what-is-the-value-of-and-how-can-combute-erf-pi-

Question Number 127214 by mohammad17 last updated on 27/Dec/20

what is the value of and how can combute? erf(π)?

$${what}\:{is}\:{the}\:{value}\:{of}\:{and}\:{how}\:{can}\:{combute}? \\ $$$$ \\ $$$${erf}\left(\pi\right)? \\ $$

Answered by Olaf last updated on 27/Dec/20

erf(x) = (2/( (√π)))∫_0 ^x e^(−t^2 ) dt (used in probabilities and statistics) erf(π) = (2/( (√π)))∫_0 ^π e^(−t^2 dt) We can compute : erf(x) = (2/( (√π)))Σ_(n=0) ^∞ (((−1)^n )/((2n+1)n!))x^(2n+1) erf(π) = (2/( (√π)))Σ_(n=0) ^∞ (((−1)^n )/((2n+1)n!))π^(2n+1) erf(π) ≈ 2(√π)−((2π^(5/2) )/3)+(π^(9/2) /5)−(π^(13/2) /(21))+o(π^(17/2) )

$$\mathrm{erf}\left({x}\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{{x}} {e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$\left(\mathrm{used}\:\mathrm{in}\:\mathrm{probabilities}\:\mathrm{and}\:\mathrm{statistics}\right) \\ $$$$\mathrm{erf}\left(\pi\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\pi} {e}^{−{t}^{\mathrm{2}} {dt}} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{compute}\:: \\ $$$$\mathrm{erf}\left({x}\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\pi}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!}{x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\mathrm{erf}\left(\pi\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\pi}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!}\pi^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\mathrm{erf}\left(\pi\right)\:\approx\:\mathrm{2}\sqrt{\pi}−\frac{\mathrm{2}\pi^{\mathrm{5}/\mathrm{2}} }{\mathrm{3}}+\frac{\pi^{\mathrm{9}/\mathrm{2}} }{\mathrm{5}}−\frac{\pi^{\mathrm{13}/\mathrm{2}} }{\mathrm{21}}+{o}\left(\pi^{\mathrm{17}/\mathrm{2}} \right) \\ $$

Commented by mohammad17 last updated on 27/Dec/20

$${thank}\:{you}\:{sir} \\ $$

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