TAOCP 6.4 Exercise 53
The error in the proposed solution occurs at the change of variables and the resulting failure to preserve the structure needed for an incomplete gamma representation.
Exercise 53. [HM20] [HM20] Prove that the function R(a,n) can be expressed in terms of the incomplete gamma function, and use the result of exercise 1.2.11.3, 9 to find the asymptotic value of R(a,n) to O(n~?) as n > 00, for fixed a <1.
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The error in the proposed solution occurs at the change of variables and the resulting failure to preserve the structure needed for an incomplete gamma representation. We restart from the correct integral representation and derive everything cleanly.
1. Correct starting representation
From Exercise 52 (and the definition used in Exercise 51), we have
$$ R(a,n)=a^{n+1}\int_{0}^{\infty} e^{-t}(1+at)^n,dt. $$
2. Transformation to incomplete gamma form
Let
$$ x = 1+at. $$
Then
$$ t=\frac{x-1}{a}, \quad dt=\frac{dx}{a}, \quad t=0 \Rightarrow x=1. $$
Substituting,
$$ R(a,n)=a^{n+1}\int_{1}^{\infty} e^{-(x-1)/a} x^n \frac{dx}{a}. $$
Hence
$$ R(a,n)=a^n e^{1/a}\int_{1}^{\infty} x^n e^{-x/a},dx. $$
Now scale to match the incomplete gamma function. Let
$$ u=\frac{x}{a}, \quad x=au, \quad dx=a,du. $$
Then the lower limit becomes $u=1/a$, and
$$ R(a,n)=a^n e^{1/a}\int_{1/a}^{\infty} (au)^n e^{-u} a,du. $$
Collecting powers of $a$,
$$ R(a,n)=a^{2n+1} e^{1/a}\int_{1/a}^{\infty} u^n e^{-u},du. $$
Using the incomplete gamma function
$$ \Gamma(n+1,x)=\int_x^{\infty} t^n e^{-t},dt, $$
we obtain the exact representation
$$ R(a,n)=a^{2n+1} e^{1/a},\Gamma(n+1,1/a). $$
3. Asymptotics of $\Gamma(n+1,1/a)$ for fixed $a<1$
Since $a<1$, the cutoff point $1/a$ is fixed while the saddle point of $t^n e^{-t}$ occurs at $t=n$. Hence
$$ \frac{1}{a} \ll n \quad (n\to\infty), $$
so the incomplete part
$$ \gamma(n+1,1/a)=\int_0^{1/a} t^n e^{-t},dt $$
is exponentially small in $n$. By Exercise 1.2.11.3–9,
$$ \gamma(n+1,1/a)=O(e^{-cn}) $$
for some $c>0$, hence
$$ \Gamma(n+1,1/a)=\Gamma(n+1)\left(1+O(e^{-cn})\right). $$
Thus,
$$ R(a,n)=a^{2n+1} e^{1/a}\Gamma(n+1)\left(1+O(e^{-cn})\right). $$
4. Apply Stirling’s formula
Using Stirling’s approximation,
$$ \Gamma(n+1)=n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n, $$
we obtain
$$ R(a,n)\sim a^{2n+1} e^{1/a}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n. $$
5. Final asymptotic form
Rewriting,
$$ R(a,n)\sim e^{1/a}\sqrt{2\pi n}, a^{2n+1}\left(\frac{n}{e}\right)^n. $$
Equivalently,
$$ R(a,n)\sim e^{1/a},a,\sqrt{2\pi n}\left(\frac{a^2 n}{e}\right)^n. $$
The relative error is exponentially small in $n$, hence much stronger than any algebraic $O(n^{-k})$ term.
Final result
Exact form:
$$ R(a,n)=a^{2n+1} e^{1/a},\Gamma(n+1,1/a). $$
Asymptotic form for fixed $a<1$:
$$ R(a,n)\sim e^{1/a},a,\sqrt{2\pi n}\left(\frac{a^2 n}{e}\right)^n, \quad n\to\infty. $$
This completes the corrected derivation.