{{n}−\mathrm{1}} \mid_{{x}=−\mathrm{1}} } \\ $$$$\mathrm{S}=\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +...+{nx}^{{n}−\mathrm{1}} \\ $$$${xS}=\:\:\:\:\:\:{x}+\mathrm{2}{x}^{\mathrm{2}} +....+\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{1}} +{nx}^{{n}} \\ $$$$\left(\mathrm{1}−{x}\right){S}=\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +..+{x}^{{n}−\mathrm{1}} \right)−{nx}^{{n}} \\ $$$${x}=−\mathrm{1} \\ $$$$\mathrm{2}{S}={n}+\mathrm{1}\:\left({n}\:{odd}\right) \\ $$$${S}=\frac{{n}+\mathrm{1}}{\mathrm{2}} \\ $$$$\delta\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +..+{x}^{{n}} \right)=\frac{\mathrm{2}\delta\left({x}+\mathrm{1}\right)}{{n}+\mathrm{1}}\:\left({n}\:{odd}\right) \\ $$$$\mathrm{So}\:\mathrm{for}\:\mathrm{odd}\:{n} \\ $$$$\int_{−\infty} ^{+\infty} {x}^{{n}+\mathrm{1}} \delta\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +...+{x}^{{n}} \right){dx}=\frac{\mathrm{2}}{{n}+\mathrm{1}} \\ $$

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