Prove $sumlimits_{n=0}^inftyfrac{x^n}{n!}=limlimits_{nrightarrowinfty}left(1+frac{x}{n}right)^n$ [duplicate]
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Validity and Equivalence of two definitions of the real exponential function
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$xinmathbb{R}$. I can prove that both sides converge, and maybe we should show
$limsuplimits_{nrightarrowinfty}left|sumlimits_{k=0}^{n}frac{x^k}{n!}-left(1+frac{x}{n}right)^n right|<varepsilon$, but the construction of the right hand side is a little tricky for me.
real-analysis sequences-and-series limits
asked Nov 18 at 22:10
The R
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Nov 18 at 22:58
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This question already has an answer here:
Validity and Equivalence of two definitions of the real exponential function
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$xinmathbb{R}$. I can prove that both sides converge, and maybe we should show
$limsuplimits_{nrightarrowinfty}left|sumlimits_{k=0}^{n}frac{x^k}{n!}-left(1+frac{x}{n}right)^n right|<varepsilon$, but the construction of the right hand side is a little tricky for me.
real-analysis sequences-and-series limits
asked Nov 18 at 22:10
The R
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Nov 18 at 22:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Show that they both solve the differential equation $f'(x) = f(x)$ with initial condition $f(0) = 1$ (the solution is unique; this isn't hard to prove). The LHS solves it by finding its Taylor series and the RHS solves it using Euler's method.
– Qiaochu Yuan
Nov 18 at 22:15
Expand the right side (binomial expansion) and match up term by term (powers of $x$) to the left side.
– herb steinberg
Nov 18 at 22:24
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up vote
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down vote
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This question already has an answer here:
Validity and Equivalence of two definitions of the real exponential function
2 answers
$xinmathbb{R}$. I can prove that both sides converge, and maybe we should show
$limsuplimits_{nrightarrowinfty}left|sumlimits_{k=0}^{n}frac{x^k}{n!}-left(1+frac{x}{n}right)^n right|<varepsilon$, but the construction of the right hand side is a little tricky for me.
real-analysis sequences-and-series limits
asked Nov 18 at 22:10
The R
516
This question already has an answer here:
Validity and Equivalence of two definitions of the real exponential function
2 answers
$xinmathbb{R}$. I can prove that both sides converge, and maybe we should show
$limsuplimits_{nrightarrowinfty}left|sumlimits_{k=0}^{n}frac{x^k}{n!}-left(1+frac{x}{n}right)^n right|<varepsilon$, but the construction of the right hand side is a little tricky for me.
This question already has an answer here:
Validity and Equivalence of two definitions of the real exponential function
2 answers
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
asked Nov 18 at 22:10
The R
516
asked Nov 18 at 22:10
The R
516
asked Nov 18 at 22:10
The R
516
asked Nov 18 at 22:10
The R
516
asked Nov 18 at 22:10
The R
516
516
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Nov 18 at 22:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Show that they both solve the differential equation $f'(x) = f(x)$ with initial condition $f(0) = 1$ (the solution is unique; this isn't hard to prove). The LHS solves it by finding its Taylor series and the RHS solves it using Euler's method.
– Qiaochu Yuan
Nov 18 at 22:15
Expand the right side (binomial expansion) and match up term by term (powers of $x$) to the left side.
– herb steinberg
Nov 18 at 22:24
add a comment |
1
Show that they both solve the differential equation $f'(x) = f(x)$ with initial condition $f(0) = 1$ (the solution is unique; this isn't hard to prove). The LHS solves it by finding its Taylor series and the RHS solves it using Euler's method.
– Qiaochu Yuan
Nov 18 at 22:15
Expand the right side (binomial expansion) and match up term by term (powers of $x$) to the left side.
– herb steinberg
Nov 18 at 22:24
1
1
Show that they both solve the differential equation $f'(x) = f(x)$ with initial condition $f(0) = 1$ (the solution is unique; this isn't hard to prove). The LHS solves it by finding its Taylor series and the RHS solves it using Euler's method.
– Qiaochu Yuan
Nov 18 at 22:15
Show that they both solve the differential equation $f'(x) = f(x)$ with initial condition $f(0) = 1$ (the solution is unique; this isn't hard to prove). The LHS solves it by finding its Taylor series and the RHS solves it using Euler's method.
– Qiaochu Yuan
Nov 18 at 22:15
Expand the right side (binomial expansion) and match up term by term (powers of $x$) to the left side.
– herb steinberg
Nov 18 at 22:24
Expand the right side (binomial expansion) and match up term by term (powers of $x$) to the left side.
– herb steinberg
Nov 18 at 22:24
add a comment |
1 Answer
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Rather than proving this directly, I'd proceed as follows:
(1) Define $exp(x)$ to be the unique solution to $y' = y$ with $y(0) = 1$. Let $log$ denote its inverse (which exists because $exp$ is increasing).
(2) Use uniqueness of $exp$ to show that $exp(x + y) = exp(x) exp(y)$ and thus $log(xy) = log(x) + log(y)$.
(3) Use the defining differential equation of $exp$ to show that $exp(x) = sum x^n/n!$.
(4) Use the definition of $log$ to show that $log'(1) = 1$, then compute the log of the right-hand limit with l'Hopital's rule.
answered Nov 18 at 22:27
anomaly
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Rather than proving this directly, I'd proceed as follows:
(1) Define $exp(x)$ to be the unique solution to $y' = y$ with $y(0) = 1$. Let $log$ denote its inverse (which exists because $exp$ is increasing).
(2) Use uniqueness of $exp$ to show that $exp(x + y) = exp(x) exp(y)$ and thus $log(xy) = log(x) + log(y)$.
(3) Use the defining differential equation of $exp$ to show that $exp(x) = sum x^n/n!$.
(4) Use the definition of $log$ to show that $log'(1) = 1$, then compute the log of the right-hand limit with l'Hopital's rule.
answered Nov 18 at 22:27
anomaly
17.1k42662
add a comment |
up vote
1
down vote
accepted
Rather than proving this directly, I'd proceed as follows:
(1) Define $exp(x)$ to be the unique solution to $y' = y$ with $y(0) = 1$. Let $log$ denote its inverse (which exists because $exp$ is increasing).
(2) Use uniqueness of $exp$ to show that $exp(x + y) = exp(x) exp(y)$ and thus $log(xy) = log(x) + log(y)$.
(3) Use the defining differential equation of $exp$ to show that $exp(x) = sum x^n/n!$.
(4) Use the definition of $log$ to show that $log'(1) = 1$, then compute the log of the right-hand limit with l'Hopital's rule.
answered Nov 18 at 22:27
anomaly
17.1k42662
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Rather than proving this directly, I'd proceed as follows:
(1) Define $exp(x)$ to be the unique solution to $y' = y$ with $y(0) = 1$. Let $log$ denote its inverse (which exists because $exp$ is increasing).
(2) Use uniqueness of $exp$ to show that $exp(x + y) = exp(x) exp(y)$ and thus $log(xy) = log(x) + log(y)$.
(3) Use the defining differential equation of $exp$ to show that $exp(x) = sum x^n/n!$.
(4) Use the definition of $log$ to show that $log'(1) = 1$, then compute the log of the right-hand limit with l'Hopital's rule.
answered Nov 18 at 22:27
anomaly
17.1k42662
Rather than proving this directly, I'd proceed as follows:
(1) Define $exp(x)$ to be the unique solution to $y' = y$ with $y(0) = 1$. Let $log$ denote its inverse (which exists because $exp$ is increasing).
(2) Use uniqueness of $exp$ to show that $exp(x + y) = exp(x) exp(y)$ and thus $log(xy) = log(x) + log(y)$.
(3) Use the defining differential equation of $exp$ to show that $exp(x) = sum x^n/n!$.
(4) Use the definition of $log$ to show that $log'(1) = 1$, then compute the log of the right-hand limit with l'Hopital's rule.
answered Nov 18 at 22:27
anomaly
17.1k42662
edited Nov 19 at 0:27
answered Nov 18 at 22:27
anomaly
17.1k42662
answered Nov 18 at 22:27
anomaly
17.1k42662
answered Nov 18 at 22:27
anomaly
17.1k42662
17.1k42662
add a comment |
add a comment |
1
Show that they both solve the differential equation $f'(x) = f(x)$ with initial condition $f(0) = 1$ (the solution is unique; this isn't hard to prove). The LHS solves it by finding its Taylor series and the RHS solves it using Euler's method.
– Qiaochu Yuan
Nov 18 at 22:15
Expand the right side (binomial expansion) and match up term by term (powers of $x$) to the left side.
– herb steinberg
Nov 18 at 22:24