Asymptotics of special functions

[5/100%]
Show
LaTeX
Problem 5360 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.7 "Approximate Integration of Differential Equations"

Find the asymptotic expression, as n→∞n \to \infty, for the Chebyshev--Hermite polynomials Hn(x)=ex22n!dne−x22dxnH_{n}(x) = \dfrac {e^{\frac{x^{2}}{2}}}{n!}\dfrac {d^{n}e^{-\frac{x^{2}}{2}}}{dx^{n}} (see problem 5359).

?
Problem 1.90 from Pólya G., Szegő G. "Problems and Theorems in Analysis" [PóSz98]More problems in Section 3 "Transformations of Sequences into Functions. Theorem of Cesàro"

If 0<k<10<k<1 and if kk converges to 11, then

∫01dx(1−x2)(1−k2x2)∼12log⁡11−k.[See Problem 202 of Part II.] \int _0^1 \frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} \sim \frac{1}{2} \log \frac{1}{1-k} . \qquad \text{[See Problem 202 of Part II.]}
?
Problem 4.73 from Pólya G., Szegő G. "Problems and Theorems in Analysis" [PóSz98]More problems in Section 4 "$\mu(r)$ and $M(r)$ under Special Regularity Assumptions"

Show that for the Bessel function of order 00, J0(z)J_{0}(z), we have as r→+∞r \to +\infty

J0(ir)=∑n=0∞1n!n!(r2)2n∼2πr∑n=0∞r2n(2n)!∼er2πr. J_{0}(ir) = \sum _{n=0}^{\infty } \frac{1}{n! n!} \left(\frac{r}{2}\right)^{2n} \sim \sqrt{\frac{2}{\pi r}} \sum _{n=0}^{\infty } \frac{r^{2n}}{(2n)!} \sim \frac{e^{r}}{\sqrt{2\pi r}}.
?
Remark.
?

See Problem 204 of Part II.

Problem 4.74 from Pólya G., Szegő G. "Problems and Theorems in Analysis" [PóSz98]More problems in Section 4 "$\mu(r)$ and $M(r)$ under Special Regularity Assumptions"

Let a1,a2,…,ap,b1,b2,…,bqa_{1}, a_{2}, \ldots , a_{p}, b_{1}, b_{2}, \ldots , b_{q} be real constants different from 0,−1,−2,−3,…0, -1, -2, -3, \ldots, where p<qp < q,

P(z)=(z+a1−1)(z+a2−1)⋯(z+ap−1), P(z) = (z+a_{1}-1)(z+a_{2}-1) \cdots (z+a_{p}-1), Q(z)=(z+b1−1)(z+b2−1)⋯(z+bq−1). Q(z) = (z+b_{1}-1)(z+b_{2}-1) \cdots (z+b_{q}-1).

As rr tends to infinity (r>0r > 0) the everywhere convergent power series

1+P(1)Q(1)r+P(1)P(2)Q(1)Q(2)r2+⋯+P(1)P(2)⋯P(n)Q(1)Q(2)⋯Q(n)rn+⋯ 1 + \frac{P(1)}{Q(1)} r + \frac{P(1)P(2)}{Q(1)Q(2)} r^{2} + \cdots + \frac{P(1)P(2) \cdots P(n)}{Q(1)Q(2) \cdots Q(n)} r^{n} + \cdots ∼Γ(b1)Γ(b2)⋯Γ(bq)Γ(a1)Γ(a2)⋯Γ(ap)(2π)1−l2l−12r2Δ+l+12lelr1l, \sim \frac{\Gamma (b_{1}) \Gamma (b_{2}) \cdots \Gamma (b_{q})}{\Gamma (a_{1}) \Gamma (a_{2}) \cdots \Gamma (a_{p})} (2\pi )^{\frac{1-l}{2}} l^{-\frac{1}{2}} r^{\frac{2\Delta +l+1}{2l}} e^{l r^{\frac{1}{l}}},

where we set l=q−pl = q-p, Δ=a1+a2+⋯+ap−b1−b2−⋯−bq\Delta = a_{1}+a_{2}+\cdots +a_{p} - b_{1}-b_{2}-\cdots -b_{q}.

?
Remark.
?

Verify that from this problem 73 follows as a special case after a change of variables. Use as a ``comparison series''

1+lll!r+l2l(2l)!r2+⋯+lnl(nl)!rn+⋯∼1lelr1l. 1 + \frac{l^{l}}{l!} r + \frac{l^{2l}}{(2l)!} r^{2} + \cdots + \frac{l^{nl}}{(nl)!} r^{n} + \cdots \sim \frac{1}{l} e^{l r^{\frac{1}{l}}}.
Problem 4.75 from Pólya G., Szegő G. "Problems and Theorems in Analysis" [PóSz98]More problems in Section 4 "$\mu(r)$ and $M(r)$ under Special Regularity Assumptions"

Obtain the asymptotic formula in problem 74, without making use of problem 72, by means of Problem 94 of Part I and Problem 207 of Part II.

?

References

[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.