On the computation of zeros of entire functions by means of series in problems of mechanics

ON THE COMPUTATION OF ZEROS OF ENTIRE FUNCTIONS BY MEANS OF SERIESIN PROBLEMS OF MECHANICS PMM Vol. 31, No. 3, 1967, pp. 489-499 V.A. BORODIN (Moscow)

(Xeceived Analytical

expressions

functions of fluid

flows,

breakup

the zeros

A considerable instances

particular in terms tive

amount

of hypergeometric of relevant again

iently

type.

general

In the present of entire

for practical

some

e.a.1.

function

functions.

In more recent is given

are not in a form which

paper

an attempt (including

application.

is made

Several

easily

[l

author’s

by elementary

only

for some

polynomials

[2] an exhaus-

own results to functions means,

of any degree)

from mechanics

by

to 121, but in

of algebraic

be applied

polynomials

examples

exists

or they are derived

with

as stability

of its parameters.

work of Belardinelli

to obtain,

algebraic

of the obtained

problem for zeros

together

could

(such

can be obtained

in terms

are not derived,

in [1] ex p ressions

functions

series

with this

transcendental

of mechanics

expressions

obtains

investigations

application

These

dealing

expressions

and of entire

problems

into a power

of literature

usable

Lakhtin

unfortunately,

practical

of fluid currents

either

polynomials

in solving

of a given

function.

review

zeros

of algebraic

find many applications

expanding most

for xeros

June 27, 1966)

which, of suffic-

expansions

of

in a form suitable

and numerical

methods

show

series.

1. Let where

a = z - 5 is an entire

one of the zeros finite

of f(z).

By the definition

of entire

function,

series

(1.1) converges

(1.1) near for any

z and [.

We know (Hurwitz some

f (2) = PO + P,U + P2U2 + **** function of I, expanded in a series at the point g lying

region

[13] ) that if ft(z),

D bounded

f(r) f 0, then the point of a set

of zeros

Therefore, at any finite

z.

lying

to the zeros represent

=

(a0

of functions

analytic

+ f(z) uniformly

in

fn D and point

function

and the series

(1.1) representing

of zeros

of partial

stuns

it converges of the expsnsfon

of f(z).

a zero of f(z) by an expansion =

of po. Let us introduce U’

if j,(z)

D is a zero of f(z) if and only if it is a limit

then the sequences

u in powers

contour,

f,(z).

if f(z) is an entire of a plane,

. . . is a sequence

fi(f),

closed

within

of the function

point

(1.1) converge We shall

by a simple

+

alp0

+

a,

+

U$o

+-a&JIpt

+

...

(1.2)

the notation %Poa

+

**Y

= 50

+

c,lpo

+

C&o”

+

...

(1.3)

Here [ 141 c so =

Q,

c,,

=

-m’n,

i

(si +

i=l

513

i -

m) aiTsm_,

for

m),

1

(1.4)

514

V.A.

It can easily

be shown

Borodin

that

*yf+“j+-+=k)

Grn Summation natural

ia performed-hek

over’all

possible

partitions

...

/f/j

of m into equal

&1.5)

or unequal

component6

iai + jaj + . . . + ka, = m Inserting

(1.2) into (1.1) and taking

(1.3) into account,

(1.6)

we obtain

j$‘k&.‘pQf =’ Equating eyetern

to zero the coefficients

of equation6

of consecutive

for the coefficients

powers

1 + 2 p,&l

we obtain

&$kl L=l

=

o

(1.8)

(t = 2, 3, . . .)

= 0

(1.9)

Eq. of (1.8) hae one zero root a,, = 0. Let ue find the coefficient8

ponding

to thin root. When u. = 0, Formula cl,;

= 2

t! al! aa!

Taking

into account

a0 =

0,

=

(1.10)

(1.10)

al+aa+a3+...+ap=t Ct, k” =

0,

=

=s. . . apCP

&za

. . a,!

when

0

for a0 = 0 we obtain,

1 + Plh

of (1.2) correa-

(1.5) becomea

l~aI+2ar+3a3+...+pa,=k, co,k”

a

h=O

6=l

First

of p,, in (1.7),

ok

pkao” = 0,

g

(1.7)

PIas +

0

i

in place Pkck.

(1.11)

t>k of (1.8) and (1.9),

so =

(s =

o

2, 3, . . .)

(1.12)

k-2

defining the coefficiente of (1.2). Defining the coefficients ok from (1.12) a0

0,

=

al

=

-$

we obtain A,p,a’P,a”.

ak =

. . PQaq

(k = 2, 3,

. . .) (1.13)

where

(1.14)

A k= Summation

in (1.13)

or unequal

natural

2)!

a,! a,!...~,!

is performed

2 (kinto equal

(2k -

-

when

kl

over all pdssible

partitiona

r+1 of a natural

number

1) = ra, + sa, + . . . + qa,

components,

under the condition

(1.15)

that

a,+a,+...+a,=k--l A eerie0 (quoted

resembling

by Bazhenov

(1.2) with coefficienta

(1.13),

wae obtained

in [s] ) f or a real root of an algebraic Pk -=qk? Pl

and inserting

(1.16)

it into (1.2) we obtain,

equation.

qO+Z=q

after some

transformations,

by Heegman

[8]

Putting (1.17)

Computation

+ 2 (_I)=‘+1 where

summation

natural

numbers

except

2

of aeros

q”m+lla’qsat

of

. . qpap

in the second

with the number

and use

-

QOdOl+ +

2. Series see

(1.18)

(1.20

-

m+1,

(1.19)

formula,

This

(1.20)

tlo

formula

-~q!15q32hJ

40%4644

in terms

O” (Fit+ 2k-l i)! ,.,t Iis t! (t + n)!

n P-l> =

-qr,6qtJQetJ +

qOsq3q4a76

can be expressed

even

components

by

into a usable

Qo3Q363,3

of consecutive, natural

al + a, + . . . + an = m

of (1.18)

I:

them to expand

u=

aums

O” (2t + 2k)! rlt = fro t! (t + n)! ’

wi, n(q)

(1.18)

4, into m possible

2m = al + sa, + . . . + pa,, the internal

$ja

f=So term is performed over all partitions

2m, beginning

Let us denote

515

entire functions

;qn7qs3a97

is +

-*

of hypergeometric

qo5q5%5

+

(1.21)

**

functions.

We can ea8ily

that

(2.1)

Gk. n (7) =

=$b’(k+l,

62h+l. n (q) =

where

F is a Gauss’

at any values

of k and n which

then be expressed Using

hypergeometric in terms

transformation

function.

given

-i

+ I’ (n +

1) I’ (2k -

41)

series

in the internal

k-

sums

in ([14],

p. 1057),

I’(n-k)l’(n

and can

x

F(-k,

n + 3/a; ‘+)+ -k+;;n-2k

nfl;

r (2k - n + S/r) (1 - 4q)u-*k-a”

2k--n+;; F(-k,

(2.3) + +;

Q-

r (n + I)

of (1.18),

- k + ‘/a) (4q)“+’

n + 1; 2k-

(4tl)n

k-

truncated

we obtain

r(ft+i)r(n--2k-*a/,)

xJ’(k+l,

(2.2)

become

functions.

n + ‘/a) (1 - 4q)n-2’-“r

r tk + 4) r (k +‘/a)

n+l;

I’(n+i)I’(n-2k--/a)

WY yj-

xF(k+l,

+;

Hypergeometric

may be encountered

of algebraic

formulas

QZh, n(rl) =

k+

I-

4?)} (2.4)

‘q)+

-k-;;n-2k-+;

I-4q)}

r (k + 1) r (k + */a) (4tP where r(x) is a gamma function and all hypergeometric functions are expressed in a finite form. Finite expressions for series CJk, . appearing in the first terms of the expansion (1.21) calculated by means of (2.3) and (2.4), are

516

V.A.

Co; = Since

18

‘1’

au algebraic

polynomial

expansions Limiting

Zi(lR~ -

__3rl”-3RMt {

Rorodin

420115 f

25218 -

6rl’ (I-

! is a particular

case

(2.5)

4tj~p

of an entire function,

can be used for determination

of the roots

ourselves

containing

to the term of (1.1)

6OR + 5

of algebraic

the above

equations.

urn r we shall

consider

an m-th order

Eq.

(4 =

f

k$Pkuk =0

(2.6)

On substitution

!?a% = @21 -qdqfl Pq. (2.6)

=

(--j)&

@&?I -**,

qo”$

=

or&, . . . . u

=

-

qoc

(2.7)

becomes f (5)

It is now easy the smallest

=

2 -

to obtain,

5 +

w&”

from (1.181,

+

o&3

+

*.. -j- 0,5”

the following

expression

= 0 (2.81 for a root of (2.8) with

modulus

.I.

In particular, 5 =

for a Snd,

501

for

(02)

3-rd, 4th

and 5th

m = 2,

degree

5 =

5

Eqs.,

$5a*,

we obtain !&?+a@a)

for

m==3

8==0

r = In practice,

5 !$ 1.GO

ble to utilize

are fixed.

i

$

htw5k.

(1.21)

now determine

and (2.10)

for

25+3r+4k+l @z)

m

5

=

. s=o

(2.9)

and (2.10)

at a point which is sufficiently

the sxpausion

3. We shall (2.9)

$

small number of terms in Formulas

that the root is expanded

(1.21),

5 r=0

and Expressions

We can write any of these

can be used,

provided

to it. This makes it possi-

(2.5).

the radii of convergence

when the values

close

of all indices

for internal except

sums crk,.

tbc indices

in (1.18),

of internal

sums,

sums as (3.1)

where,

by our assumption,

k and s are constants.

found using a well known formula

Radius

of convergence

of (3.1)

can be

Computation

(2t

1 -P = This also

yields

the value

that (1.17)

of zeros

,1E

of entire

517

functions

I--2 + /i)! t! (t + s)!

(t + I)! (1 + 1 -i- s)! (2t --t /i)!

p = l/4,

hence

and (2.7) are taken

the sums

listed

above

converge,

if (provided

into account) (3.2)

holds. Let us find the region Using

of convergence

(1.1) we can write

for the considered

the inequality

sum,

on the t-plane.

(3.2) as (3.3)

Now

let us put 2

2f(E)f”(E) = where

m is the degee

Using sums

2

(3.4)

h-0

of a polynomial

(3.4) we can obtain

in expressions

(m-l)

f(z).

from (3.3) an equation

for roots

of an algebraic

of regions

of convergence

of internal

Eq. a (m-1)-k

F(R,

0) =

$

+

a(&pd3

=

2

(dk = 2

0

k=l

It can be assumed

4. Expansion

obtained

lying

general

in radicals series

[

(2.9) converges

within

a region

(2.9) of a root of an algebraic

is of sufficiently soluble

that the series

at the point

form(*),

or in terms

is feasible.

)

(3.5)

4 4

IAj Bi I

Dij = expanded

Dl,s+kRaa+k

s-0

m-th degree

and can be used

of some

We shall

when the polynomial

of convergence

special

equation

to oktain

functions

now consider

some

(1.1)

is

(3.5).

exact

in cases

obtained solutions when

previously of equations

summation

of the

examples.

a) For m = 2 f (2) = By (2.10)

22 + pz +

Q= 0

(4.1)

and (2.5) we have p=bo,1(02)=

i -

rl=qIp27

Or

J$“”

(4.2)

where

c=-_(PIQ)Z,

z=--‘/zP$

Vqz=-Y~

(4.3)

b) For m = 3 f (z) By (2.10)

and (1.20)

28 + pz +

q= 0

(4.4)

we have

where

+$z,

+) Equation which does not include transformed using a displacement

oa=o,

w8=-g

the unknown in the first power (PI = 01, lhoald transformation, before (2.9) can be applied.

be.

V.A. Borodin

518

Substitution

of (4.6)

into (4.5) yields (4.7)

Since

(4.8) Using a tranaformatios

formula (given

$-

ZZZ-_

Applying

to it the summation

in [14],

I)--“‘F($

(@-f-

p. 1057, Formula 3.131~1) we obtain

, ; ; ; ; t= _2___) . +1

formula igiven in [14], p. 1054, Formnla 9.121.4)

(4.9) we have

for n - lf3r Z=-which on substitution

e) Expansion

$[1

i_(P

+ l)L$L-

[--t

of t from (4.8) into it, becomes

+(ta

+ #P]““)

(4.10)

the well known Cardan’s

formula

of a root of p+Zy--~=-_

is of interest,

Applying

(2.9) we obtain (4.13)

A necessary

and sufficient

condition

for (4.13)

or (4.14)

to converge,

is

(4.15) We know that (4.121 has an expansion

due to Meliin 123

co t-

y=f

fWl(f-tal/n)

22r(a+l)r(l~+[i-(n--I)a]/r~)~a

t4.16)

a=0

its condition

of convergence

We see that (4.13) obtain

a solution

supplements

the expansion

obtained

by Mellin and enables

us to

at any value of parameter n.

5. Several

examples

and numerical

methods,

a) In fl5l

being

of application

of obtained

results

to some problems

follow.

8 characteristic

Eq. H9 +

$52HZ t

p,H

-I- PO =

0

of mechanics

Computation

A = I,$ (I-

ewm),

of zeros

of entire

functions

M = @,

D

Mwm3,w

_

519

p&Fe Q

=

PI was

obtained,

where m is a dimensionless

fluids

and W is the Weber number

layer,

V ia velocity

dependence

(pz

and o is surface

of the oscillatory

wave

tension

increment

on the wave number for various In [15] a graphical solution

number,

M is the ration

h is the thickness

is the density,

coefficient).

i.e.

We shall

of the imaginary

(5.11 of densities

of the

of the boundary

try to determine

part of a complex

the

root of (5.11,

values of the Weber number W, was obtained. From (5.11, we have

z.3 +

brz +

Ii=

b, = 0,

Z-V&

(5.21

and - 9plpa .-i_ 27po P2 b. =5Pr3 (5.31 3 ’ 27 It can easily be shown that the second and third root of (5.2) can be bi = 3P1-

Also,

Im&)

expressed

= Im(zl. in terms

of the first (real) root, by za,a = - I/! .x1 F r/2 i f-5

(5.41 (5.51

+ 4bl

Zi = l/x 1/32i’

we can apply one of the Formulas To determine t transformed into ( b’.12) by means of a substitution

(4.13)

or (4,161.

Eqs.

(5.2) can be (5.6)

z = y (&Jlz’~ Subsequent cal method degree

calculation

of solution

confirmed

utilised

the results

by us makes

obtsined

it possible

in flS].

We see

to achieve

that the snalyti-

any predetermined

of accuracy.

b) In [16] a problem

of the breakup

of a current

of viscous

fluid.

Characteristic

equation

has the form ~2+~~[2~~~~)

---

2kl

11 (kc) I,~(zat

ke + 24’II (la)

a _ 1

11 (kn) IskW) _ -

ck (I Pa”

k$

IO (ka) 1%+ ka (5.7)

(12 = k’J+ a/v) where u denotes density

radius

of the fluid,

of the current,

cr is surface

ko is a dimensionless

tension

coefficient,

wave

v is kinematic

number,

p is the

viscosity,

and Q is

the complex frequency of oscillations while Io&a) and ii&o) are Bessel functions of an imaginary argument. Author of [16] states that the equation is too complicated to be solved

by analytical

Let us introduce

methods

m= into (5.7).

and only considers

dimensionless

ka,

Then,

the limiting

cases.

paremeters

z = a a% ‘it ( -ii- i ’

(5.81

(5.9) and Eq. (5.71 assumes

the form

2ma.z

z2

2m jfmz 4 AZ 2m2 + Al

C

= 3

AZ

=

m(’ - m2) 2mZ +

11(m)

Expanding Bessel functions and their derivatives we obtain from (S,lOf, after some traMformations, i

t=o where

(5.10)

AZ ZO(m)

M,c’ = 0

into series

in zeros

of their arguments

(5.111

520

V.A.

6 = mz + .4z,

1czo

Borodin

m4Zo(m) -

=

6m3Z1 (m) -

m (1 -

m?) II (m) 1”

10 (ml M1

m2 (16 -

=

3m?) IO (m) -

810

M

I

= [16(r-l)r(r+l)+8t(~

(1 -- ~12) II (m) A2

2m (8 + 7m*) II (m) -m (ml

l)b+2(3+4t)tn3]

+l)m2+m4]Z~(m)-[88t(t-+

It(m)

_

22’t! (t + I)! lo (m) m(l-m2)ZI(m).~2

-

for 22’1! (t + l)! lo(m) Let us find the oscillation increment for the case solution

of glycerine.

our computation.

In this case

Broken curve is obtained 2 =

which results equation

-

Fig.

+

Fig. Values

of characteristic means of (5.12)

+

of oscillation

increments

the equation

of convergence

2 shows

Let us compute inside

regions

of internal

q5

+

6q4 _t

2.

by means of a more exact times as large as those

gives

solution

obtained

by

much better idea of the length of the

of a circle

-

z5 -

a =

sums in (1.21)

(64a2 +

of convergence

of convergence

differ from

current.

of dissection

9) =

for the case

We see that the curves

obtained

0

(5.13)

is, by (3.5),

defined

by Eq.

39 R 10) -/- 128aR5 c& 56 = 0 of series

the real root of the equation,

the region

(5.8) into a simplified

of maximum increment

one-and-half

of course,

(5.12)

m2):j”’

parameters

fluid (Rayleigh).

are, roughly,

quoted in 1161. This,

F (R, Fig.

m2 (1 -

the position

f (2) = Region

+

Fig.

part of the considered

c) Consider

from

1.

equation

aqueous

= 33.5. Fig. 1 shows the result of

of dimensionless

of ideal

. .)

of flow of a current of a 75%

A 2 = tp

(2

1 also shows

when the current is composed

unbroken

‘$

from the substitution

given in [la].

each other.

we have

(t = 2, 3,

for each of five roots

expanding

of the real root. Then,

f(z)

at the point

in place

(5.14)

of (5.13)

for (I = 2.

5 = 1.2 lying

of (5.13).

we obtain

14.4q3 + 17.28112 + 10.368q + 0.48832 = 0

(5.15)

where q =

z-l.2 on assumption

We shall use the first two terms of (1.21). small enough to be neglected.

We obtain

(5.16) that the remaining

terms are

Computation

of zeros

q(J= 0.047099, u’~ = qoqz = 0.0784976,

of entire functions

qz

q:$ = 1.38889,

1.66667,

E

521

cro,I = 1.093897,

us, 3 = 1.56252

and finally 11 = with high degree

of accuracy.

o = 2, is z = 20.2 = 1.148698

(z) =

Its roots lo-’

zi-

& .!

0.05130andc can easily

(see

d) Let us find the smallest

f

-

This

using

2

1181

any of the numerous computations

qa = -

root.

0.00494670,

q3 = 109.73202, 3.

Using

Graph of f(z)

3, permits

us to choose

Taking

qz = -

(5.19) equation

17.42303

11 = q,qz = 0.0861865

again only the first two terms of (1.21)

obtain o.c(‘r44’4 ‘C C.

a c while [18] gives

methods.

u = L - 0.02 the coefficients of a transformed

we obtain

.3

existing

shown on Fig.

near to the required

-Ii’ I4Y Fig.

(5.17) for

(5.18)

a point sufficiently

0.2

as the real root of (5.13)

e.g. [17] ).

root of the equation

baaed on approximate

0.1

1.14856

9 -:-

can be located

f

=

be checked,

or

L=

we

0.0254454

(5.20)

1 = 0

(5.21)

its value as z = 0.02544604.

e) Let us find a real zero of the function f(z) Approximate

values

of zeros

= cos scoehz

of f(z)

-

are given by [19]

a n = t/a (Zn + where n denotes Expansion

the n-th zero, when the trivial

of f(z)

=

5

(4’

XSS ---1:-o (Zs)!

$5

k-3

on application

series,

becomes f (6)=

of a formula ([14],

$j

p. 18, Fonnol~

(5.23)

s=a

czk6k = 6

p. 29, Formula

(cza =2$$

h=l

or ([14],

(5.22)

at the point z = 0 has the form f (X)

which,

1) 3t

value z = 0 is disregarded.

0.316)

(2f)f((;;Jl)‘tl),

for multiplication

+

&&

of power

(5.24)

t=n

0.153.1

and 0.153.3)

cz,=(--1) Let uz dirregzrd the trivial zolotion the fnzction f(c) into a zeriem

f (b) = 5

‘2 (k+&

B 2zs (4s)I’

b=ti

5 = 0. To compute

= 2

k=o

PI,)Ik,

(5.25) the zero x, S we zhall expand

q={--iZ*4

h=o

W26)

obtaining

Ps = g

f(‘) (an”) = f

i

k (k -

1) . . . (k -

s + 1) cz o,+1)an4 (‘-‘)

(5.n)

k=S To obtain the fimt zero x 1, we zhall again nze fimt two termz of (1.21). The coefficientz now will be

522

V.A.

Q,, = and they

yield

-

7.40712,

rl =

3.1418

the value

Substituting

into

zero of (5.21) f) We shall

Borodin

qz = ‘10-3,

4.2416

uu,i =

of 7 = z4 - at4

it the value

of at

according

as rt = 4.73004, while [I91 quotes find a real zero of the function

quoted

problem

solved

by Eufer

*IO-a,

1.0158,

to (5.22).

we obtain

the first

x1 = 4.73. = 0

who expanded

(5.28)

the zero

into

an infinite

product;

it is

in f2Of.

L.et us use Fig.

was

c~a,a =

= 7.43052.

calculated

1 (z) := J* (:! v’;, This

q3 = 4.16592

‘10-4,

l.Ou3162,

(3.3)

4 shows

to construct

the regions

the regions

of convergence

of convergence of series

F

0.50

straight

line

to the abscissa

ordinate

equal

to ?/, gives

for three

given

zeros

can

parallel

zeros

of (5.28).

three

(5.28)

a.25

for the expansion

for the first

of (5.28).

regions

into

the

of convergence

on the real

be expanded

A

with

axis.

Function

a series

at a zero,

as follows m

i?

20

IO Fig.

j(z)=

4.

of convergence

the region

q,, = -

(Fig.

yields,

&

-0

(5.29)

for the first

zero

us expand

f(t)at

the point

t =

1.4 lying

4). We obtain

qz = -

0.045085370, i =

which

(--1)”

1,z-0 Let

within

2

of (5.28),

q3 = 0.043352828,

0.34104094,

O.O1537395i,

QoQx =

the value

of a t = 1.4457964.

In [20]

it is given

as Q t = 1.445796.

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1.

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I..K.,

Moscow, 2.

Algebraic

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Soluble

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hyperg&omktriques

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algkbriques

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L.M.,

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Translated

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