Question Number 101104 by student work last updated on 30/Jun/20

(d/dx)(x!)=?

Answered by smridha last updated on 30/Jun/20

x!=𝚪(x+1)=xΓx  =lim_(n→∞) ((n!)/((x+1)(x+2)......(x+n)))n^x [Euler definition]  take ln both sides we get  ln[𝚪(x+1)]=lim_(n→∞) [ln(n!)+xln(n)−ln(x+1)−ln(x+2)−.....−ln(x+n)]  now diff: both sides wrt x  (([𝚪(x+1)]^′ )/(𝚪(x+1)))≡𝚿(x+1)[Digamma f^n ]                      =lim_(n→∞) [ln(n)−(1/((x+1)))−(1/((x+2)))−...(1/((x+n)))]     so 𝚿(x+1)=lim_(n→∞) [(ln(n)−H_n )−Σ_(m=1) ^n ((1/((x+m)))−(1/m))]                     =−𝛄+Σ_(m=1) ^∞ (x/(m(m+x)))  so,[𝚪(x+1)]^′ =𝚪(x+1).𝚿(x+1).  where...  𝛄(Euler−Mascheroni constant)  defind by  𝛄=lim_(n→∞) (H_n −ln(n))  H_n (harmonic number)=Σ_(m=1) ^n (1/m).

Commented byMJS last updated on 30/Jun/20

Γ(x+1)=∫_0 ^∞ t^x e^(−t) dt  (d/dx)[Γ(x+1)]=∫_0 ^∞ t^x e^(−t) ln t dt  which is not easier or harder to calculate  than your answer

Commented bysmridha last updated on 30/Jun/20

my way is the standard way   and the result is also standard..  your method is just a application  of my method.many of the   conclutions can be derived by  the way I approched.  I  have an another interesting way  infinite product of gamma f^n   Weierstrass form           (1/(𝚪(x)))≡xe^(𝛄x) Π_(n=1) ^∞ (1+(x/n))e^(−(x/n))   the same result can be evaluated  from this.  I always show my favourite method  among the methods I know!!!

Commented byShazneen last updated on 01/Jul/20

simple to understand

Commented byTawa11 last updated on 06/Nov/21

Good

Answered by  M±th+et+s last updated on 30/Jun/20

∵x!=(√(2πx))((x/e))^x   ⇒(d/dx)(x!)=(√(2πx)).e^(x(ln(x)−)) .(x.(1/x)+ln(x)−1)+e^(x(ln(x)−1)) ((2π)/( (√(2πx))))  =e^(x(ln(x)−1)) ((√(2πx)) ln(x)+((2π)/( (√(2πx)))))  ∴(d/dx)(x!)=((2π.e^(x(ln(x)−1)) )/( (√(2πx))))(xln(x)+1)

Commented bysmridha last updated on 30/Jun/20

stirling′s formula very good!!  i know also but it is approximate  result:  because  𝚪(x+1)≈(√(2𝛑x)).x^x .e^(−x)

Commented bymathmax by abdo last updated on 30/Jun/20

x! ∼ (√(2πx))((x/e))^x  is approximation of x! at +∞(not equality..!)

Commented byTinku Tara last updated on 30/Jun/20

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Commented byMJS last updated on 01/Jul/20

thank you