The real zeros of the transcendental function y = ax + becx are characterized, and the results should alleviate the difficulty of determining their existence, location, and number

Hedgley, D. R.

English

NASA Technical Reports Server

A Characterization of the Real Zeros 
of a Particular Transcendental Function 
David R. Hedgley 
MARCH 1979 
https://ntrs.nasa.gov/search.jsp?R=19790010508 2020-03-21T23:16:20+00:00Z
TECH LIBRARY KAFB, NM 
NASA Technical Paper 1420 
A Characterization of the Real Zeros 
of a Particular Transcendental Function 
David R. Hedgley 
Dryderr  Flight Research Cerrter 
Edwards, Califorrria 
National Aeronautics 
and Space Administration 
Scientific  and  Technical 
Information  Office 
1979 
INTRODUCTION 
L i t e r a t u r e  d i s c u s s i n g  t h e  c h a r a c t e r i z a t i o n  o f  t h e  real  zeros  of  t ranscen-  
d e n t a l   f u n c t i o n s  is  consp icuous ly   absen t   ( r e f .  1). A s  a r e s u l t ,  s c i e n t i s t s  and 
eng inee r s  who w i s h  t o  d e t e r m i n e  t h e  z e r o s  o f  s u c h  f u n c t i o n s  are a t  a s e v e r e  
d i sadvan tage  un le s s  t hey  have  some p r io r  knowledge  conce rn ing  the  loca t ion  o f  
t h e  z e r o s .  All t h e  i t e r a t ive  schemes a v a i l a b l e  r e q u i r e  a t  least  one estimate 
i n  o r d e r  t o  i n i t i a t e  t h e  a l g o r i t h m .  I f  t h e  estimate i s  n o t  s u f f i c i e n t l y  c l o s e  
t o  a real  z e r o ,  o r  i f  no rea l  z e r o  e x i s t s ,  t h e  i t e r a t i o n  may d i v e r g e  o r  l e a d  
t o  t h e  "wrong" z e r o   ( r e f .   2 ) .  
This  recur ren t  problem is  t h e  m o t i v a t i o n  f o r  t h i s  p a p e r ,  w h i c h  c h a r a c -  
t e r i z e s  t h e  real  z e r o s   o f   t h e   t r a n s c e n d e n t a l   f u n c t i o n  y = a x  + beCX  (and 
equivalent   forms)   where a,  b y  and c are real  numbers  and e = 2.71828. 
T h i s  t r a n s c e n d e n t a l  f u n c t i o n  w a s  chosen because i t  i s  t h e  s o l u t i o n  o f  many 
f i r s t - o r d e r  d i f f e r e n t i a l  e q u a t i o n s  a n d  s o m e t i m e s  a p p e a r s  i n  t h e  n u m e r i c a l  s o l u -  
t i o n  o f  n o n l i n e a r  d i f f e r e n t i a l  e q u a t i o n s .  T h u s ,  t h i s  p a p e r  s h o u l d  f a c i l i t a t e  
t h e  s o l u t i o n  o f  many everyday problems, as  w e l l  as h a v e  h e u r i s t i c  v a l u e .  
The f o l l o w i n g  d i s c u s s i o n  a d d r e s s e s  t h e  a b o v e  p r o b l e m  w i t h  r e s p e c t  t o  t h i s  
p a r t i c u l a r  t r a n s c e n d e n t a l  f u n c t i o n  by way of  theorems. The theorems  speak   to  
t h e   q u e s t i o n s   o f   t h e   e x i s t e n c e ,   b o u n d s ,   a n d  number of  rea l  zeros .  The answers  
t o  t h e s e  q u e s t i o n s  are p a r t i c u l a r l y  i m p o r t a n t  i n  v i e w  o f  t h e  v a l u e  o f  c o m p u t e r  
r e s o u r c e s ,  s i n c e  t h e y  remove t h e  i n e f f i c i e n c y  i n v o l v e d  i n  s t a r t i n g  a n  i t e r a t i o n  
from a p o o r  i n i t i a l  estimate o r  i n  p u r s u i n g  s o l u t i o n s  t h a t  do n o t  e x i s t .  It  i s  
hoped t h a t  t h i s  d i s c u s s i o n  w i l l  a f f o r d  i n s i g h t  i n t o  o t h e r  t y p e s  o f  t r a n s c e n -  
d e n t a l  f u n c t i o n s  as w e l l .  
DISCUSSION 
Proposition I :  The t r a n s c e n d e n t a l   f u n c t i o n  y = a x  + beCX  has a t  most 
two d i s t i n c t  rea l  ze ros .  
Proof: L e t  x1 and  x2 be two ze ros   o f  y and l e t  x1 < x2. Then,  by 
R o l l e ' s   t h e o r e m ,   t h e r e   e x i s t s  a v a l u e  xt s u c h   t h a t  x1 < xt < x2,  where 
y '   (x t )  = a + cbe  CXt = 0 and   y '   deno tes   t he   de r iva t ive   o f   y .  Now suppose 
t h a t  t h e r e  are more  than two ze ros  of y.  They may be   o rde red  so  t h a t  
x1 < x2 < x3, .. ., 5. From R o l l e ' s   t h e o r e m ,   t h e r e   e x i s t s  xt s u c h   t h a t  
x1 < xt  < x2, x2 < ym < x3,  and so  on,  where  xt # x, and   y ' (x t ) ,  y'(x,) = 0. 
C l e a r l y ,  t h e n ,  a + cbe  CXt = a + cbeCXm = 0, wh ich  impl i e s  t ha t  x t  = x m' 
b cxo C bceCXO 
a Proof: L e t  c > 0 and x. > 0. Then x. = --e and 7x0 = -5 
C 
C bc > 0. Thus, e > -- Since  2 ~ o  > 0, i t  f o l l o w s   t h a t  -7 2a 
TxO bceCXO and 
cxO - 
2 2a 0 
bc 
cx 
" 
bc 
2a > e , o r   I n  -- > -  2 9 o r  
I C 1  
> / x o  I . From t h e  o r i g i n a l  
2a 
cxo > In l i d  I$[; t h a t  is ,  < lxol < b equat ion,   lx0 I = I$/ e I C 1  
Consider now t h e   c a s e   i n   w h i c h  c > 0 and x. < 0 such   tha t   cxo  < 0. 
-cxo 
Then w e  may write -axo + be  = 0 where i n   t h i s  form > 0 and  c > 0. xO - 
b -cxo b  b -cxo < Thus  x = -e where - > 0 and  lxol = l--le 1:l . Fur the r ,  O a  a 
b -cxo 
x. = ae i m p l i e s  t h a t  
C l e a r l y ,  e > 2 e  
-cxO -cxO and t h a t  2 / x o  I = 2e  
I cxo I 2 I b / I  a'x O 1  + I cx O /  
Simpl i fy ing ,   /xo  I > 
o r  e > 2 and 
I n  2 . That is ,  
Now l e t  c < 0 and -cxO x. > 0. Then w e  may write axo + be  = 0 where 
b -cxO where -- > 0 and b x > 0 and c > 0 i n   t h i s  form.  Thus,  x0 =--e 
0 a a 
] x o  I = 1$1 - < [:I. For the same reasons as  i n  t h e  c a s e  w h e r e  1 
C X t  
x. < 0 
P -
and c > 0, i t  fo l lows   tha t   lxo l  > I n  2 
I n   t h e   f i n a l   c a s e ,  c < 0, and x < 0. Again, we may write 0 
cxO b cxO 
-axo + be = 0 where x. > 0 and c > 0 i n   t h i s  form.  Thus, x. = ;e 
2 
where -- > 0. A s  i n   t h e  case where c > 0 and > 0, i t  f o l l o w s  t h a t  b a xO 
Theorem 2: I f   t h e   f u n c t i o n  y = ax + be   has  two real  ze ros ,   bo th  cx 
Proof: L e t  c > 0 and l e t  x1 and  x2 be two zeros  where x1 < x2. 
cx2 cx 1 
Then  ax, + be = 0 and ax, + be = 0. From Rol le ' s   theorem,   there  exis t  
L 
x s u c h   t h a t  x1 < xt  < x2 a:d y ' ( x t )  = 0; t h a t  i s ,  a + bceCXt = 0, o r  
a = -bce . S u b s t i t u t i n g   i n  a x 2  + be = 0 y i e l d s  (-cbe t, x2 + be = 0,  
t 
cx t cx2 cx2 
c (X2-Xt) c (x l -x t )  
cx 
e 
O r  x2 - 
- , which i m p l i e s  t h a t  x > 0. S i m i l a r l y ,  x1 - e 
C 2 C 
and x1 > 0. Thus, x2 and x1 have   t he   s ign  of c y  and  hence cxl > 0 
and  cx > 0.  From t h e   x p r e s s i o n s   f o r  x1 and x a n d   t h e   f a c t   t h a t  2 2 
From 1 1 x1 < xt  < x2,  i t  fo l lows   t ha t   Ix l  I < - and  tha t   Ix2  I > - 
I C 1  I C Y  
b 1 1 theorem 1, i t  f o l l o w s  t h a t  
Fu r the r  , x1 < x2 implies t h a t  I x21 > I x1 I .  
/ a 1  < l X l l  < IC/ and t h a t  - < / x 2  
I C  I 
2 I n  
I <"-- 
I C  
I n   t he   ca se   where  c < 0, t h e  same argument   l eads   to   the   fo l lowing  
expres s ions  : 
and 
where x1 < xt  < x2. We may rewrite t h e   e x p r e s s i o n s  as fo l lows:  
"c (X1-Xt) 
e - 
x1 - -C 
2a 
I 
3 
and 
“c (X2-Xt> 
e 
2 “c x =  
c (Xt-X1> c (Xt-X2> 
where c > 0. Thus, x1 - 
e e - and  x2 - - . Th 
“c  “c 
si  
and 
From theorem 1, 
es .ons 
a n d   t h e   c o n d i t i o n s   t h a t  x1 < 0 and  x2 < 0 imply  tha t   Ix l  I > ] x 2  1 ,  
Theorem 3: L e t  abc < 0. Then -- > e i f  a n d   o n l y   i f   t h e   f u n c t i o n  a bc 
y = a x  + beCX  has two d i s t i n c t  z e r o s .  
Proof: L e t  y have two d i s t i n c t   z e r o s ,  x1 and  x2.  Then,  by  theorem 2 ,  
one  zero l i e s  i n   t h e   i n t e r v a l  (0, :). Also, [y(O)-! [y(+,] < 0, b e c a u s e   i f  
t h i s  were n o t  t h e  c a s e ,  x l ,  a ze ro ,  would a l so  be  an  extremum  such  that  
y ’ ( ~ l )  = 0 .  But  by Rolle’s t h e o r e m  t h e r e  e x i s t s  a n o t h e r  p o i n t ,  
X s u c h   t h a t  x < x < x and   t ha t  
t ’  1 t 2  
f o r   t h i s   p a r t i c u l a r   f u n c t i o n .  Hence 
than  zero.   That i s ,  - + e < 0 o r  -- a a bc  bc > e. 
12 I n  
I C  I Conversely,  l e t  - 
-a > e. Then 0 < I f i  < 
bc 
. Observe   tha t  
. But t h e  s i g n  of i s  t h a t  of [y(O,] [y(:)]. 
b2 
4 
b(: + be)  a 
bc That is, b2 
= - + e < 0 by  hypothesis .   Thus,   by  ' the   intermediate  
I ., 
va lue   t heo rem,   t he re  exists a t  least o n e  z e r o  i n  t h e  i n t e r v a l  (0, $). I n  
a d d i t i o n ,  
2 I n  - -2a [.($)I [y ( bc i] = ( z  + be)[?(ln -2a bc  + ")3 bc  
But t h e  s i g n  o f  t h i s  p r o d u c t  d e p e n d s  o n  t h e  s i g n  of (: + be)($) s i n c e  
I n  - -2a  2a bc  bc + - < 0. But (: + be)($) h a s  t h e  same s i g n  as 
e q u a l s  1 + -e. From t h i s   h y p o t h e s i s ,  i t  f o l l o w s   t h a t  1 +-e > 0 and bc  bc a a 
is a z e r o  i n  
two. 
- Y (2 1 3 1  < 0. Thus, by the  in t e rmed ia t e  va lue  theo rem,  the re  
2 I n  - - 2a 
t h e   i n t e r v a l  ($, "). By p r o p o s i t i o n  1, t h e r e  are  e x a c t l y  
Theorem 4: L e t  abc < 0. Then - = e i f  a n d   o n l y   i f   t h e r e  i s  e x a c t l y  -a bc 
one   ze ro   fo r  y = a x  + b e  . c x  
Proof: L e t  - = e. Then  observe  that   y(+)  = + be.   But   the  xpres-  
s i o n   b e  = - i m p l i e s   t h a t  = 0. That is, - i s  a z e r o   f o r  y ,  and 
from theorem 3 i t  is t h e  o n l y  z e r o .  
-a 
bc 
-a 
( l )  
1 
C yc C 
Conversely,  l e t  x. b e   t h e   o n l y   z e r o   f o r   y .  Then  axo + beCXo = 0 
and cxo = -- bcecxo. But t h i s  s a y s  t h a t  c x o  > 0, and  from  theorem 1, a 
-2a 
1x01 < . That i s ,  x. b e l o n g s   t o   t h ei t e r v a l  
But [ y(0)l  [ y f ~~~ I;?)] - 2zb [In -2a bc + - bc 'a] > 0. Thus, t h i s   i m p l i e s   t h a t  x. 
-a I n  - 
i s  a n  extremum,  and  hence  y'(xo) = a + beCXo = 0 and x. = - bc  
C 
. Upon 
s u b s t i t u t i o n ,  w e  have 
5 
I 
a I n  - -a bc I n  - + be C ( ;') = 2 I n  2 + b ( 2 )  c bc 
= :(In -a - 9 
= o  
This  i m p l i e s  t h a t  I n  - = 1, which   imp l i e s   t ha t  - - e.  -a -a bc  bc 
Theorem 5: L e t  abc < 0. Then -- a bc < e i f  a n d   o n l y   i f   t h e r e  are no 
z e r o s   f o r  y = ax  + be . cx  
Proof: The proof   of   this   theorem  fol lows  immediately  f rom  theorems 3 
and 4 .  
Theorem 6 :  abc  > 0 i f  and   on ly   i f  y = a x  + be   has   exac t ly   one   ze ro ,  cx 
xo,   such  that   cxo < 0. 
Proof: L e t  y have   xac t ly   one   ze ro   such   t ha t   cxo  < 0. Then 
-b cxO -bc cxO 
x. = -e a o r   c x o  =  e . But cxo < 0 i m p l i e s   t h a t   a b c  > 0. a 
Conversely,  l e t  abc > 0. Then -f(x)]  = c x  + -eCX, which is  z e r o   i f  C bc 
a a 
and o n l y   i f   y ( x )  = 0. I f  v = cx ,   then   g(v)  = v + y e  . Since i t  is g iven  bc v 
t h a t  - > 0, any   va lue   o f  v t h a t  would s a t i s f y   g ( v )  = 0 must  be less than  
zero.   That  i s ,  vo = cxo < 0 i f   s u c h  a v e x i s t s .   I f   v o  is a z e r o   f o r  g, 
then,  from  theorem 1, Ivo I < . From theorem 2, t h e r e  is a t  most  one  zero. 
Observe  tha t  
bc 
a 
2 -  -bc 
Thus, by the   i n t e rmed ia t e   va lue   t heo rem,   t he re   ex i s t s  a z e r o   f o r  g(v)  and 
hence   for   y (x) .  
6 
EXAMPLES 
The theorems above address a l l  t r anscenden ta l  func t ions  wh ich  can  be  
e x p r e s s e d  i n  t h e  f o r m  y = ax + becx.  The fo l lowing  examples  shou ld  c l a r i fy  
the  app l i ca t ion  o f  t hese  theo rems .  
ExampZe I: F i n d   t h e   i n t e r v a l s   o f   t h e   z e r o s   o f  y = -6x + 3e2x. 
SoZution: From theorem 5, t h i s  f u n c t i o n  h a s  no  zero.  
ExampZe 2 :  F i n d   t h e   i n t e r v a l s   o f   t h e   z e r o s   o f  y = -4x + e . X 
Solution: From theorem  3, y h a s  two z e r o s .   I f  y has  two ze ros  , t hen  
from  theorem 2 they  must  be  posit ive.   Also  from  theorem 2,  t h e  f i r s t  z e r o  l i es  
i n  t h e  i n t e r v a l  (-$ 1) and  the  second l i e s  i n   t h e   i n t e r v a l   ( 1 ,   4 . 1 6 ) .  
lfxampze 3 :  F i n d   t h e   i n t e r v a l   c o n t a i n i n g   t h e   z e r o s   f o r  y = 4x - 2e . -2x 
Solution: y has  a s ingle   zero   ( f rom  theorem 6) ,  and i t s  s i g n  i s  p o s i t i v e .  
From theorem 1, the   ze ro  i s  i n  t h e  i n t e r v a l  ( i, v), which  equals  (+, 0.12).  
ALTERNATE FORMS 
Example 4:  Show t h a t   t h e   z e r o s   o f  y = aert + bteSt  (where r ,  t ,  
and s are a r b i t r a r y  real  numbers) are  t h e   z e r o s   o f  a f u n c t i o n  g = b t  + ae . d t  
SoZution: I f  y h a s  a zero  to, w e  can write ae = -btoestO.  Dividing 
st ( r - s )  t 
b o t h   s i d e s  by e O, ae = -bt, o r  aedto + b t O  = 0. That  i s ,  t o  i s  
a zero  of  g = aedt + b t  and  hence 0 ;  y. 
Example 5 :  Show t h a t  by a s u i t a b l e   t r a n s f o r m a t i o n ,   t h e   f u n c t i o n  
y = a x  + be + d can  be  reduced t o  a f u n c t i o n  g = t u  + re a n d   t h a t   h u s  
the   ze ros   o f  g l e a d   d i r e c t l y   t o  t hose   o f   y .  
Solution: S e t  u = c x  + - so  t h a t  
c x  U 
cd 
a 
which   equals   g (u)  , o r  
7 
. .  . 
o r  
cd 
au  a U- - 
g(U) = - + be 
C 
= t u  + re u 
cd 
a 
-- 
where r = be and t = -. a 
C 
L e t  uo b e  a zero   for   g ;   then ,  by using  the  above  theorems  on  g,  
w e  can   l oca t e   uo .   Thus ,   i f  uo is  a zero,   then 
uO 
and - - - c a  i s  a zero  for  y .  But  x. = - uo - - d.  Thus, by t h e  i n v e r s e  t r a n s -  c a  
formation,   one  can  f ind x. d i rec t ly   f rom  uo .  
Example 6:  Show t h a t   t h e   f u n c t i o n  y = a x  + b I n   ( c x  + d)  + p (where p 
is a n  a r b i t r a r y  real number)  can be reduced to  a func t ion  g (x )  = c x  + keSX + d 
P -- 
where k = -e and s = -- whose z e r o s  are those   o f   y .  
Solut ion:  L e t  x. be  a z e r o   f o r   y .  Then 
axo + b I n  ( c x o  + d) + p = 0 
o r  
o r  
-axe - 
b = I n  ( c x o  + d) 
-axo-p 
b 
e = cxo + d 
and 
e = CXO + d 
Hence , 
P a  -- -"x0 
cxo - e e + d = O  
8 
or 
cxO + keSXo + d = 0 
Thus, a  zero for g is  one for y. 
Dryden FZight Research Center 
NationaZ Aeronautics and Space Administration 
Edwards, CaZif.,  October 25, 1978 
REFERENCES 
1. Scarborough, J. B.: Numerical Mathematical Analysis. Fifth ed. The 
Johns  Hopkins Press, 1962, pp. 192-222. 
2. Burington, Richard Stevens: Handbook of Mathematical Tables and Formulas. 
Fifth ed. McGraw-Hill Book Co.,  c.1973, pp. 186-187. 
9 
~~ 
1. Report No. I NASA TP-1420 2. Government Accession No. 
" 
"~p[3Recipient's Cata log No. 
. . " . - . 
" . . 
A CHARACTERIZATION OF THE REAL ZEROS OF A 
PARTICULAR  TRANSCENDENTAL  FUNCTION 
-1- "~
5. Report Date 
6. Performing Organization Code 
~ - - .. ." ~ " - .~. ." 
7. Author(s1 8. Performing Organlzation Report NO. 
David  R.  Hedgley 
H-1065 
~ 10.  Work  Unit No. 
.. ~ "~ 
"= - - " ~ i- 
9. Performing Organization Name and Address 
NASA  Dryden  Flight  Research  Center l l .  Contract or Grant No. 
P.0. Box  273 
Edwards,  CA 93523 
~. .. . . . ".. . ~ .. . _- ~ . . ~  "_ 
12. Sponsoring Agency Name and Address Technical  Paper 
National  Aeronautics  and  Space  Administration 
Washington, DC 20546 
- 
14. Sponsoring Agency Code 
~ ~ = -  . 
15. Supplementary Notes 
.~ 
16. Abstract 
~~ ~~~~ . ~~ .  ~ " 
The  real  zeros  of  the  transcendental  function y = ax + beCX are  characterized, 
and  the  results  should  alleviate  the  difficulty  of  determining  their  existence, 
17.  Key Words (Suggested by Author(s1) 
Numerical  analysis 
~___- ~" . . "~ 
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Unclassified - Unlimited 
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A characterization of the real zeros of a particular transcendental function

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