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Chapter 11: Q11.10P (page 554)

Use Stirling’s formula to evaluate limn→∞(n!)1nn.

Short Answer

Expert verified

The value of the function is1e.

Step by step solution

01

Given Information

The function islimn→∞(n!)1nn.

02

Definition ofthe sterling’s formula.

Sterling’s formula is used to simplify formulas involving factorial. n!~nne-n2πn.

03

Find the value of the function.

The function is limn→∞(n!)1nn.

The sterling’s formula is n!~nne-n2πn.

Substitute above values in the equation mentioned below.

limn→∞(n!)1nn=limn→∞(nne-n2πn)1nnlimn→∞(n!)1nn=limn→∞e-12πn12nlimn→∞(n!)1nn=1e

Hence, the value of the function is 1e.

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Most popular questions from this chapter

The figure is part of a cycloid with parametric equations x=a(θ+sinθ),y=a(1-cosθ) (The graph shown is like Figure 4.4 of Chapter 9 with the origin shifted to P2.) Show that the time for a particle to slide without friction along the curve from (x1, y1) to the origin is given by the differential equation for θ(t) is t=ayg∫01dyy(y1-y).

Hint: Show that the arc length element is ds=2aydy. Evaluate the integral to show that the time is independent of the starting height y1 .

Show that ∫-∞∞e-y22dt=2π

(a)by using (9.5) and (9.2a) .

(b) by reducing it to a Γ function and using (5.3) .

Prove that B(p,q)=B(q,p). Hint:Putx=1-yin Equation (6.1).

Use the term 1/(12p)in (11.5) to show that the error in Stirling’s formula (11.1) is < 10%for p > 1; < 1%for p > 10; < 0.1%for p > 100; < 0.01%for p > 1000.

Sometimes you may find the notation F(ϕ,k) in (12.2)used when k> 1 . Allowing this notation, show that13F(sin-135,43)=14F(sin-145,34). Hints: Using the Jacobi form of F in (12.2), write the integral which is equal to13F(sin-135,43). Follow Example 3 to make a change of variable, write the corresponding integral, and verify that it is equal to14F(sin-145,34).
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