Find the Maclaurin series for $f(x) = e^{-2x}$, and find the interval of convergence for the series.
Find the Maclaurin series for $f(x) = e^{-2x}$, and find the interval of convergence for the series.
I got the maclaurin series to be this
$e^x = 1 - 2frac{x}{1!} + 2^2frac{x^2}{2!} + cdots$
$$sum_{n=0}^{infty} frac{(-1)^n (2)^n x^n}{n!}$$
Using the ratio test to find the interval of convergence:
$$lim_{ntoinfty} left| frac{(-1)^{n+1}2^{n+1}x^{n+1}}{(n+1)!} frac{n!}{(-1)^n 2^n x^n} right| = 2|x| lim_{ntoinfty} frac{1}{n+1} = 2|x|(0) < 1$$
Therefore for any value of x is in the interval of convergence. Is this right?
sequences-and-series
asked Apr 19 '17 at 18:52
TinlerTinler
517312
add a comment |
Find the Maclaurin series for $f(x) = e^{-2x}$, and find the interval of convergence for the series.
I got the maclaurin series to be this
$e^x = 1 - 2frac{x}{1!} + 2^2frac{x^2}{2!} + cdots$
$$sum_{n=0}^{infty} frac{(-1)^n (2)^n x^n}{n!}$$
Using the ratio test to find the interval of convergence:
$$lim_{ntoinfty} left| frac{(-1)^{n+1}2^{n+1}x^{n+1}}{(n+1)!} frac{n!}{(-1)^n 2^n x^n} right| = 2|x| lim_{ntoinfty} frac{1}{n+1} = 2|x|(0) < 1$$
Therefore for any value of x is in the interval of convergence. Is this right?
sequences-and-series
asked Apr 19 '17 at 18:52
TinlerTinler
517312
Yes, you're right
– caverac
Apr 19 '17 at 18:57
Do I need to check the endpoints at x = 0, or no? Im a bit confused when the ratio test gives $infty$ or zero. If its $infty$ instead of 0, so like $2|x|infty > 1$ Can I say there is no point in the interval that is convergence?
– Tinler
Apr 19 '17 at 18:57
The limit is 0 regardless of the value of $x$, so no need to check anything else
– caverac
Apr 19 '17 at 18:58
There are no endpoints to check. The interval of convergence is $(-infty,infty)$.
– John Wayland Bales
Apr 19 '17 at 18:59
What happens if its $2|x|infty > 1$ ?. As in instead of 0 i put infinity.
– Tinler
Apr 19 '17 at 19:05
add a comment |
Find the Maclaurin series for $f(x) = e^{-2x}$, and find the interval of convergence for the series.
I got the maclaurin series to be this
$e^x = 1 - 2frac{x}{1!} + 2^2frac{x^2}{2!} + cdots$
$$sum_{n=0}^{infty} frac{(-1)^n (2)^n x^n}{n!}$$
Using the ratio test to find the interval of convergence:
$$lim_{ntoinfty} left| frac{(-1)^{n+1}2^{n+1}x^{n+1}}{(n+1)!} frac{n!}{(-1)^n 2^n x^n} right| = 2|x| lim_{ntoinfty} frac{1}{n+1} = 2|x|(0) < 1$$
Therefore for any value of x is in the interval of convergence. Is this right?
sequences-and-series
asked Apr 19 '17 at 18:52
TinlerTinler
517312
Find the Maclaurin series for $f(x) = e^{-2x}$, and find the interval of convergence for the series.
I got the maclaurin series to be this
$e^x = 1 - 2frac{x}{1!} + 2^2frac{x^2}{2!} + cdots$
$$sum_{n=0}^{infty} frac{(-1)^n (2)^n x^n}{n!}$$
Using the ratio test to find the interval of convergence:
$$lim_{ntoinfty} left| frac{(-1)^{n+1}2^{n+1}x^{n+1}}{(n+1)!} frac{n!}{(-1)^n 2^n x^n} right| = 2|x| lim_{ntoinfty} frac{1}{n+1} = 2|x|(0) < 1$$
Therefore for any value of x is in the interval of convergence. Is this right?
sequences-and-series
sequences-and-series
asked Apr 19 '17 at 18:52
TinlerTinler
517312
asked Apr 19 '17 at 18:52
TinlerTinler
517312
asked Apr 19 '17 at 18:52
TinlerTinler
517312
asked Apr 19 '17 at 18:52
TinlerTinler
517312
asked Apr 19 '17 at 18:52
TinlerTinler
517312
517312
Yes, you're right
– caverac
Apr 19 '17 at 18:57
Do I need to check the endpoints at x = 0, or no? Im a bit confused when the ratio test gives $infty$ or zero. If its $infty$ instead of 0, so like $2|x|infty > 1$ Can I say there is no point in the interval that is convergence?
– Tinler
Apr 19 '17 at 18:57
The limit is 0 regardless of the value of $x$, so no need to check anything else
– caverac
Apr 19 '17 at 18:58
There are no endpoints to check. The interval of convergence is $(-infty,infty)$.
– John Wayland Bales
Apr 19 '17 at 18:59
What happens if its $2|x|infty > 1$ ?. As in instead of 0 i put infinity.
– Tinler
Apr 19 '17 at 19:05
add a comment |
Yes, you're right
– caverac
Apr 19 '17 at 18:57
Do I need to check the endpoints at x = 0, or no? Im a bit confused when the ratio test gives $infty$ or zero. If its $infty$ instead of 0, so like $2|x|infty > 1$ Can I say there is no point in the interval that is convergence?
– Tinler
Apr 19 '17 at 18:57
The limit is 0 regardless of the value of $x$, so no need to check anything else
– caverac
Apr 19 '17 at 18:58
There are no endpoints to check. The interval of convergence is $(-infty,infty)$.
– John Wayland Bales
Apr 19 '17 at 18:59
What happens if its $2|x|infty > 1$ ?. As in instead of 0 i put infinity.
– Tinler
Apr 19 '17 at 19:05
Yes, you're right
– caverac
Apr 19 '17 at 18:57
Yes, you're right
– caverac
Apr 19 '17 at 18:57
Do I need to check the endpoints at x = 0, or no? Im a bit confused when the ratio test gives $infty$ or zero. If its $infty$ instead of 0, so like $2|x|infty > 1$ Can I say there is no point in the interval that is convergence?
– Tinler
Apr 19 '17 at 18:57
Do I need to check the endpoints at x = 0, or no? Im a bit confused when the ratio test gives $infty$ or zero. If its $infty$ instead of 0, so like $2|x|infty > 1$ Can I say there is no point in the interval that is convergence?
– Tinler
Apr 19 '17 at 18:57
The limit is 0 regardless of the value of $x$, so no need to check anything else
– caverac
Apr 19 '17 at 18:58
The limit is 0 regardless of the value of $x$, so no need to check anything else
– caverac
Apr 19 '17 at 18:58
There are no endpoints to check. The interval of convergence is $(-infty,infty)$.
– John Wayland Bales
Apr 19 '17 at 18:59
There are no endpoints to check. The interval of convergence is $(-infty,infty)$.
– John Wayland Bales
Apr 19 '17 at 18:59
What happens if its $2|x|infty > 1$ ?. As in instead of 0 i put infinity.
– Tinler
Apr 19 '17 at 19:05
What happens if its $2|x|infty > 1$ ?. As in instead of 0 i put infinity.
– Tinler
Apr 19 '17 at 19:05
add a comment |
2 Answers
2
active
oldest
votes
The series representation for the exponential function converges everywhere (uniformly as well), as so any related sum will also converge everywhere.
answered Apr 19 '17 at 18:55
Sean RobersonSean Roberson
6,39031327
add a comment |
This Maclaurin series is valid for $|x|<infty$.
You know this because the series is valid over that domain. There are no points where the function "blows up".
By convention, the Maclauren series in centered about $x=0$. A power series allows three possibilities:
- Series converges only at $x=0$.
- Series converges on an interval centered at $x=0$.
- Series converges for all values of $xinmathbb{R}$.
answered Apr 19 '17 at 18:56
dantopadantopa
6,42932142
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The series representation for the exponential function converges everywhere (uniformly as well), as so any related sum will also converge everywhere.
answered Apr 19 '17 at 18:55
Sean RobersonSean Roberson
6,39031327
add a comment |
The series representation for the exponential function converges everywhere (uniformly as well), as so any related sum will also converge everywhere.
answered Apr 19 '17 at 18:55
Sean RobersonSean Roberson
6,39031327
add a comment |
The series representation for the exponential function converges everywhere (uniformly as well), as so any related sum will also converge everywhere.
answered Apr 19 '17 at 18:55
Sean RobersonSean Roberson
6,39031327
The series representation for the exponential function converges everywhere (uniformly as well), as so any related sum will also converge everywhere.
answered Apr 19 '17 at 18:55
Sean RobersonSean Roberson
6,39031327
answered Apr 19 '17 at 18:55
Sean RobersonSean Roberson
6,39031327
answered Apr 19 '17 at 18:55
Sean RobersonSean Roberson
6,39031327
answered Apr 19 '17 at 18:55
Sean RobersonSean Roberson
6,39031327
6,39031327
add a comment |
add a comment |
This Maclaurin series is valid for $|x|<infty$.
You know this because the series is valid over that domain. There are no points where the function "blows up".
By convention, the Maclauren series in centered about $x=0$. A power series allows three possibilities:
- Series converges only at $x=0$.
- Series converges on an interval centered at $x=0$.
- Series converges for all values of $xinmathbb{R}$.
answered Apr 19 '17 at 18:56
dantopadantopa
6,42932142
add a comment |
This Maclaurin series is valid for $|x|<infty$.
You know this because the series is valid over that domain. There are no points where the function "blows up".
By convention, the Maclauren series in centered about $x=0$. A power series allows three possibilities:
- Series converges only at $x=0$.
- Series converges on an interval centered at $x=0$.
- Series converges for all values of $xinmathbb{R}$.
answered Apr 19 '17 at 18:56
dantopadantopa
6,42932142
add a comment |
This Maclaurin series is valid for $|x|<infty$.
You know this because the series is valid over that domain. There are no points where the function "blows up".
By convention, the Maclauren series in centered about $x=0$. A power series allows three possibilities:
- Series converges only at $x=0$.
- Series converges on an interval centered at $x=0$.
- Series converges for all values of $xinmathbb{R}$.
answered Apr 19 '17 at 18:56
dantopadantopa
6,42932142
This Maclaurin series is valid for $|x|<infty$.
You know this because the series is valid over that domain. There are no points where the function "blows up".
By convention, the Maclauren series in centered about $x=0$. A power series allows three possibilities:
- Series converges only at $x=0$.
- Series converges on an interval centered at $x=0$.
- Series converges for all values of $xinmathbb{R}$.
answered Apr 19 '17 at 18:56
dantopadantopa
6,42932142
answered Apr 19 '17 at 18:56
dantopadantopa
6,42932142
answered Apr 19 '17 at 18:56
dantopadantopa
6,42932142
answered Apr 19 '17 at 18:56
dantopadantopa
6,42932142
6,42932142
add a comment |
add a comment |
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Yes, you're right
– caverac
Apr 19 '17 at 18:57
Do I need to check the endpoints at x = 0, or no? Im a bit confused when the ratio test gives $infty$ or zero. If its $infty$ instead of 0, so like $2|x|infty > 1$ Can I say there is no point in the interval that is convergence?
– Tinler
Apr 19 '17 at 18:57
The limit is 0 regardless of the value of $x$, so no need to check anything else
– caverac
Apr 19 '17 at 18:58
There are no endpoints to check. The interval of convergence is $(-infty,infty)$.
– John Wayland Bales
Apr 19 '17 at 18:59
What happens if its $2|x|infty > 1$ ?. As in instead of 0 i put infinity.
– Tinler
Apr 19 '17 at 19:05