Show that exp(x) is of the form $e^x$?












1














Let me define the function $exp(x)$ as that which satisfies:
$$ frac{d}{dx}exp(x) = exp(x) $$
We can show that
$$ exp(x) = sum_{n=0}^{infty} frac{x^n}{n!}.$$



So far so good, we have just defined the exponential by the differential equation it satisfies.



How can we show that the exponential, as defined above, is of the form:
$$exp(x) = e^x,$$
where $e$ is a real number.



In other words, how can we show, for example that:
$$ e^a = prod_{m=1}^{a}e^1$$
$$sum_{n=0}^infty frac{a^n}{n!} = prod_{m=1}^a left[ sum_{l=0}^{infty} frac{1}{n!} right]$$
for any positive integer $a$.






algebra-precalculus differential-equations functions special-functions







share|cite|improve this question
























  • An interesting question. This is so often considered trivial that it is often treated as a definition.
    – The Count
    Nov 22 at 21:25






  • 2




    "We can show that ..." You need an initial condition for the differential equation.
    – user587192
    Nov 22 at 21:32












  • The first identity after "how can we show, for example that:" does not make sense.
    – user587192
    Nov 22 at 21:34










  • Usually, the limit definition is used to prove this property
    – Yuriy S
    Nov 22 at 21:37










  • @user587192 Could you expand on why it doesn't make sense?
    – Daniel Duque
    Nov 22 at 21:41
















1














Let me define the function $exp(x)$ as that which satisfies:
$$ frac{d}{dx}exp(x) = exp(x) $$
We can show that
$$ exp(x) = sum_{n=0}^{infty} frac{x^n}{n!}.$$



So far so good, we have just defined the exponential by the differential equation it satisfies.



How can we show that the exponential, as defined above, is of the form:
$$exp(x) = e^x,$$
where $e$ is a real number.



In other words, how can we show, for example that:
$$ e^a = prod_{m=1}^{a}e^1$$
$$sum_{n=0}^infty frac{a^n}{n!} = prod_{m=1}^a left[ sum_{l=0}^{infty} frac{1}{n!} right]$$
for any positive integer $a$.






algebra-precalculus differential-equations functions special-functions







share|cite|improve this question
























  • An interesting question. This is so often considered trivial that it is often treated as a definition.
    – The Count
    Nov 22 at 21:25






  • 2




    "We can show that ..." You need an initial condition for the differential equation.
    – user587192
    Nov 22 at 21:32












  • The first identity after "how can we show, for example that:" does not make sense.
    – user587192
    Nov 22 at 21:34










  • Usually, the limit definition is used to prove this property
    – Yuriy S
    Nov 22 at 21:37










  • @user587192 Could you expand on why it doesn't make sense?
    – Daniel Duque
    Nov 22 at 21:41














1












1








1







Let me define the function $exp(x)$ as that which satisfies:
$$ frac{d}{dx}exp(x) = exp(x) $$
We can show that
$$ exp(x) = sum_{n=0}^{infty} frac{x^n}{n!}.$$



So far so good, we have just defined the exponential by the differential equation it satisfies.



How can we show that the exponential, as defined above, is of the form:
$$exp(x) = e^x,$$
where $e$ is a real number.



In other words, how can we show, for example that:
$$ e^a = prod_{m=1}^{a}e^1$$
$$sum_{n=0}^infty frac{a^n}{n!} = prod_{m=1}^a left[ sum_{l=0}^{infty} frac{1}{n!} right]$$
for any positive integer $a$.






algebra-precalculus differential-equations functions special-functions







share|cite|improve this question















Let me define the function $exp(x)$ as that which satisfies:
$$ frac{d}{dx}exp(x) = exp(x) $$
We can show that
$$ exp(x) = sum_{n=0}^{infty} frac{x^n}{n!}.$$



So far so good, we have just defined the exponential by the differential equation it satisfies.



How can we show that the exponential, as defined above, is of the form:
$$exp(x) = e^x,$$
where $e$ is a real number.



In other words, how can we show, for example that:
$$ e^a = prod_{m=1}^{a}e^1$$
$$sum_{n=0}^infty frac{a^n}{n!} = prod_{m=1}^a left[ sum_{l=0}^{infty} frac{1}{n!} right]$$
for any positive integer $a$.






algebra-precalculus differential-equations functions special-functions




algebra-precalculus differential-equations functions special-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 21:52

























asked Nov 22 at 21:23









Daniel Duque

267




267












  • An interesting question. This is so often considered trivial that it is often treated as a definition.
    – The Count
    Nov 22 at 21:25






  • 2




    "We can show that ..." You need an initial condition for the differential equation.
    – user587192
    Nov 22 at 21:32












  • The first identity after "how can we show, for example that:" does not make sense.
    – user587192
    Nov 22 at 21:34










  • Usually, the limit definition is used to prove this property
    – Yuriy S
    Nov 22 at 21:37










  • @user587192 Could you expand on why it doesn't make sense?
    – Daniel Duque
    Nov 22 at 21:41


















  • An interesting question. This is so often considered trivial that it is often treated as a definition.
    – The Count
    Nov 22 at 21:25






  • 2




    "We can show that ..." You need an initial condition for the differential equation.
    – user587192
    Nov 22 at 21:32












  • The first identity after "how can we show, for example that:" does not make sense.
    – user587192
    Nov 22 at 21:34










  • Usually, the limit definition is used to prove this property
    – Yuriy S
    Nov 22 at 21:37










  • @user587192 Could you expand on why it doesn't make sense?
    – Daniel Duque
    Nov 22 at 21:41
















An interesting question. This is so often considered trivial that it is often treated as a definition.
– The Count
Nov 22 at 21:25




An interesting question. This is so often considered trivial that it is often treated as a definition.
– The Count
Nov 22 at 21:25




2




2




"We can show that ..." You need an initial condition for the differential equation.
– user587192
Nov 22 at 21:32






"We can show that ..." You need an initial condition for the differential equation.
– user587192
Nov 22 at 21:32














The first identity after "how can we show, for example that:" does not make sense.
– user587192
Nov 22 at 21:34




The first identity after "how can we show, for example that:" does not make sense.
– user587192
Nov 22 at 21:34












Usually, the limit definition is used to prove this property
– Yuriy S
Nov 22 at 21:37




Usually, the limit definition is used to prove this property
– Yuriy S
Nov 22 at 21:37












@user587192 Could you expand on why it doesn't make sense?
– Daniel Duque
Nov 22 at 21:41




@user587192 Could you expand on why it doesn't make sense?
– Daniel Duque
Nov 22 at 21:41










3 Answers
3






active

oldest

votes


















2














First note that apart from $exp'=exp$, you need to postulate $exp(0)=1$. Then it is the uniqueness of the solution to the differential equation that determines the desired properties:



The derivative of $f(x)= bcdot exp(a+x)+cexp(x)$ apparently is $$f'(x)=bexp'(a+x)+cexp'(x)=bexp(a+x)+cexp(x)=f(x).$$ So if we adjust $b,c$ to ensure $f(0)=1$, we conclude that $fequiv exp$.



If $exp(a)=0$, we take $b=c=1$ and infer $exp(x)=exp(a+x)+exp(x)$ for all $x$, i.e., $expequiv 0$, contradiction.



Hence we may assume $exp(a)ne 0$, can take $c=0$ and $b=frac 1{exp(a)}$, and find
$$ exp(x)=frac{exp(a+x)}{exp(a)}$$
for all $a,xinBbb R$. In other words,
$$ exp(a+b)=exp(a)exp(b)$$
for all $a,binBbb R$.
From this, we learn among others (by induction) that $$exp(n)=exp(1)^n$$ for all $ninBbb Z$.






share|cite|improve this answer





















  • Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
    – Daniel Duque
    Nov 22 at 22:02



















1














It suffices to show that $$exp(x+y)=exp (x)exp(y)$$But this can be easily shown since $$exp(x)exp(y){=sum_{n=0}^{infty}{x^nover n!}sum_{m=0}^{infty}{y^mover m!}\=sum_{n,m=0}^{infty}{x^ny^mover n!m!}\=sum_{k=0}^infty{1over k!}sum_{m+n=k}{k!x^ny^{k-n}over n!(k-n)!}\=sum_{k=0}^infty {(x+y)^kover k!}\=exp(x+y)}$$






share|cite|improve this answer





























    0














    One way is through $$e^xequivlim_{nrightarrowinfty}left(1+frac{x}{n}right)^n
    $$



    You can show through the binomial expansion that this recipe gives the Taylor series for $exp x$ and thence that it solves the differential equation; or by differentiating the expression with 'lim' in it directly; and also showing that this expression is also the natural extension of exponentiation into the domain of reals from its definition for integer argument in terms of iterated multiplication. If you ply these various relations backward & forward in every conceivable direction, bringing in also the fundamental definition of differentiation in terms of limits, it becomes clear that it all constitutes a beautifully internally consistent system of relations. And you can also get



    $$frac{dy}{dx}equiv afrac{y}{x}$$ &



    $$x=1implies y=1$$



    as a definition of $x^a$ for general real $a$.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      First note that apart from $exp'=exp$, you need to postulate $exp(0)=1$. Then it is the uniqueness of the solution to the differential equation that determines the desired properties:



      The derivative of $f(x)= bcdot exp(a+x)+cexp(x)$ apparently is $$f'(x)=bexp'(a+x)+cexp'(x)=bexp(a+x)+cexp(x)=f(x).$$ So if we adjust $b,c$ to ensure $f(0)=1$, we conclude that $fequiv exp$.



      If $exp(a)=0$, we take $b=c=1$ and infer $exp(x)=exp(a+x)+exp(x)$ for all $x$, i.e., $expequiv 0$, contradiction.



      Hence we may assume $exp(a)ne 0$, can take $c=0$ and $b=frac 1{exp(a)}$, and find
      $$ exp(x)=frac{exp(a+x)}{exp(a)}$$
      for all $a,xinBbb R$. In other words,
      $$ exp(a+b)=exp(a)exp(b)$$
      for all $a,binBbb R$.
      From this, we learn among others (by induction) that $$exp(n)=exp(1)^n$$ for all $ninBbb Z$.






      share|cite|improve this answer





















      • Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
        – Daniel Duque
        Nov 22 at 22:02
















      2














      First note that apart from $exp'=exp$, you need to postulate $exp(0)=1$. Then it is the uniqueness of the solution to the differential equation that determines the desired properties:



      The derivative of $f(x)= bcdot exp(a+x)+cexp(x)$ apparently is $$f'(x)=bexp'(a+x)+cexp'(x)=bexp(a+x)+cexp(x)=f(x).$$ So if we adjust $b,c$ to ensure $f(0)=1$, we conclude that $fequiv exp$.



      If $exp(a)=0$, we take $b=c=1$ and infer $exp(x)=exp(a+x)+exp(x)$ for all $x$, i.e., $expequiv 0$, contradiction.



      Hence we may assume $exp(a)ne 0$, can take $c=0$ and $b=frac 1{exp(a)}$, and find
      $$ exp(x)=frac{exp(a+x)}{exp(a)}$$
      for all $a,xinBbb R$. In other words,
      $$ exp(a+b)=exp(a)exp(b)$$
      for all $a,binBbb R$.
      From this, we learn among others (by induction) that $$exp(n)=exp(1)^n$$ for all $ninBbb Z$.






      share|cite|improve this answer





















      • Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
        – Daniel Duque
        Nov 22 at 22:02














      2












      2








      2






      First note that apart from $exp'=exp$, you need to postulate $exp(0)=1$. Then it is the uniqueness of the solution to the differential equation that determines the desired properties:



      The derivative of $f(x)= bcdot exp(a+x)+cexp(x)$ apparently is $$f'(x)=bexp'(a+x)+cexp'(x)=bexp(a+x)+cexp(x)=f(x).$$ So if we adjust $b,c$ to ensure $f(0)=1$, we conclude that $fequiv exp$.



      If $exp(a)=0$, we take $b=c=1$ and infer $exp(x)=exp(a+x)+exp(x)$ for all $x$, i.e., $expequiv 0$, contradiction.



      Hence we may assume $exp(a)ne 0$, can take $c=0$ and $b=frac 1{exp(a)}$, and find
      $$ exp(x)=frac{exp(a+x)}{exp(a)}$$
      for all $a,xinBbb R$. In other words,
      $$ exp(a+b)=exp(a)exp(b)$$
      for all $a,binBbb R$.
      From this, we learn among others (by induction) that $$exp(n)=exp(1)^n$$ for all $ninBbb Z$.






      share|cite|improve this answer












      First note that apart from $exp'=exp$, you need to postulate $exp(0)=1$. Then it is the uniqueness of the solution to the differential equation that determines the desired properties:



      The derivative of $f(x)= bcdot exp(a+x)+cexp(x)$ apparently is $$f'(x)=bexp'(a+x)+cexp'(x)=bexp(a+x)+cexp(x)=f(x).$$ So if we adjust $b,c$ to ensure $f(0)=1$, we conclude that $fequiv exp$.



      If $exp(a)=0$, we take $b=c=1$ and infer $exp(x)=exp(a+x)+exp(x)$ for all $x$, i.e., $expequiv 0$, contradiction.



      Hence we may assume $exp(a)ne 0$, can take $c=0$ and $b=frac 1{exp(a)}$, and find
      $$ exp(x)=frac{exp(a+x)}{exp(a)}$$
      for all $a,xinBbb R$. In other words,
      $$ exp(a+b)=exp(a)exp(b)$$
      for all $a,binBbb R$.
      From this, we learn among others (by induction) that $$exp(n)=exp(1)^n$$ for all $ninBbb Z$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 22 at 21:40









      Hagen von Eitzen

      275k21268495




      275k21268495












      • Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
        – Daniel Duque
        Nov 22 at 22:02


















      • Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
        – Daniel Duque
        Nov 22 at 22:02
















      Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
      – Daniel Duque
      Nov 22 at 22:02




      Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
      – Daniel Duque
      Nov 22 at 22:02











      1














      It suffices to show that $$exp(x+y)=exp (x)exp(y)$$But this can be easily shown since $$exp(x)exp(y){=sum_{n=0}^{infty}{x^nover n!}sum_{m=0}^{infty}{y^mover m!}\=sum_{n,m=0}^{infty}{x^ny^mover n!m!}\=sum_{k=0}^infty{1over k!}sum_{m+n=k}{k!x^ny^{k-n}over n!(k-n)!}\=sum_{k=0}^infty {(x+y)^kover k!}\=exp(x+y)}$$






      share|cite|improve this answer


























        1














        It suffices to show that $$exp(x+y)=exp (x)exp(y)$$But this can be easily shown since $$exp(x)exp(y){=sum_{n=0}^{infty}{x^nover n!}sum_{m=0}^{infty}{y^mover m!}\=sum_{n,m=0}^{infty}{x^ny^mover n!m!}\=sum_{k=0}^infty{1over k!}sum_{m+n=k}{k!x^ny^{k-n}over n!(k-n)!}\=sum_{k=0}^infty {(x+y)^kover k!}\=exp(x+y)}$$






        share|cite|improve this answer
























          1












          1








          1






          It suffices to show that $$exp(x+y)=exp (x)exp(y)$$But this can be easily shown since $$exp(x)exp(y){=sum_{n=0}^{infty}{x^nover n!}sum_{m=0}^{infty}{y^mover m!}\=sum_{n,m=0}^{infty}{x^ny^mover n!m!}\=sum_{k=0}^infty{1over k!}sum_{m+n=k}{k!x^ny^{k-n}over n!(k-n)!}\=sum_{k=0}^infty {(x+y)^kover k!}\=exp(x+y)}$$






          share|cite|improve this answer












          It suffices to show that $$exp(x+y)=exp (x)exp(y)$$But this can be easily shown since $$exp(x)exp(y){=sum_{n=0}^{infty}{x^nover n!}sum_{m=0}^{infty}{y^mover m!}\=sum_{n,m=0}^{infty}{x^ny^mover n!m!}\=sum_{k=0}^infty{1over k!}sum_{m+n=k}{k!x^ny^{k-n}over n!(k-n)!}\=sum_{k=0}^infty {(x+y)^kover k!}\=exp(x+y)}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 22:11









          Mostafa Ayaz

          13.7k3836




          13.7k3836























              0














              One way is through $$e^xequivlim_{nrightarrowinfty}left(1+frac{x}{n}right)^n
              $$



              You can show through the binomial expansion that this recipe gives the Taylor series for $exp x$ and thence that it solves the differential equation; or by differentiating the expression with 'lim' in it directly; and also showing that this expression is also the natural extension of exponentiation into the domain of reals from its definition for integer argument in terms of iterated multiplication. If you ply these various relations backward & forward in every conceivable direction, bringing in also the fundamental definition of differentiation in terms of limits, it becomes clear that it all constitutes a beautifully internally consistent system of relations. And you can also get



              $$frac{dy}{dx}equiv afrac{y}{x}$$ &



              $$x=1implies y=1$$



              as a definition of $x^a$ for general real $a$.






              share|cite|improve this answer


























                0














                One way is through $$e^xequivlim_{nrightarrowinfty}left(1+frac{x}{n}right)^n
                $$



                You can show through the binomial expansion that this recipe gives the Taylor series for $exp x$ and thence that it solves the differential equation; or by differentiating the expression with 'lim' in it directly; and also showing that this expression is also the natural extension of exponentiation into the domain of reals from its definition for integer argument in terms of iterated multiplication. If you ply these various relations backward & forward in every conceivable direction, bringing in also the fundamental definition of differentiation in terms of limits, it becomes clear that it all constitutes a beautifully internally consistent system of relations. And you can also get



                $$frac{dy}{dx}equiv afrac{y}{x}$$ &



                $$x=1implies y=1$$



                as a definition of $x^a$ for general real $a$.






                share|cite|improve this answer
























                  0












                  0








                  0






                  One way is through $$e^xequivlim_{nrightarrowinfty}left(1+frac{x}{n}right)^n
                  $$



                  You can show through the binomial expansion that this recipe gives the Taylor series for $exp x$ and thence that it solves the differential equation; or by differentiating the expression with 'lim' in it directly; and also showing that this expression is also the natural extension of exponentiation into the domain of reals from its definition for integer argument in terms of iterated multiplication. If you ply these various relations backward & forward in every conceivable direction, bringing in also the fundamental definition of differentiation in terms of limits, it becomes clear that it all constitutes a beautifully internally consistent system of relations. And you can also get



                  $$frac{dy}{dx}equiv afrac{y}{x}$$ &



                  $$x=1implies y=1$$



                  as a definition of $x^a$ for general real $a$.






                  share|cite|improve this answer












                  One way is through $$e^xequivlim_{nrightarrowinfty}left(1+frac{x}{n}right)^n
                  $$



                  You can show through the binomial expansion that this recipe gives the Taylor series for $exp x$ and thence that it solves the differential equation; or by differentiating the expression with 'lim' in it directly; and also showing that this expression is also the natural extension of exponentiation into the domain of reals from its definition for integer argument in terms of iterated multiplication. If you ply these various relations backward & forward in every conceivable direction, bringing in also the fundamental definition of differentiation in terms of limits, it becomes clear that it all constitutes a beautifully internally consistent system of relations. And you can also get



                  $$frac{dy}{dx}equiv afrac{y}{x}$$ &



                  $$x=1implies y=1$$



                  as a definition of $x^a$ for general real $a$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 23 at 3:24









                  AmbretteOrrisey

                  56710




                  56710






























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                      Puebla de Zaragoza

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                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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                      Musa

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                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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