Mathematical problem page - Journal of Chemical Education (ACS

Educ. , 1935, 12 (8), p 392. DOI: 10.1021/ed012p392. Publication Date: August 1935. Note: In lieu of an abstract, this is the article's first page. Cl...
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MATHEMATICAL PROBLEM PAGE Directed by EDWARD L. HAENISCH Montana State College, Bozeman, Montana 2.

PROBLEMS

k t a = 8 = x.

From (2) we have:

O S ~ sina% - (1 - COS%)

COS2z = C

1. Hermitian polynomials are often defined by the equation : n(n - 1 ) ( 2 ~ ) " - ~ Hn(x) = (2.x)" 1

= cosZx

'/dcos 2x

+ 1) = COS?~

since, sin%

Show that this definition is equivalent to the one given in equatiou (119), p. 383. Obtain the value of

+ +easar = 1.

for sin 0 = sin 2 s = 0. I n a similar fashion s i n 4 = 1/$(1 - cos 2%)and the inte-

Rely)

--a,--,.

2. Use the above definition of the Hermitian polynomial to prove equations (120) and (121), p. 383. 3. Show that +.(x) = e"'IZ. H&), for n = 1,isan actual solution of the S. equation used at the start of the article:

gral also evaluates to

and

r. 2

sin mx cos nrdx =

f *

sin (m

+ n)&x +

This demonstrates that the form determined from the arbitrary, = eP'/2 # ( x ) , actually satisfies the original A cos (m - 4% Z(m - n) equation. =-1+1-1+1=0. 4. Trace the curve corresponding to &, i. e., y = The other integrals can he handle? hy a like pmedure. 6 = e4'/' (he- 2). (Hint: Refer to the Problem .. Page for May in which "curve tracing" is discussed.) mw*. mrx = 2A; sin - ; + A = -2Ai sin By what factor does the curve you have obtained differ 4. d d from the one for n = 2 in Figure 20, p. 384? Wby was mwr = 4A' sin' this factor introduced? d 5. Evaluate the integral, A' e4'&, by the expansion of the function into a seqies and the subsequent integration of each term. 6. Find the center of gravity of a Bemisphere whose density varies as the distance from the base.

+

1.

SOLUTrONS TO PRORLEMS IN JULY x (a) Let ar = B = - From (I),* sin 2' X

X

+ s i n - cos2 2

4As 2

5.

X X M S ~ ~ X = ~ S ~ - C O S -

2

(1

+ sinx 2lC

- cosa 2E)

= iuAeh.* - iaJlc-imz

_

- a2Bc-im~

= -a%Ash~

dz'

= 1, we have cos x = cos*

or. 2 cosx

*

!?

X

case -X2

~f

dr

2

(h) Let ar = 8 = - From (2),cm 2' Since

4A' If - d = 1, then 2A = 2 + = Aeim + ~ ~ - i o =

= - d.

ISSUE

4% d

= -a2(Aeiez + BL-im%)

Substituting in the difierential equation: Bk-iw) 8r'p E(A& + Be-i= ) = 0 -af(Aei.x

X

-2

+

+

: = 1 + COS x 2

( c ) I f x = a + B a n d y = a - B,wehavea='/r(x+y)and = '/& - y). From (1) sin x = sin l/*(x y) cos '/l(x - y) cos '/dz y) sin I/&-y)

a

sin y = sin 1/.(x sin x

+

+

+ Y)cos ' / d r - y) - cos '/dx

+ sin y = 2-sin '/& + y) cos

+

+ Y)sin V ~ X - Y )

dN N

6. d~ = - y ~ d t ; - = --,at;

- y)

In N = -yt

+c

N when t = 0. N = NOor C = In NO. Thus In - = -yt Ne

* This number refers to a formula given on the July page. 392

393 N = Nae-TI; d N = -yNoe-%t. Using the formula for average life.

A table of definite integrals gives, F o r n = l a n d z = t , r = - 7 . - 1= - 1

r2

Y'