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Question Number 117086 by mnjuly1970 last updated on 09/Oct/20

       ...nice  mathematics...   evaluate ...         lim_(n→∞) (Π_(k=1) ^n sin(((kπ)/(4n))))^(1/n) =?                    ...m.n.1970...

...nicemathematics...evaluate...limn(nk=1sin(kπ4n))1n=?...m.n.1970...

Answered by Ar Brandon last updated on 09/Oct/20

Commented by mnjuly1970 last updated on 09/Oct/20

thank you very much..

thankyouverymuch..

Commented by Ar Brandon last updated on 09/Oct/20

You're welcome Sir ��

Answered by Dwaipayan Shikari last updated on 09/Oct/20

lim_(n→∞) (Π_(k=1) ^n sin(((kπ)/(4n))))^(1/n) =y  lim_(n→∞) (1/n)Σ_(k=1) ^n log(sin(((kπ)/(4n))))=logy  ∫_0 ^1 log(sin(((πx)/4)))=logy  (4/π)∫_0 ^(π/4) log(sint)dt  −1.27=logy  (Wolfram)  y=e^(−1.27)

limn(nk=1sin(kπ4n))1n=ylimn1nnk=1log(sin(kπ4n))=logy01log(sin(πx4))=logy4π0π4log(sint)dt1.27=logy(Wolfram)y=e1.27

Commented by Olaf last updated on 10/Oct/20

∫_0 ^(π/4) ln(sint)dt = −(K/2)−(π/4)ln2 exactly.  (with K = β(2) : constant of Catalan)  So limit is (1/2)e^(−((2β(2))/π))

0π4ln(sint)dt=K2π4ln2exactly.(withK=β(2):constantofCatalan)Solimitis12e2β(2)π

Commented by mnjuly1970 last updated on 09/Oct/20

thank you so much.   answer :=e^((((−2G)/π)−log(2)))  ✓  recall: ∫_0 ^( (π/4)) log(sin(x))dx                   =((−G)/2) −(π/4) log(2)     m.n.1970    sincerely  yours...

thankyousomuch.answer:=e(2Gπlog(2))recall:0π4log(sin(x))dx=G2π4log(2)m.n.1970sincerelyyours...

Commented by mnjuly1970 last updated on 10/Oct/20

you are right.  your solution is  simplified...

youareright.yoursolutionissimplified...

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