Mixing
Question about Exponential Series Definition and Convergence
$begingroup$
I've seen in many textbooks that the following is just a definition: $$e^x = sum_{n=0}^{infty} frac{x^{n}}{n!}$$
And then many textbooks just go ahead to prove the absolute convergence of the infinite series using, for example, ratio test, to conclude that the infinite sum actually makes sense. But even if the infinite series is absolutely convergent, how do I know the series does converge to $e^x$ instead of some other functions? The series convergence tests never mention about the limit that the series converges to. How do people come up with such a definition at the first place?
calculus sequences-and-series convergence exponential-function
asked Feb 2 at 19:51
xf16xf16
937
$endgroup$
add a comment |
$begingroup$
I've seen in many textbooks that the following is just a definition: $$e^x = sum_{n=0}^{infty} frac{x^{n}}{n!}$$
And then many textbooks just go ahead to prove the absolute convergence of the infinite series using, for example, ratio test, to conclude that the infinite sum actually makes sense. But even if the infinite series is absolutely convergent, how do I know the series does converge to $e^x$ instead of some other functions? The series convergence tests never mention about the limit that the series converges to. How do people come up with such a definition at the first place?
calculus sequences-and-series convergence exponential-function
asked Feb 2 at 19:51
xf16xf16
937
$endgroup$
1
$begingroup$
If they're using it as a definition, then they're defining the function $e^x$ to be this series.
$endgroup$
– Dave
Feb 2 at 20:04
add a comment |
$begingroup$
I've seen in many textbooks that the following is just a definition: $$e^x = sum_{n=0}^{infty} frac{x^{n}}{n!}$$
And then many textbooks just go ahead to prove the absolute convergence of the infinite series using, for example, ratio test, to conclude that the infinite sum actually makes sense. But even if the infinite series is absolutely convergent, how do I know the series does converge to $e^x$ instead of some other functions? The series convergence tests never mention about the limit that the series converges to. How do people come up with such a definition at the first place?
calculus sequences-and-series convergence exponential-function
asked Feb 2 at 19:51
xf16xf16
937
$endgroup$
I've seen in many textbooks that the following is just a definition: $$e^x = sum_{n=0}^{infty} frac{x^{n}}{n!}$$
And then many textbooks just go ahead to prove the absolute convergence of the infinite series using, for example, ratio test, to conclude that the infinite sum actually makes sense. But even if the infinite series is absolutely convergent, how do I know the series does converge to $e^x$ instead of some other functions? The series convergence tests never mention about the limit that the series converges to. How do people come up with such a definition at the first place?
calculus sequences-and-series convergence exponential-function
calculus sequences-and-series convergence exponential-function
asked Feb 2 at 19:51
xf16xf16
937
asked Feb 2 at 19:51
xf16xf16
937
asked Feb 2 at 19:51
xf16xf16
937
asked Feb 2 at 19:51
xf16xf16
937
asked Feb 2 at 19:51
xf16xf16
937
937
1
$begingroup$
If they're using it as a definition, then they're defining the function $e^x$ to be this series.
$endgroup$
– Dave
Feb 2 at 20:04
add a comment |
1
$begingroup$
If they're using it as a definition, then they're defining the function $e^x$ to be this series.
$endgroup$
– Dave
Feb 2 at 20:04
1
1
$begingroup$
If they're using it as a definition, then they're defining the function $e^x$ to be this series.
$endgroup$
– Dave
Feb 2 at 20:04
$begingroup$
If they're using it as a definition, then they're defining the function $e^x$ to be this series.
$endgroup$
– Dave
Feb 2 at 20:04
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
To see that your equation is true, you could take the derivative of the right side and see that you get the same thing. This shows that the sum is a solution to the differential equation
$$y'=y.$$
You can the check that plugging in $x=0$ on both sides yields the same number, so that you know the left and right sides both solve the initial value problem
$$y' = y, y(0) =1.$$
Since solutions to linear IVP's are unique, you know the left side equals the right side.
And note that there are different ways to define $e^x.$ If you use one of the other ways, then it's not too hard to show that it's Maclauren series is your sum.
answered Feb 2 at 20:02
B. GoddardB. Goddard
20.2k21543
$endgroup$
$begingroup$
Hi Goddard, is this logic correct: the infinite sum is actually the Maclauren series which is Taylor series at 0, of the exponential function, so based on the theory of taylor series, if the series does converge, it must converge to the exponential function, which is justified by showing the series is absolutely convergent everywhere. Thanks a lot.
$endgroup$
– xf16
Feb 2 at 20:19
$begingroup$
Yeah, I guess that works. But without stating which definition of $e^x$ you're using, it seems a bit...um...disconnected.
$endgroup$
– B. Goddard
Feb 2 at 21:06
add a comment |
$begingroup$
Once you know that the series is absolutely convergent you can deduce from it all the characteristics that define the $e^x$ function, for instance you can take derivative:
$$
frac{d}{dx}sum_{i=0}^inftyfrac{x^n}{n!}= sum_{i=0}^inftyfrac{x^n}{n!}
$$
showing you that $(e^x)'=e^x$.
More interestingly you can use this definition to extend exponential to any square matrix $X$. The series is always absolutely convergent. Obviously you have $XX^k=X^kX$, therefore, from the series definition you can see that:
$$
Xe^X=e^XX
$$
With a little more work you can also see that:
$$
e^{(A+B)}=e^Ae^B
$$
if $A$ and $B$ commutes: $AB=BA$.
answered Feb 2 at 20:14
Picaud VincentPicaud Vincent
1,434310
$endgroup$
add a comment |
$begingroup$
Hint: Use Taylor's theorem and $(e^x)'=e^x$.
answered Feb 2 at 20:25
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To see that your equation is true, you could take the derivative of the right side and see that you get the same thing. This shows that the sum is a solution to the differential equation
$$y'=y.$$
You can the check that plugging in $x=0$ on both sides yields the same number, so that you know the left and right sides both solve the initial value problem
$$y' = y, y(0) =1.$$
Since solutions to linear IVP's are unique, you know the left side equals the right side.
And note that there are different ways to define $e^x.$ If you use one of the other ways, then it's not too hard to show that it's Maclauren series is your sum.
answered Feb 2 at 20:02
B. GoddardB. Goddard
20.2k21543
$endgroup$
$begingroup$
Hi Goddard, is this logic correct: the infinite sum is actually the Maclauren series which is Taylor series at 0, of the exponential function, so based on the theory of taylor series, if the series does converge, it must converge to the exponential function, which is justified by showing the series is absolutely convergent everywhere. Thanks a lot.
$endgroup$
– xf16
Feb 2 at 20:19
$begingroup$
Yeah, I guess that works. But without stating which definition of $e^x$ you're using, it seems a bit...um...disconnected.
$endgroup$
– B. Goddard
Feb 2 at 21:06
add a comment |
$begingroup$
To see that your equation is true, you could take the derivative of the right side and see that you get the same thing. This shows that the sum is a solution to the differential equation
$$y'=y.$$
You can the check that plugging in $x=0$ on both sides yields the same number, so that you know the left and right sides both solve the initial value problem
$$y' = y, y(0) =1.$$
Since solutions to linear IVP's are unique, you know the left side equals the right side.
And note that there are different ways to define $e^x.$ If you use one of the other ways, then it's not too hard to show that it's Maclauren series is your sum.
answered Feb 2 at 20:02
B. GoddardB. Goddard
20.2k21543
$endgroup$
$begingroup$
Hi Goddard, is this logic correct: the infinite sum is actually the Maclauren series which is Taylor series at 0, of the exponential function, so based on the theory of taylor series, if the series does converge, it must converge to the exponential function, which is justified by showing the series is absolutely convergent everywhere. Thanks a lot.
$endgroup$
– xf16
Feb 2 at 20:19
$begingroup$
Yeah, I guess that works. But without stating which definition of $e^x$ you're using, it seems a bit...um...disconnected.
$endgroup$
– B. Goddard
Feb 2 at 21:06
add a comment |
$begingroup$
To see that your equation is true, you could take the derivative of the right side and see that you get the same thing. This shows that the sum is a solution to the differential equation
$$y'=y.$$
You can the check that plugging in $x=0$ on both sides yields the same number, so that you know the left and right sides both solve the initial value problem
$$y' = y, y(0) =1.$$
Since solutions to linear IVP's are unique, you know the left side equals the right side.
And note that there are different ways to define $e^x.$ If you use one of the other ways, then it's not too hard to show that it's Maclauren series is your sum.
answered Feb 2 at 20:02
B. GoddardB. Goddard
20.2k21543
$endgroup$
To see that your equation is true, you could take the derivative of the right side and see that you get the same thing. This shows that the sum is a solution to the differential equation
$$y'=y.$$
You can the check that plugging in $x=0$ on both sides yields the same number, so that you know the left and right sides both solve the initial value problem
$$y' = y, y(0) =1.$$
Since solutions to linear IVP's are unique, you know the left side equals the right side.
And note that there are different ways to define $e^x.$ If you use one of the other ways, then it's not too hard to show that it's Maclauren series is your sum.
answered Feb 2 at 20:02
B. GoddardB. Goddard
20.2k21543
answered Feb 2 at 20:02
B. GoddardB. Goddard
20.2k21543
answered Feb 2 at 20:02
B. GoddardB. Goddard
20.2k21543
answered Feb 2 at 20:02
B. GoddardB. Goddard
20.2k21543
20.2k21543
$begingroup$
Hi Goddard, is this logic correct: the infinite sum is actually the Maclauren series which is Taylor series at 0, of the exponential function, so based on the theory of taylor series, if the series does converge, it must converge to the exponential function, which is justified by showing the series is absolutely convergent everywhere. Thanks a lot.
$endgroup$
– xf16
Feb 2 at 20:19
$begingroup$
Yeah, I guess that works. But without stating which definition of $e^x$ you're using, it seems a bit...um...disconnected.
$endgroup$
– B. Goddard
Feb 2 at 21:06
add a comment |
$begingroup$
Hi Goddard, is this logic correct: the infinite sum is actually the Maclauren series which is Taylor series at 0, of the exponential function, so based on the theory of taylor series, if the series does converge, it must converge to the exponential function, which is justified by showing the series is absolutely convergent everywhere. Thanks a lot.
$endgroup$
– xf16
Feb 2 at 20:19
$begingroup$
Yeah, I guess that works. But without stating which definition of $e^x$ you're using, it seems a bit...um...disconnected.
$endgroup$
– B. Goddard
Feb 2 at 21:06
$begingroup$
Hi Goddard, is this logic correct: the infinite sum is actually the Maclauren series which is Taylor series at 0, of the exponential function, so based on the theory of taylor series, if the series does converge, it must converge to the exponential function, which is justified by showing the series is absolutely convergent everywhere. Thanks a lot.
$endgroup$
– xf16
Feb 2 at 20:19
$begingroup$
Hi Goddard, is this logic correct: the infinite sum is actually the Maclauren series which is Taylor series at 0, of the exponential function, so based on the theory of taylor series, if the series does converge, it must converge to the exponential function, which is justified by showing the series is absolutely convergent everywhere. Thanks a lot.
$endgroup$
– xf16
Feb 2 at 20:19
$begingroup$
Yeah, I guess that works. But without stating which definition of $e^x$ you're using, it seems a bit...um...disconnected.
$endgroup$
– B. Goddard
Feb 2 at 21:06
$begingroup$
Yeah, I guess that works. But without stating which definition of $e^x$ you're using, it seems a bit...um...disconnected.
$endgroup$
– B. Goddard
Feb 2 at 21:06
add a comment |
$begingroup$
Once you know that the series is absolutely convergent you can deduce from it all the characteristics that define the $e^x$ function, for instance you can take derivative:
$$
frac{d}{dx}sum_{i=0}^inftyfrac{x^n}{n!}= sum_{i=0}^inftyfrac{x^n}{n!}
$$
showing you that $(e^x)'=e^x$.
More interestingly you can use this definition to extend exponential to any square matrix $X$. The series is always absolutely convergent. Obviously you have $XX^k=X^kX$, therefore, from the series definition you can see that:
$$
Xe^X=e^XX
$$
With a little more work you can also see that:
$$
e^{(A+B)}=e^Ae^B
$$
if $A$ and $B$ commutes: $AB=BA$.
answered Feb 2 at 20:14
Picaud VincentPicaud Vincent
1,434310
$endgroup$
add a comment |
$begingroup$
Once you know that the series is absolutely convergent you can deduce from it all the characteristics that define the $e^x$ function, for instance you can take derivative:
$$
frac{d}{dx}sum_{i=0}^inftyfrac{x^n}{n!}= sum_{i=0}^inftyfrac{x^n}{n!}
$$
showing you that $(e^x)'=e^x$.
More interestingly you can use this definition to extend exponential to any square matrix $X$. The series is always absolutely convergent. Obviously you have $XX^k=X^kX$, therefore, from the series definition you can see that:
$$
Xe^X=e^XX
$$
With a little more work you can also see that:
$$
e^{(A+B)}=e^Ae^B
$$
if $A$ and $B$ commutes: $AB=BA$.
answered Feb 2 at 20:14
Picaud VincentPicaud Vincent
1,434310
$endgroup$
add a comment |
$begingroup$
Once you know that the series is absolutely convergent you can deduce from it all the characteristics that define the $e^x$ function, for instance you can take derivative:
$$
frac{d}{dx}sum_{i=0}^inftyfrac{x^n}{n!}= sum_{i=0}^inftyfrac{x^n}{n!}
$$
showing you that $(e^x)'=e^x$.
More interestingly you can use this definition to extend exponential to any square matrix $X$. The series is always absolutely convergent. Obviously you have $XX^k=X^kX$, therefore, from the series definition you can see that:
$$
Xe^X=e^XX
$$
With a little more work you can also see that:
$$
e^{(A+B)}=e^Ae^B
$$
if $A$ and $B$ commutes: $AB=BA$.
answered Feb 2 at 20:14
Picaud VincentPicaud Vincent
1,434310
$endgroup$
Once you know that the series is absolutely convergent you can deduce from it all the characteristics that define the $e^x$ function, for instance you can take derivative:
$$
frac{d}{dx}sum_{i=0}^inftyfrac{x^n}{n!}= sum_{i=0}^inftyfrac{x^n}{n!}
$$
showing you that $(e^x)'=e^x$.
More interestingly you can use this definition to extend exponential to any square matrix $X$. The series is always absolutely convergent. Obviously you have $XX^k=X^kX$, therefore, from the series definition you can see that:
$$
Xe^X=e^XX
$$
With a little more work you can also see that:
$$
e^{(A+B)}=e^Ae^B
$$
if $A$ and $B$ commutes: $AB=BA$.
answered Feb 2 at 20:14
Picaud VincentPicaud Vincent
1,434310
edited Feb 3 at 7:17
answered Feb 2 at 20:14
Picaud VincentPicaud Vincent
1,434310
answered Feb 2 at 20:14
Picaud VincentPicaud Vincent
1,434310
answered Feb 2 at 20:14
Picaud VincentPicaud Vincent
1,434310
1,434310
add a comment |
add a comment |
$begingroup$
Hint: Use Taylor's theorem and $(e^x)'=e^x$.
answered Feb 2 at 20:25
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
Hint: Use Taylor's theorem and $(e^x)'=e^x$.
answered Feb 2 at 20:25
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
Hint: Use Taylor's theorem and $(e^x)'=e^x$.
answered Feb 2 at 20:25
Chris CusterChris Custer
14.4k3827
$endgroup$
Hint: Use Taylor's theorem and $(e^x)'=e^x$.
answered Feb 2 at 20:25
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 20:25
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 20:25
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 20:25
Chris CusterChris Custer
14.4k3827
14.4k3827
add a comment |
add a comment |
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$begingroup$
If they're using it as a definition, then they're defining the function $e^x$ to be this series.
$endgroup$
– Dave
Feb 2 at 20:04