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Question Number 26223 by abdo imad last updated on 22/Dec/17

let put ξ(x)= Σ_(n=1) ^∝ (1/n^x ) with x>1 and δ(x) =Σ_(n=1) ^∝ (((−1)^n )/n^x ) find a relation between ξ(x) and δ(x).

$${let}\:{put}\:\xi\left({x}\right)=\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{{n}^{{x}} }\:\:{with}\:{x}>\mathrm{1} \\ $$ $${and}\:\:\delta\left({x}\right)\:\:=\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} }\:\:\:{find}\:{a}\:{relation} \\ $$ $${between}\:\xi\left({x}\right)\:{and}\:\delta\left({x}\right). \\ $$

Commented byabdo imad last updated on 23/Dec/17

δ(x) = Σ_(p=1) ^∝ (1/((2p)^x )) − Σ_(p=0) ^∝ (1/((2p+1)^x ))= 2^(−x) ξ(x)− Σ_(p=0) ^∝ (1/((2p+1)^x )) but ξ(x) = Σ_(p=1) ^∝ (1/((2p)^x )) + Σ_(p=0) ^∝ (1/((2p+1)^x )) ⇒ Σ_(p=0) ^∝ (1/((2p+1)^x )) = ξ(x) −2^(−x) ξ(x) =(1−2^(−x) )ξ(x) ⇒ δ(x)= 2^(−x) −(1−2^(−x) )ξ(x)=( 2^(1−x) −1)ξ(x)

$$\delta\left({x}\right)\:=\:\:\sum_{{p}=\mathrm{1}} ^{\propto} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}\right)^{{x}} }\:\:−\:\sum_{{p}=\mathrm{0}} ^{\propto} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{{x}} }=\:\mathrm{2}^{−{x}} \xi\left({x}\right)−\:\sum_{{p}=\mathrm{0}} ^{\propto} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{{x}} } \\ $$ $${but}\:\:\xi\left({x}\right)\:\:=\:\:\sum_{{p}=\mathrm{1}} ^{\propto} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}\right)^{{x}} }\:\:+\:\sum_{{p}=\mathrm{0}} ^{\propto} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{{x}} } \\ $$ $$\Rightarrow\:\:\:\sum_{{p}=\mathrm{0}} ^{\propto} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{{x}} }\:\:=\:\xi\left({x}\right)\:−\mathrm{2}^{−{x}} \xi\left({x}\right)\:=\left(\mathrm{1}−\mathrm{2}^{−{x}} \right)\xi\left({x}\right) \\ $$ $$\Rightarrow\:\:\delta\left({x}\right)=\:\mathrm{2}^{−{x}} −\left(\mathrm{1}−\mathrm{2}^{−{x}} \right)\xi\left({x}\right)=\left(\:\mathrm{2}^{\mathrm{1}−{x}} \:−\mathrm{1}\right)\xi\left({x}\right) \\ $$

Answered by prakash jain last updated on 22/Dec/17

δ(x)=Σ_(n=1) ^∞ (((−1)^n )/n^x ) =−(Σ_(n=1) ^∞ (1/n^x )−2Σ_(n=1) ^∞ (1/((2n)^x ))) −1+(1/2^x )−(1/3^x )+..=−(1+(1/2^x )+(1/3^x ))..+2[(1/2^x )+(1/4^x )+..] =−(Σ_(n=1) ^∞ (1/n^x )−2Σ_(n=1) ^∞ (1/((2n)^x ))) δ(x)=−ζ(x)+(2/2^x )ζ(x) δ(x)=(2^(x−1) −1)ζ(x)

$$\delta\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} } \\ $$ $$=−\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{x}} }−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{{x}} }\right) \\ $$ $$−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{{x}} }−\frac{\mathrm{1}}{\mathrm{3}^{{x}} }+..=−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{{x}} }+\frac{\mathrm{1}}{\mathrm{3}^{{x}} }\right)..+\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}^{{x}} }+\frac{\mathrm{1}}{\mathrm{4}^{{x}} }+..\right] \\ $$ $$=−\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{x}} }−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{{x}} }\right) \\ $$ $$\delta\left({x}\right)=−\zeta\left({x}\right)+\frac{\mathrm{2}}{\mathrm{2}^{{x}} }\zeta\left({x}\right) \\ $$ $$\delta\left({x}\right)=\left(\mathrm{2}^{{x}−\mathrm{1}} −\mathrm{1}\right)\zeta\left({x}\right) \\ $$

Commented byprakash jain last updated on 22/Dec/17

δ(x)=−η(x) (eta function)

$$\delta\left({x}\right)=−\eta\left({x}\right)\:\left(\mathrm{eta}\:\mathrm{function}\right) \\ $$

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