Power series are aalytic Horia Corea 1 1 Fubii s theorem for double series Theorem 1.1. Let {α m }, be a real sequece idexed by two idices. Assume that the series α m is coverget for all ad C := ( α m ) <. (1.1) The we have that α m coverges for all m ad: ( α m ) = C. (1.) Moreover, Fially, N ( α m ) = >N M ( α m ) = 0. (1.3) m>m ( ) ( ) α m = α m R. (1.4) Proof. We recall a few fudametal results. If a 0 is a oegative sequece, we defie s N = N =0 a to be a icreasig sequece of partial sums. The a = N s N exists ad is fiite if ad oly if the sequece {s N } N 0 is bouded from above. Moreover, if s N coverges the it is Cauchy, hece for all ɛ > 0 there exists N ɛ 0 such that 0 s N+k s N < ɛ for all k 1 ad N N ɛ. This implies: 0 s N+k s N = N+k =N+1 a < ɛ, k 1. Takig the supremum over k we get 0 N+1 a ɛ for every N N ɛ. I other words: N >N If N ad M are fiite atural umbers, the we have: M N α m = N a = 0. (1.5) M α m N α m C. (1.6) =0 I the last two iequalities we employed the assumptio (1.1). Hece 1 IMF, AAU, March 9, 013 M N α m C <, N, M 0. (1.7) 1
I particular, N α m C <, N, m 0. =0 This shows that α m is coverget for all m 0. Now we ca take the it N i (1.7) ad obtai: M α m C <, M 0. m=0 But this shows that the sequece of the partial sums geerated by a m := α m is bouded, hece D := ( α m ) C. Now usig agai the first idetity i (1.6) we have: or N M α m = N M N α m M α m D m=0 M α m D, N, M 0. Our hypothesis guaratees that M M m=0 α m exists ad is fiite, hece: Thus by takig N we get: N α m D, N 0. =0 C = α m D which proves that C = D. Now we have to prove (1.3). Defie β m = α m if > N, ad β m = 0 if 0 N. The we have: β m = β m or α m = ( α m ). >N >N Deotig by a = α m we see that (use (1.5)): >N α m = a 0 whe N. >N I a similar way we ca prove the other it i (1.3). Now we have to prove (1.4). First of all, because α m α m, 0 we have that ( α m) is absolutely coverget. The same holds true for the series i the right had side of (1.4). Thus we oly eed to prove that the two double series are equal. If N ad M are fiite atural umbers we have: M N α m = N M α m, (1.8)
which implies: M N M N α m α m = α m α m, (1.9) m=0 =0 m=0 >N =0 m>m which leads to: M N α m m=0 =0 α m α m + α m. (1.10) >N m>m Now we use (1.3) i (1.10): take both M ad N to ifiity, ad obtai: α m α m 0 which eds the proof. The biomial idetity Theorem.1. Let a, b R ad N. The: (a + b) = k=0 k!( k)! ak b k. Proof. Let P : R R give by P (x) = (x + b). We have that P (x) = (x + b) 1, P (x) = ( 1)(x + b), ad by iductio we ca prove: P (k) (x) = ( 1)... ( k + 1)(x + b) k = ( k)! (x + b) k, 0 k. Moreover, P (k) (x) = 0 if k >. The Taylor formula with remaider provides us with some c betwee 0 ad x such that: P (x) = P (0) + k=1 The fial result is obtaied by replacig x with a. P (k) (0) x k + P (+1) (c) k! ( + 1)! xk = P (0) + 3 The expoetial ad the logarithm For every x R we defie k=1 exp(x) := 1 + x + x + + x + = x, P (k) (0) x k. k! assumig for the momet that it coverges. If x = 0 we have exp(0) = 1. If x 0 cosider the series α give by α = x. Usig the ratio criterio we have: α +1 α = x 0 < 1 whe, + 1 which shows that the series defiig exp(x) coverges absolutely. 3
We wat to prove that the expoetial is everywhere differetiable. Fix a R ad let h R. Defie the fuctio F (h) := (h + a),. (3.11) The Taylor formula with remaider provides us with a c = c,a,h betwee 0 ad h such that F (h) = F (0) + F (0)h + F (c)h /, or: which leads to: Thus if h 0: (h + a) a = a 1 h + ( 1) h (c,a,h + a), (3.1) (h + a) a = a 1 ( 1)! h + (c,a,h + a) h,. (3.13) ( )! exp(h + a) exp(a) h = 1 + a 1 ( 1)! + h (c,a,h + a). ( )! Note first that 1 + a 1 ( 1)! = exp(a). Moreover, sice c,a,h + a a + h we may write: exp(h + a) exp(a) exp(a) h h ( a + h ) h exp( a + h ) h exp( a + 1) =, ( )! which holds for every 0 < h 1. It follows that the expoetial fuctio is differetiable at a ad exp (a) = exp(a). Theorem 3.1. We have that exp( x) exp(x) = 1 ad exp(x) > 0 for all x R. exp(a + b) = exp(a) exp(b) for all a, b R. Defie the logarithm fuctio l(x) := x 1 1 dt, x > 0. t The we have l(exp(x)) = x for all x R, ad exp(l(x)) = x for all x > 0. Moreover, Proof. We kow that exp(0) = 1 ad exp (x) = exp(x) holds o R. Defie the fuctio f(x) = exp( x) exp(x). The f is differetiable ad f(0) = 1, f (x) = 0, x R. Hece f(x) = 1 o R, which proves that exp( x) exp(x) = 1 for all x R. The same idetity shows that exp(x) ca ever be zero. Now sice exp(0) = 1 > 0 ad because exp is cotiuous (beig differetiable), it caot chage sig because it would have to go through a zero (remember the itermediate value theorem). Hece exp(x) > 0 o R. Now defie the fuctio g(x) = exp( x b) exp(x) exp(b) for some fixed b. We agai have g(0) = 1 ad g (x) = 0 for all x R, hece exp( x b) exp(x) exp(b) = 1 o R. Multiply with exp(x + b) o both sides ad obtai exp(x) exp(b) = exp(x + b) o R. The logarithm fuctio is defied to be a primitive of 1/x, i.e.: l (x) = 1, l(1) = 0. x Defie f(x) = l(exp(x)) x o R, which is possible because exp(x) > 0. We have f(0) = 0 ad f (x) = 0 for all x R, hece l(exp(x)) = x o R. If x > 0, cosider the fuctio f(x) = 1 x exp(l(x)). We have that f(1) = 1 ad f (x) = 0 for all x > 0, hece exp(l(x)) = x for all x > 0. We have just proved that the expoetial ad the logarithm are iverses to each other. 4
Corollary 3.. We have l(ab) = l(a) + l(b) for all a, b > 0. Moreover, l(y x ) = x l(y) for all y > 0 ad x R. Thus if y > 0 ad x R, we have y x = exp(x l(y)). Proof. Sice exp(l(ab)) = ab = exp(l(a)) exp(l(b)) = exp(l(a) + l(b)), we must have l(ab) = l(a) + l(b) due to the ijectivity of exp. If ab = 1 we have 0 = l(a) + l(a 1 ), or l(a 1 ) = l(a). Now if a = b we get l(a ) = l(a). By iductio, we obtai that l(a ) = l(a) for all N. Replacig a i the last idetity with b 1/ we obtai l(b 1/ ) = 1 l(b). Thus l(b m ) = m l(b). Moreover, l(b m ) = m l(b). Thus we have just proved that for every ratioal umber r ad for every positive umber y > 0 we have l(y r ) = r l(y). This implies y r = exp(r l(y)) for every ratioal umber r. Fially, we use that every real umber x is the it of a sequece of ratioal umbers, together with the cotiuity of exp. Corollary 3.3. Let α, β, γ > 0. We have that x x α exp(βx) = l(x) x x γ = 0. (3.14) Proof. Let N be a iteger such that α < N. We have the iequality: The: exp(βx) 1 + βx + + βn x N x α N! βn x N, x > 0. N! 0 exp(βx) N! β N 0 whe x. xn α Now if γ > 0 ad x > 0 we have x γ = exp(γ l(x)). Deote by y = l(x). The we have: l(x) y x x γ = y exp(γy) = 0. 4 Power series are aalytic fuctios Let {a } R such that sup a 1/ <. Defie r = 1/{ sup a 1/ } if sup a 1/ > 0 ad r = if sup a 1/ = 0. Let 0 < R < r ad defie f : (x 0 R, x 0 + R) R give by: f(x) := a (x x 0 ). The series is absolutely coverget because sup a (x x 0 ) 1/ = x x0 r < 1. Theorem 4.1. Let b (x 0 R, x 0 + R) be a arbitrary poit. The f is idefiitely differetiable at b, ad for every t (x 0 R, x 0 + R) with t b < R b x 0 we have: f(t) = where the Taylor series is absolutely coverget. f (m) (b) (t b) m, 5
Proof. Deote by α m := ( 1)... ( m + 1)a if m 1. Note that if > k we have: ( ) ( ) l( k) l() + l(1 k/) ( k) 1/ = exp(l[( k) 1/ ]) = exp = exp ad usig (3.14): ( l() exp + l(1 k/) ) exp(0) = 1 whe. It follows that sup α m 1/ = 1, m 1. r Thus the series m α mt m is absolutely coverget for all t < R. Give x such that x x 0 ρ < R < r, there exists some h 0 > 0 such that x + h x 0 (R + ρ)/ < R for all h h 0. Usig (3.1) with a = x x 0 ad h h 0 we have: f(x + h) f(x) = h 1 a (x x 0 ) 1 + h ( 1)a (x + c,a,h x 0 ), where c,a,h lies betwee 0 ad h. Note that both series o the right had side coverge absolutely because: a (x x 0 ) 1 α 1 ρ 1, ( 1)a (x + c,a,h x 0 ) α [(R + ρ)/]. We coclude that f (x) = 1 a (x x 0 ) 1 for all x x 0 < R. By iductio, we obtai: f (m) (x) = m α m (x x 0 ) m, m 1. It follows that we have the idetity: f (m) (x) h m = a ( m)! hm (x x 0 ) m m which holds true for all m 0. Now defie β m = 0 if m > ad β m = a ( k)!k! hm (x x 0 ) m if m. We see that β m = β m a m=0 m=0 ( m)! h m x x 0 m = a ( h + x x 0 ) where we used the biomial idetity i the last equality. Now if h < R x x 0 it follows that a ( h + x x 0 ) <, hece: β m <. The coditios of Theorem 1.1 are satisfied, hece Now we observe that β m = β m = β m = m=0 6 β m. a (x + h x 0 ) = f(x + h),
while β m = β m = m m a ( m)! hm (x x 0 ) m = f (m) (x) h m. I other words, f(x + h) = f (m) (x) h m. Now replace x + h = t ad x = b ad the theorem is proved. 7