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find-e-1-ln2-




Question Number 68068 by mhmd last updated on 04/Sep/19
find e^(1/ln2)   =?
finde1/ln2=?
Commented by peter frank last updated on 04/Sep/19
e^x =1+x+(x^2 /(2!))+(x^3 /(3!))+(x^4 /(4!))  substitute x=(1/(ln2))
ex=1+x+x22!+x33!+x44!substitutex=1ln2
Commented by mathmax by abdo last updated on 04/Sep/19
for x≠0  we have e^(1/x)  =Σ_(n=0) ^∞    (1/(n!x^n )) =1+(1/x) +(1/(2!x^2 )) +(1/(3!x^3 )) +...  let take x =ln(2) ⇒e^(1/(ln2))  =1+(1/(ln(2))) +(1/(2!(ln2)^2 )) +(1/(3!(ln(2))^3 )) +...  and if we want a spproximate value we can take 10 terms  for this serie.
forx0wehavee1x=n=01n!xn=1+1x+12!x2+13!x3+lettakex=ln(2)e1ln2=1+1ln(2)+12!(ln2)2+13!(ln(2))3+andifwewantaspproximatevaluewecantake10termsforthisserie.

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