Mixing
Special case in power series
$begingroup$
I know how to write $e^x$ as power series. How can I write $e^{-x}$ as power series? Is it
$$
sum_{n=0}^infty (-1)^n frac{x^n}{n!}?
$$
power-series fourier-series
edited Jan 27 at 14:29
Daniele Tampieri
2,5772922
asked Jan 27 at 13:21
Jade DJade D
31
$endgroup$
add a comment |
$begingroup$
I know how to write $e^x$ as power series. How can I write $e^{-x}$ as power series? Is it
$$
sum_{n=0}^infty (-1)^n frac{x^n}{n!}?
$$
power-series fourier-series
edited Jan 27 at 14:29
Daniele Tampieri
2,5772922
asked Jan 27 at 13:21
Jade DJade D
31
$endgroup$
add a comment |
$begingroup$
I know how to write $e^x$ as power series. How can I write $e^{-x}$ as power series? Is it
$$
sum_{n=0}^infty (-1)^n frac{x^n}{n!}?
$$
power-series fourier-series
edited Jan 27 at 14:29
Daniele Tampieri
2,5772922
asked Jan 27 at 13:21
Jade DJade D
31
$endgroup$
I know how to write $e^x$ as power series. How can I write $e^{-x}$ as power series? Is it
$$
sum_{n=0}^infty (-1)^n frac{x^n}{n!}?
$$
power-series fourier-series
power-series fourier-series
edited Jan 27 at 14:29
Daniele Tampieri
2,5772922
asked Jan 27 at 13:21
Jade DJade D
31
edited Jan 27 at 14:29
Daniele Tampieri
2,5772922
asked Jan 27 at 13:21
Jade DJade D
31
edited Jan 27 at 14:29
Daniele Tampieri
2,5772922
edited Jan 27 at 14:29
Daniele Tampieri
2,5772922
edited Jan 27 at 14:29
Daniele Tampieri
2,5772922
2,5772922
asked Jan 27 at 13:21
Jade DJade D
31
asked Jan 27 at 13:21
Jade DJade D
31
asked Jan 27 at 13:21
Jade DJade D
31
31
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $$ e^x = sum frac{ x^n }{n!} $$
then
$$ e^{-x} = sum frac{ (-x)^n }{n!} = sum frac{ (-1)^n x^n }{n!} $$
answered Jan 27 at 13:22
Jimmy SabaterJimmy Sabater
2,560325
$endgroup$
add a comment |
$begingroup$
Substitute $-x$ for $x$ in the power series for $e^x$: $e^{-x}=sum_{n=0}^infty frac{(-x)^n}{n!}=sum_{n=0}^inftyfrac{(-1)^nx^n}{n!}$.
answered Jan 27 at 13:30
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Since $$ e^x = sum frac{ x^n }{n!} $$
then
$$ e^{-x} = sum frac{ (-x)^n }{n!} = sum frac{ (-1)^n x^n }{n!} $$
answered Jan 27 at 13:22
Jimmy SabaterJimmy Sabater
2,560325
$endgroup$
add a comment |
$begingroup$
Since $$ e^x = sum frac{ x^n }{n!} $$
then
$$ e^{-x} = sum frac{ (-x)^n }{n!} = sum frac{ (-1)^n x^n }{n!} $$
answered Jan 27 at 13:22
Jimmy SabaterJimmy Sabater
2,560325
$endgroup$
add a comment |
$begingroup$
Since $$ e^x = sum frac{ x^n }{n!} $$
then
$$ e^{-x} = sum frac{ (-x)^n }{n!} = sum frac{ (-1)^n x^n }{n!} $$
answered Jan 27 at 13:22
Jimmy SabaterJimmy Sabater
2,560325
$endgroup$
Since $$ e^x = sum frac{ x^n }{n!} $$
then
$$ e^{-x} = sum frac{ (-x)^n }{n!} = sum frac{ (-1)^n x^n }{n!} $$
answered Jan 27 at 13:22
Jimmy SabaterJimmy Sabater
2,560325
answered Jan 27 at 13:22
Jimmy SabaterJimmy Sabater
2,560325
answered Jan 27 at 13:22
Jimmy SabaterJimmy Sabater
2,560325
answered Jan 27 at 13:22
Jimmy SabaterJimmy Sabater
2,560325
2,560325
add a comment |
add a comment |
$begingroup$
Substitute $-x$ for $x$ in the power series for $e^x$: $e^{-x}=sum_{n=0}^infty frac{(-x)^n}{n!}=sum_{n=0}^inftyfrac{(-1)^nx^n}{n!}$.
answered Jan 27 at 13:30
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Substitute $-x$ for $x$ in the power series for $e^x$: $e^{-x}=sum_{n=0}^infty frac{(-x)^n}{n!}=sum_{n=0}^inftyfrac{(-1)^nx^n}{n!}$.
answered Jan 27 at 13:30
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Substitute $-x$ for $x$ in the power series for $e^x$: $e^{-x}=sum_{n=0}^infty frac{(-x)^n}{n!}=sum_{n=0}^inftyfrac{(-1)^nx^n}{n!}$.
answered Jan 27 at 13:30
Chris CusterChris Custer
14.2k3827
$endgroup$
Substitute $-x$ for $x$ in the power series for $e^x$: $e^{-x}=sum_{n=0}^infty frac{(-x)^n}{n!}=sum_{n=0}^inftyfrac{(-1)^nx^n}{n!}$.
answered Jan 27 at 13:30
Chris CusterChris Custer
14.2k3827
answered Jan 27 at 13:30
Chris CusterChris Custer
14.2k3827
answered Jan 27 at 13:30
Chris CusterChris Custer
14.2k3827
answered Jan 27 at 13:30
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
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