Solution to Problem 79-15: An identity

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Solution to Problem 79-15: An identity

Citation for published version (APA):

Lossers, O. P. (1980). Solution to Problem 79-15: An identity. SIAM Review, 22(3), 372-373. https://doi.org/10.1137/1022068

DOI:

10.1137/1022068

Document status and date: Published: 01/01/1980

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372 PROBLEMS AND SOLUTIONS

When 0 is small,

M

is a slowly-varyingfunction of v, its Fourier harmonics are small, and theseries for

Pn

converges rapidly.

We

expand

M

in a powerseries in 0"

-0

sin22

cos2u -sin2

1/20-sin

2

(0

sin

v)

sin[1/20(1-sin

v)]

sin

[1/20(1

+sin

v)].

Letv 90

-

2 w, p 0sin2w, q 0 cos2w.Then,

sin22w sin

210

2 4pqsin

10

2

M

2

sinpsinq 02sinp sin

q"

NOW, SO sinpsinq 1- sin4 w+. 1

cos4

w "+-" pc/

--02

1--(3

+cos 4w)+.

,

2sinl

_[

02 _]

M 0 1

+

_--(3

-cos

2v)+ 0(0

4)

Through 02there arethereforeno termswith2k

>

2,while the coefficientof

_J2

is 4

f

,n-/2 02 02

rJ0 48

cOS2

2v dv

48’

andourapproximationis

Pn(cos 0)

g sin

_{Jo(x)---J2(x).}

The functions

J0

and

-J2

areroughlyinphasewitheachother,andtheybothhave peaks thatdecrease withincreasing x,or n. Thisproves the conjecture so long as 0 is smallenoughfortheapproximationtobevalid.Butcalculationshowsthat,at 0 45

,

the two Besseltermsgive about 0.1% error inthe firstfewpeaks and in thepeaks at n=80.5 and n =4000.5. The hypergeometric formula says successive peaks are

r/0

=4 apartin n, so theiramplitude ratio is

(n/(n

+4))

1/2.

Forthis to be assmallas eventwice our0.1%, n mustbeover 1000and the "large

n"

proofis applicable.

We

conclude theconjectureis trueinthe range 0

<

0

<

45

,

andthereforefor all 0.

REFERENCE

[1] M.ABRAMOWITZANDI.STEGUN, HandbookofMathematical Functions,U.S. GovernmentPrinting Office, NationalBureau of Standards, 1965.

An

Identity

Problem 79-15, by

J.

D. LOVE(AustralianNationalUniversity, Canberra,Australia).

In

ananalysis oftheelectrostaticpotential fortwocharged dielectric spheres,the following identityarises’

n-1

csch x P(cosh x)Q,,(cosh

x)+

O,(cosh

x)

_Z

Pm(cosh

x)

e m=0

+P(coshx)

Om

(cosh

x e

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PROBLEMS AND SOLUTIONS 373 where Pn(cosh

x)

and Qn(cosh

x)

are modified Legendre functions of the first and second kinds, respectively.Provethe identity forrealx

>

0and nonnegative integersn.

Solution by O. P. LOSSERS (Eindhoven University of Technology, Eindhoven, The

Netherlands).

Letthe right-handside oftheidentitybedenoted by$.Thenbyuseoftheintegral

[1,

p.

₁₅₄₃

1

f_

P(t)Pm(t)

dt, n,<_m,

P, (cosh x)O, (cosh

x)

cosh x Scanbe reducedto 1

f]

,.(t)

(

)

S=-e

,

Pm(t)

e dr.

lCOShx-t

m=0

The infiniteseriescanbe evaluated by means of the generatingfunction forLegendre polynomials

[1,

p.

154],

viz., -2x 1/2

_2-1/2

1/2x -1/2 P,,(t)e

"x=(1-2e

+e

)-

e

(coshx-t)

m=0 thus leadingto S 2-3/2e(n+l/2)x

If

)3/2

dt. Pn(t) (coshx Becauseof[2,p. 822,7.225

(4)],

Pn(t)

24-

-(2/1)(x/2) (coshx t)1/2dt 2n

+

1 e

Itfollows bydifferentiation with respectto x that

f

Pn(t)t)3/2

at

2/-

-(n+l/2)x

e.

(coshx sinh x

which showsthat

S csch x.

Also solved by H. E. FETTIS (MountainView,California),whoused an induction argument andby D.

K. Ross (La

TrobeUniversity, Australia), who usedrecurrence relations satisfiedby the Legendre functions.

REFERENCES

[1] A. ERDtLYI, W. MAGNUS, F. OBERHETTINGER AND F. G. TRICOMI, Higher Transcendental

Functions, Vol.I,McGraw-Hill,NewYork, 1953.

[2] I.S. GRADSTEYNANDI.M.RYZHIK,Tables_ofIntegrals,etc., AcademicPress, NewYork,1965.

ERRATUM:

An

Integral Inequality

Problem 77-6 by J.E. WILKINS, JR.

Theerratumgiven for thisproblem (this Review,

22(1980)

p. 102) is in error. It

Solution to Problem 79-15: An identity