数理解析研究所講究録1090 巻 1999 年 116-125

116

On asymptotic estimates for coefficients of divergent solutions tosecond order non-homogeneous linear ordinary differential
equations.
Nakamura Yayoi
Ochanomizu Univercity
中村弥生 (お茶の水女子大)

1

Introduction

Conseder a second order linear ordinary differential equation(1)$\frac{d^{2}y}{dx^{2}}-p.(x)y=0$$p(x)=x+a1X-1+m\ldots+ma_{m}$This equation has a irregular singular point at infinity with irregularity $m+2$ .
It is known that this equation has integral solutions in following four cases:(i) $p(x)=x+a_{1}$(i.e. $m=1$ )(ii) $p(x)=x^{2}+a_{1}x+a_{2}$(i.e. $m=2$ )(i\"u) $p(x)=x^{m}$(iv) $p(x)=X^{2p}+CX^{\mathrm{p}-1}$
In these cases, each equation will be transformed into confluent hypergeometric differential equation.
Especially, the solution of the equation (1) with coefficient (i) is Airy function and (ii) parabolic cylinderfunction.
Let and be the ring of convergent power series and the ring of formal power series in , respectively.
Then, for a linear ordinary differential operator $P$ with coefficients in holomorphic function, we see thefollowing isomorphism of linear spaces due to Deligne;$\hat{\mathcal{O}}$$\mathcal{O}$$x$$H^{1}(S^{1}, \kappa_{er}(P:A\mathrm{o}))\cong \mathrm{K}\mathrm{e}\mathrm{r}(P;\hat{O}/\mathcal{O}arrow\hat{O}/O)$

whereis sheaf of germs of functions asymptotically developable to the formal power series on thecircle .
To pay attension to this isomorphism, by vanishing theorem due to [3] in asymptotic analysis, we
obtain the asymptotic estimates for coefficients of divergent solutions to non-homogeneous linear ordinary$A_{0}$$0$$S^{1}$

differential equations about the differential operators$P= \frac{d^{2}}{dx^{2}}-p(x)$

.

117

In this note, we see the numerical calculation about the estimation about the case of (\"ui), (iv). By
these calculation, we can see the estimate and the asymptotic estimate for coefficients of a solution of
the on-homogeneous differential equation are almost the same.
$.\mathrm{n}$About the operator

2$P= \frac{d^{2}}{dx^{2}}-x^{m}$

The differential equation(2)$Py=( \frac{d^{2}}{dx^{2}}-X^{m})y=0$

has irregular singular point at $x=\infty$ . The irregularity of$P$

at $x=\infty$ equals $i_{\infty}(P)=m+2$ . Put$U_{k}=\{x\in \mathrm{C};|x|>0,$ $\frac{2k-3}{m+2}\pi<\arg x<\frac{2k+1}{m+2}\pi\}$, $k=0,1,$$\ldots,$$m+1$ .

Then $\{U_{k}, k=0,1, \ldots, m+1\}$ forms an open sectorial covering at $x=\infty$ , and put$S_{k}=U_{k}\cap U_{k+1}=\{x\in \mathrm{C};|x|>0,$ $| \arg x-\frac{2k}{m+2}\pi|<\frac{\pi}{m+2}\}$, $k=0,1,$$\ldots,$$m+1$ ,

where $U_{m+2}=U_{0}$ .
We have now a unique entire solution $y=y_{k}$ of the differential equation (2) with the asymptotic
representation:$y_{k} \sim\omega^{\mathrm{p}}4X^{-\mathrm{n}}4(k1+\sum^{\infty}B_{N}\omega 2-2xp\{(-1)N=1\mathrm{A}N_{X)e}\simeq k+1_{\frac{2}{m+2}}x^{\frac{m\neq 2}{2}}\}$

astends to infinity in any closed subsector of the open sector$N=0,1,$ , the quantities
are decided by$x$$U_{k}$. Where$\omega=e\overline{m}\overline{+2}\lrcorner\pi i$$B_{(m+2)(1}N+)= \prod_{l=0}\frac{-1}{(m+2)(\ell+1)}\{\frac{m}{4}(\frac{m}{4}+1N)+\frac{tm(m+2)}{4}+\frac{l(m+2)}{2}(\frac{\ell(m+2\rangle}{2}+1)\}$On the other hand, we have the solution with above properties:$y_{2k}= \frac{(\frac{4}{m+2})^{\iota_{+}arrow}2\overline{n}+2}{\Gamma(\frac{1}{2}-\frac{1}{m+2})}\int_{0}^{+\infty}e-x\frac{n\neq 2}{2}(\zeta+\mathrm{z}_{\overline{2}})2n"+2(\zeta+\frac{4}{m+2}\overline{m}+\zeta^{-\iota_{-}})-1" dz^{-}n+2\zeta$,.

$y_{2k+1}= \frac{(-1)n+2(\frac{4}{m+2})*+arrow n}{\Gamma(\frac{1}{2}-\frac{1}{m+2})}\int_{0}^{+\infty}e(\zeta++m2)_{\zeta^{-}(}:_{-}m+_{\mathrm{z}}\zeta \mathrm{g}^{\frac{m\neq 2}{2}}\frac{4}{m+2}+)-\mathrm{A}2-n+_{2}d\zeta$

Then each

and, for$B_{N}$$\ldots$$y_{k}$

gives$\mathrm{s}\mathrm{u}\dot{\mathrm{b}}$dominant solution in$S_{k}$

.

.

118

2.1

Coefficient and its asymptotic estimation of the solution

By using above representations of solutions, we can choose a basis ofway:
Put 1-cocycle about $\{U0, U1, \ldots, Um+1\}$ by$H^{1}(s^{1},\kappa_{e}r(P : A\mathrm{o}))$$\{u_{i}^{(k)}j’(i,j)=(\mathrm{O}, 1),\cdot(1,2), \ldots , (m, m+1), (m+1,0)\}$$u_{ij}^{(k)}=\{$$y$:$x\in S_{k}$$i=k$ ,

$0$$x\in S_{k}$$i\neq k$$m+1$ .
for $k=0,1,$
In this situation, the pair of cohomology classes ofin the folowing,,$\ldots,$$H^{1}(s^{1}, \kappa er(P:A0))$$\{u_{ij}^{(0)}\},$$\{u_{j}^{(1)}.\}|’\ldots,$ $\{u!_{j}^{m+1)}.\}$.

By vanishing theorem in asymptotic analysis, we.have -cochain, such that$\{v_{0’ 1}^{(k)(k)}v, \ldots , v_{m+1}^{(\mathrm{k})}\},$

$0$

$1$

$u_{1j}^{(k)}.=-v_{:}^{(k)}+v_{j}^{(k)},$ $(i,j)=(\mathrm{O}, 1),$

where eachpower-series$v_{\mathrm{j}}^{(k)}$forms a basis of

are defined in$U_{j}$$(1,2),$$k=0,1,$$\ldots$, $m+$$(m,m+1),$ $(m+1,0)$ ,$\cdot,$for $j=0,1,$ $\ldots,m+1$ and asymptotically developable to formal$m+1$ , respectively. Then
$k=0,1,$$\hat{v}^{(k)}(x)=\sum_{r=0}^{\infty}v(m, k, r)X^{-\Gamma},$$\ldots,$$Pu_{i}^{(k)}j=-Pv_{1}^{(\mathrm{k})}.+Pv_{j}^{(k)}$In $S:$ , we have
$Pv_{ij}^{(k)}=Pv^{(k)}$

.

.

Put
$Pv^{(k)}$:$x\in U$$g^{(k)}(_{X})=\{$$Pv_{j}^{(k)}i$$x\in U_{j}$.

Then$g^{(k)}(_{X)}$$x\in \mathrm{C}$$f^{(k)}(_{X})=\{$$\lim_{xarrow\infty}g^{(k)}(X)$$x=\infty$

define holomorphic function, and$P\hat{v}_{1}.((k)x)=f(k)(x)$.

Hence, by the vanishing theorem due to [3] in asymptotic analysis, {is of.
Moreover, we can have estimates for the coefficients $v(m, k, r)$ .

$\mathrm{b}\mathrm{a}s$$[\hat{v}^{(0)}],$$[\hat{v}^{(m+)}]1\rangle$$[\hat{v}^{(1)}],$$\ldots,$$\mathrm{K}\mathrm{e}\mathrm{r}(_{d}d=_{x}-x;\hat{O}m/\mathit{0})2$$v(m, 2k+1, r)$$=$$(-1)^{A\mathrm{L}}m+ \overline{2}^{+}1\frac{1}{2\pi i}\mathrm{r}(\frac{2r}{m+2})(\frac{4}{m+2})^{-\mp_{2}}n(\frac{2}{m+2})1-m+\overline{2}\simeq \mathrm{L}$$\cross\frac{\Gamma(\frac{1}{2}-\frac{1}{m+2})\Gamma(\frac{2r+2}{m+2})}{\Gamma(\frac{2r+1}{m+2}\dagger\frac{1}{2})}F(\frac{2r}{m+2}, \frac{1}{2}-\frac{1}{m+2}, \frac{2r+1}{m+2}+\frac{1}{2};-1)$.

forms a119$v(m,2k, \mathrm{r})$$=$$(-1)^{1_{-}}2n \#_{\frac{1}{2\pi i}}\Gamma(\frac{2\mathrm{r}}{m+2})(\frac{4}{m+2})^{-}narrow_{(}\frac{2}{m+2})^{1-\#}m$$\cross\frac{\Gamma(\frac{1}{2}-\frac{1}{m+2})\Gamma(\frac{2\mathrm{r}+2}{m+2})}{\Gamma(\frac{2r+1}{m+2}+\frac{1}{2})}F(\frac{2\mathrm{r}}{m+2}, \frac{1}{2}-\frac{1}{m+2}, \frac{2\gamma+1}{m+2}+\frac{1}{2}; -1)$.

And we can have asymptotic estimates for any sufficiently large number ,$f$$v(m, k,’)$$\sim$$v_{M}(m, k, r)$$=$$\frac{1}{2\pi i}(-\omega^{\gamma++})1k\sum_{N=0}^{M-1}B_{m,N(}\frac{m+2}{2})n+_{2()}’-4\mathrm{n}_{-_{2}}\alpha-1\mathrm{r}(\frac{2}{m+2}(\gamma-\frac{1}{4}m-\frac{N}{2}))$$+O( \Gamma(\frac{2}{m+2}(\gamma-\frac{m}{4}-\frac{M}{2})))$,

provided $1\leq M<r$ .

2.2

About the operator $P= \frac{d^{2}}{dx^{2}}-(x^{2p}+cx^{p-1})$

The differential equation(3)$Py=( \frac{d^{2}}{dx^{2}}-(_{X}2\mathrm{p}+cx^{p-1}))y=0$

has irregular singular point at $x=\infty$ . The irregularity of$P$

at $x=\infty$ equal $si_{\infty}(P)=2p+2$ . Put$U_{k}=\{x\in \mathrm{C};|x|>0,$ $\frac{2k-3}{2p+2}\pi<\arg X<\frac{2k+1}{2p+2}\pi\},$$k=0,1,$$\ldots,$$2p+1$.

Then $\{U_{k}, k=0,1, \ldots, 2p+1\}$ forms an open sectorial covering at $x=\infty$ , and put$S_{k}=U_{k}\cap U_{k+1}=\{x\in \mathrm{C};|x|>0,$ $| \arg_{X-\frac{k}{p+1}\pi}|<\frac{\pi}{2p+2}\},k=0,1,$$\ldots,$$2p+1$ ,where $U_{2p+2}=U_{0}$ .
We have now a unique entire solution $y=y_{k}$ of the differential equation (3) with the asymptotic
representation:$y_{k} \sim\omega^{\mathrm{z}\pm\underline{1}}2X-R\pm\underline{1}(k12+\sum^{\infty}B_{2}N(p+1)(\omega^{-})kc)\omega^{k}X^{-})NN-e\mathrm{r}N=1(p+1+_{\iota}(\omega-kx)\mathrm{r}+1$

astends to infinity in any closed subsector of the open sector$N=0,1,$, the quantities
are decided by$x$$\ldots$$U_{k}$. Where$\omega=e\mathrm{r}+" 1\pi i$$B_{N}$$B_{2N(1)}( \mathrm{P}+c)=\frac{1}{8(p+1))^{N}}\frac{1}{k!}\prod_{l=0}(p+C+2\ell(p+1))(p+c+2\ell(p+1)-1)N-1$,

and, for120$B_{l}(c)=0,t\neq 2N(p+1)$.

On the other hand, we have the solution with above properties:,$y_{2k}= \frac{(\frac{2}{p+1})^{\mathrm{A}}2(\frac{\mathrm{c}*1}{p+1}+1)}{\Gamma(\frac{1}{2}(\frac{\mathrm{c}-1}{p+1}+1))}\int_{0}^{+\infty}e^{-x’(}p+_{1})\zeta^{\iota}\zeta+(2\frac{e-1}{\mathrm{P}+1}-1)(\zeta+\frac{2}{\mathrm{p}+1})+1-*(\frac{e\cdot\cdot 1}{\mathrm{p}+1}‘+1)_{d\zeta}$.

$y_{2k+1}= \frac{(-1)^{*}(\prime*\mathrm{e}1)+1(+1\frac{2}{\mathrm{p}+1})^{\delta \mathrm{T}}(11-^{\mathrm{C}-1}\mathrm{r}\urcorner)}{\Gamma(-\frac{1}{2}(\frac{\mathrm{c}+1}{p+1}+1)+1)}\int_{0}^{+\infty}e^{x}+’+1(C\frac{1}{\mathrm{p}+1})\zeta-;(j\frac{+1}{+1}+1)(\zeta+\frac{2}{p+1}):(^{\mathrm{C}-}\mathrm{p}\urcorner^{1}T-1)d\zeta$

Then each2.3$y_{k}$

gives subdoninant solution in the sector$S_{k}$.

Coefficient and its asymptotic estimation of the solution

By using above representations of solution, we can.
way:
$U_{2+1}p$
Put 1-cocycle about} by$\mathrm{t}U0,$$U_{1},$$\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{C}s\mathrm{e}$a basis of$H^{1}(S^{1},\mathcal{K}e\mathrm{r}(P:A\mathrm{o}))$$\ldots,$$\{u_{i,j}^{(k)}, (i,j)=(\mathrm{O}, 1), (1,2), \ldots,(2p,2p+1), (2p+1,0)\}$$y_{j}$$x\in S_{j}$$j=k$ ,

$x\in S_{j}$$j\neq k$$u_{i,j}^{(k\rangle}=\{$$0$

for $k=0,1$ , ldots, $2p+1$ .
In this situation, the pair of the cohomology class of$H^{1}(S^{1}, \kappa e’(P : A\mathrm{o}))$$\{u_{\mathrm{j},j+1}^{(0)}\},$,,$\{u_{j,j+}^{(1)}\}2p+1$$\{u_{j1}^{(1)}\mathrm{j}+1\},$$\ldots,$.

forms a basis of

By vanishing theorem in as ymptotic analysis, we have -cochain
such that$u_{i,i}^{(k)(}\mathrm{j}=-vk)(+vjk),$$(i,j)=(0,1),$ $(1,2),$$(2p, 2p+1),$ $(2p+1,0)$ ,$\{v_{0^{k}’ 1}^{()}v^{(k)}, \ldots , v_{2p}^{(k)}\}+1’ k=0,1,$

$0$

$1$

with following$.,$ $.,$ $2p+$$\ldots,$

where eachpower-series$v_{j}^{(k)}$are defined in$U_{j}$, for $j=0,1,$$\hat{v}^{(k)}(x)=\sum_{\Gamma 0^{v}}^{\infty}=(2p+1, c, k, \gamma)X^{-r},$$\ldots,$$2p+1$ and as ymptotically developable to formal$k=0,1,$$\ldots,$$2p+1$ , respectively. Then

$Pu_{i}^{(k)}j=-Pv_{i}^{(k)}+Pv_{j}^{(k)}$In$S_{i}$.

, we have$Pv_{ij}^{(k)}=Pv^{(k)}$.

Put$Pv_{i}^{(k)}$$x\in U_{i}$$Pv_{j}^{(k)}$$x\in U_{j}$$g^{(k)}(x)=\{$.

Then$g^{(k)}(_{X)}$$x\in \mathrm{C}$$f^{(k)}(_{X})=\{$$\lim_{xarrow\infty}g^{()}(kx)$$s=\infty$

121

define holomorphic function, and$P\hat{v}_{i}((k)x)=f(\mathrm{k})(x)$.

Hence, by the vanishing theorem due to [3] in asymptotic analysis,
basis of.
Moreover, we can have estimates for the coefficients $v(2p+1, c, k, t)$ .

$([\hat{v}^{(0})],$$[\hat{v}^{(1)}],$$[\hat{v}^{(1})]2_{\mathrm{P}+}\rangle$$\ldots,$

forms a$\mathrm{K}\mathrm{e}\mathrm{r}(_{dae}^{d^{2}}\neg-(x^{2p}+CX^{p-1});\hat{\mathcal{O}}/O)$$v(2p+2, C,2k+1,’)$$=$$(-1)^{\frac{r}{p+1}++}1 \frac{1}{2\pi i}\Gamma(\frac{f}{p+1})(\frac{2}{p+1})-\mathrm{p}1(\frac{1}{p+1})^{\frac{\prime}{p+1}}+1$$\cross\frac{\Gamma(-\frac{1}{2}(\frac{\mathrm{c}+1}{p+1}-1))\Gamma(\frac{-r+1}{\mathrm{p}+1})}{\Gamma(\frac{r}{p+1}-\frac{1}{2}(\frac{c-1}{p+1}-1))}F(\frac{f}{p+1}, -\frac{1}{2}(\frac{c+1}{p+1}-1),$$\frac{f}{p+1}-\frac{1}{2}(\frac{c-1}{p+1}-1);-1)$.

$v(2p+2, C,2k, \gamma)$$=$$\frac{1}{2\pi i}(-1)-?+1^{++-+}\mathrm{r}(1\frac{r}{p+1})(p+1)-\frac{\prime}{\mathrm{P}+1}1(\frac{2}{p+1})$’$\cross\frac{\Gamma(_{2}^{\iota_{(_{p}^{\mathrm{g}}}}-+\lrcorner 1+1))\Gamma(\frac{1\neq r}{p+1}\rangle}{\Gamma(\frac{r}{p+1}+\frac{1}{2}(\frac{\mathrm{c}+1}{p+1}+1))}F(\frac{f}{p+1}, \frac{1}{2}(\frac{c-1}{p+1}+1),$$\frac{f}{p+1}+\frac{1}{2}(\frac{c+1}{p+1}+1);-1)$.

And we can have asymptotic estimates for any sufficiently large number ,$f$$v(2p+2, c, k, ’)$$\sim$$v_{M}(2p+2, C,k, r)$$=$$- \frac{1}{2\pi i}\omega^{kr}\sum_{N=0}^{M}B_{2N(_{P}}1)(+\omega^{-})p+k)(1(Cp+1)-1+p(r-(\mathrm{P}+1)(N+_{2}\mathrm{A}))-1$$\cross\Gamma(\frac{1}{p+1}(f-(p+1)(N+\frac{1}{2})))+O(\mathrm{r}(\frac{1}{p+1}(\gamma-(p+1)(M+\frac{1}{2}))))$,

provided $1\leq M<f$ .

3numerical calculation

We shall confirm that the estimate and the asymptotic estimate for a coefficient are almost the same bynumerical calculation by using symbolic computation system Mathematica.

3.1$3.1.1$

About the operator$\frac{d^{2}}{dx^{2}}-x^{5}$

and $r=200$

Put $m=5$ and $f=200$ .

$P= \frac{d^{2}}{dx^{2}}-x^{m}$

122

When

$k$

is even number, we can have$v(5,2k, 200)$

5.78091954996038444069056453-1.31945716325290011156745834$\cross 10^{103}$$=$$\cross 10^{103}10\cross i$.

If we take $M$ equals 50, we have$v\mathrm{s}\mathrm{o}(5,2k, 200)$

5.780919549960384440690564229$=$$\cross 10^{1\mathrm{o}\mathrm{s}}$$-1.319457163252900111567458271\cross 10^{103}10\cross i$.

IIence we have the module $v(5,2k, 200)$ of $v_{50}(5,2k, 200)$$v(5,2k, 200)/v\mathrm{s}\mathrm{o}(5,2k, 200)=1.000000000000000000000000053$

This shows that $v(5,2k, 200)$ and $v_{50}(5,2k, 200)$ are almost the
When is odd number, we can have

$8\mathrm{a}\mathrm{m}\mathrm{e}$

.

.

$k$

$v(5,2k+1,200)$$=$$-2.572751234767002988887166675\cross 10^{103}$$-5.34237298705158039109245417\cross 10^{103}10\cross i$.

If we take $M$ equals 50, we have$v_{50}(5,2k+1,200)$$=$$-2.57275123476700298888716654\cross 10^{103}$$-5.342372987051580391092453892\cross 10^{103}10\cross i$

Hence we have the module $v(5,2k+1,200)$ of $v_{50}(5,2k+1,200)$$v(5,2k+1,200)/v_{50}(5,2k+1,200)=1.000000000000000000000000053$ .

This shows that $v(5,2k+1,200)$ and $v_{50}(5,2k+1,200)$ are almost the same.

3.1.2$\frac{d^{2}}{dx^{2}}-x^{19}$and $r=400$

Put $m=19$ and $r=400$ .
When is even number, we can have$k$$v(19,2k,4\mathrm{o}\mathrm{o})$$=$$3.453077065783857557918438705\cross 10^{78}$$-7.170388411113402004415825685\cross 10^{78}\cross i$.

If we take $M$ equals 37, we have$v_{37}(19,2k,4\mathrm{o}\mathrm{o})$$=$$3.453077065783893137203967499\cross 10^{78}$$-7.170388411113475885563500981\cross 10^{78}\cross i$

Hence we have the module $v(19,2k, 400)$ of$v_{37}(19,2k, 4\mathrm{o}\mathrm{o})$$v(19,2k,4\mathrm{o}\mathrm{o})/v_{37}(19,2k,4\mathrm{o}\mathrm{o})=0.99999999999998969635346939$

.

.

.

123

and
This shows that
odd number, we can haveWhen$v(19,2k,4\mathrm{o}\mathrm{o})$$v_{37}(19,2k,4\mathrm{o}\mathrm{o})$are almost the same.

$k\mathrm{i}_{8}$$v(19,2k+1,400)$

6.89228977853905057503399197$=$$\cross 10^{78}$$-3.979265358972426994974539153\cross 10^{78}\cross i$.

If we take $M$ equals 37, we have$v_{37}(19,2k+1,400)$$6.892289778539121590751656574\cross 10^{78}$$=$$-3.979265358972467995918249507\cross 10^{78}\cross i$.

Hence we have the module $v(19,2k+1,400)$ of $v_{37}(19,2k+1,400)$$v(19,2k+1,400)/v_{37}(19,2k+1,400)=0.99999999999998969635346939$.

This shows that $v(19,2k+1,400)$ and $v_{37}(19,2k+1,400)$ are almost the same.

3.2
3.2.1

$P= \frac{d^{2}}{dx^{2}}-(x^{2p}+cx^{p-1})$$\frac{d^{2}}{dx^{2}}-(x^{8}+2x^{3})$and $f=100$

Put $p=4$ and $f=100$ .
When is even number, we can have$k$$v(10,2,2k, 100)$$-3.767003417074743939251665848\cross 10^{2}8$$=$$-5.184835397614809175077327119\cross 10^{2}8\cross i$.

If we take $M$ equals 19, we have$v_{19}(10,2,2k, 100)$$-3.767\mathrm{o}\mathrm{o}3420044648200708650666\cross 10^{2}8$$=$$-5.184835401702531706077014179\cross 10^{2}8\cross i$.

Hence we have the module $v(10,2,2k, 100)$ of $v_{19}(10,2,2k, 100)$.

$v(10,2,2k, 100)/v_{19}(10,2,2k, 100)=0.999999999211600327821899299+0.10^{-38}\cross i$

This shows that $v(10,2,2k, 100)$ and $v_{19}(10,2,2k, 100)$ are almost the same.
When is odd number, we can have$k$$v(10,2,2k+1,100)$$=$$-2.358251967861247883193952833\cross 10^{2}9$$+3.24585537247978857874772\tau 395\cross 10^{2}9\cross i$.

If we take $M$ equals 19, we have$v_{19}(10,2,2k+1,100)$$=$$-2.3582519540410331435767629207\cross 10^{2}9$$+3.2458553534578948741094\mathrm{o}\mathrm{o}\mathrm{o}\mathrm{o}1\cross 10^{2}9\cross i$

.

124

Hence we have the module $v(10,2,2k+1,100)$ of $v_{19}(10,2,2k+1,100)$$v(10,2,2k+1,100)/v_{19}(10,2,2k+1,100)=1.000000005860363951330673108$ .

This shows that $v(10,2,2k+1,100)$ and $v_{19}(10,2,2k+1,100)$ are almost the same.

3.2.2$\frac{d^{2}}{dx^{2}}-(x^{40}+10x^{19})$

and $r=500$

Put $p=20$ and $r=500$ .
When is even number, we can have$k$$v(42,10,2k, 500)$$=$-1.033415566729374674610622875-2.358703594391783129486533442$\cross 10^{5}0$$\cross 10^{4}9\cross i$.

If we take $M$ equals 19, we have$v_{19}(42,10,2k, 500)$$-1.033415566623428208513677288\cross 10^{5}0$$=$$-2.358703594149967234273282553\cross 10^{4}9\cross i$.

Hence we have the module $v(42,10,2k, 500)$ of $v_{19}(42,10,2k, 500)$$v(42,10,2k, 500)/v_{19}(42,10,2k, 500)=1.0000000\mathrm{o}\mathrm{o}102520679500807233+0.10-37\cross i$.

This shows that $v(42,10,2k, 500)$ and $v_{19}(42,10,2k, 500)$ are almost the same.
When is odd number, we can have$k$$v(42,10,2k+1,500)$$=$-3.018188293136985920927825471$\cross 10^{5}0$.

$+2.00243709397029281822076316\cross 10^{5}1\cross i$

If we take $M$ equals 19, we have$v_{19}(42,10,2k+1,500)$$=$$-3.018188293136015310277848565\cross 10^{5}0$$+2.0024370939696488601304956088\cross 10^{5}1\cross i$.

Hence we have the module $v(42,10,2k+1,500)$ of $v_{19}(42,10,2k+1,500)$.

$v(42,10,2k+1,500)/v_{19}(42,10,2k+1,500)=1.00000\mathrm{o}\mathrm{o}\mathrm{o}\mathrm{o}000321587176050042+0.10^{-3\mathrm{s}}\cross i$

This shows that $v(42,10,2k+1,500)$ and $v_{19}(42,10,2k+1,500)$ are almost the same.

125

References[1] Inui. T. : Special functions (Tokusyukansuu written in japanese), Iwanami-shoten (1962)[2] Majima. H.: Resurgent Equations and Stokes Multipliyers for Generalized Confluent HypergeometricDifferential Equations of the Second Order, in the Proceedings of Hayashibara Forum’90 International
Symposium on Special Functions, ICM Satellite Conference Proceedings, Springer-Verlag (1991),

pp.222–233.

[3]. H., Howls, C. J. and Olde Daalhuis, A. B.: Vanishing Theorem in Asymptotic Analysis III(to appear in “Structure of Solutions of Differential Equations” edited by T.Kawai and M. Morimoto,World Science, May 1996)$\mathrm{M}\mathrm{a}\mathrm{j}_{\dot{\mathrm{i}}^{\mathrm{m}}}\mathrm{a}$[4] Mizuno. T.: On estimates of coefficients of formal solutions to inhomogeneous linear ordinary differential equations, Master’s Dissertation, Ochanomizu University, (1996), (in japanese).

[5] Sibuya. Y.: Global Theory of a Second Order Linear Differential Equation with a Polynomial
Coefficient,North-Holland Mathematics studies,No 18.