Does $f(x) = sum_{i = 1}^{infty} e^{-xi}y(1-y)^{i-1}$ have a closed form?
Does $f(x) = sum_{i = 1}^{infty} e^{-xi}y(1-y)^{i-1}$ have a closed form? $y$ is assumed to be in $[0,1]$ and $x >0$.
What I tried: I tried differentiating/integrating the function term by term to see whether those are something that can be calculated, but it seems that this method did not work.
real-analysis analysis
asked Nov 24 at 22:24
penny
635
add a comment |
Does $f(x) = sum_{i = 1}^{infty} e^{-xi}y(1-y)^{i-1}$ have a closed form? $y$ is assumed to be in $[0,1]$ and $x >0$.
What I tried: I tried differentiating/integrating the function term by term to see whether those are something that can be calculated, but it seems that this method did not work.
real-analysis analysis
asked Nov 24 at 22:24
penny
635
add a comment |
Does $f(x) = sum_{i = 1}^{infty} e^{-xi}y(1-y)^{i-1}$ have a closed form? $y$ is assumed to be in $[0,1]$ and $x >0$.
What I tried: I tried differentiating/integrating the function term by term to see whether those are something that can be calculated, but it seems that this method did not work.
real-analysis analysis
asked Nov 24 at 22:24
penny
635
Does $f(x) = sum_{i = 1}^{infty} e^{-xi}y(1-y)^{i-1}$ have a closed form? $y$ is assumed to be in $[0,1]$ and $x >0$.
What I tried: I tried differentiating/integrating the function term by term to see whether those are something that can be calculated, but it seems that this method did not work.
real-analysis analysis
real-analysis analysis
asked Nov 24 at 22:24
penny
635
asked Nov 24 at 22:24
penny
635
asked Nov 24 at 22:24
penny
635
asked Nov 24 at 22:24
penny
635
asked Nov 24 at 22:24
penny
635
635
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Observe
begin{align}
f(x) = frac{y}{1-y}sum^infty_{i=1} [e^{-x}(1-y)]^i
end{align}
then use geometric series.
answered Nov 24 at 22:26
Jacky Chong
17.6k21128
(The case $y=1$ obviously has to be dealt with separately.)
– Clement C.
Nov 24 at 23:35
@ClementC. Yes. I will leave it as an exercise. =)
– Jacky Chong
Nov 24 at 23:43
^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
– Chase Ryan Taylor
Nov 24 at 23:44
@ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
– Clement C.
Nov 25 at 0:00
@ClementC. Which term is that?
– Chase Ryan Taylor
Nov 25 at 0:12
|
show 2 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Observe
begin{align}
f(x) = frac{y}{1-y}sum^infty_{i=1} [e^{-x}(1-y)]^i
end{align}
then use geometric series.
answered Nov 24 at 22:26
Jacky Chong
17.6k21128
(The case $y=1$ obviously has to be dealt with separately.)
– Clement C.
Nov 24 at 23:35
@ClementC. Yes. I will leave it as an exercise. =)
– Jacky Chong
Nov 24 at 23:43
^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
– Chase Ryan Taylor
Nov 24 at 23:44
@ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
– Clement C.
Nov 25 at 0:00
@ClementC. Which term is that?
– Chase Ryan Taylor
Nov 25 at 0:12
|
show 2 more comments
Observe
begin{align}
f(x) = frac{y}{1-y}sum^infty_{i=1} [e^{-x}(1-y)]^i
end{align}
then use geometric series.
answered Nov 24 at 22:26
Jacky Chong
17.6k21128
(The case $y=1$ obviously has to be dealt with separately.)
– Clement C.
Nov 24 at 23:35
@ClementC. Yes. I will leave it as an exercise. =)
– Jacky Chong
Nov 24 at 23:43
^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
– Chase Ryan Taylor
Nov 24 at 23:44
@ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
– Clement C.
Nov 25 at 0:00
@ClementC. Which term is that?
– Chase Ryan Taylor
Nov 25 at 0:12
|
show 2 more comments
Observe
begin{align}
f(x) = frac{y}{1-y}sum^infty_{i=1} [e^{-x}(1-y)]^i
end{align}
then use geometric series.
answered Nov 24 at 22:26
Jacky Chong
17.6k21128
Observe
begin{align}
f(x) = frac{y}{1-y}sum^infty_{i=1} [e^{-x}(1-y)]^i
end{align}
then use geometric series.
answered Nov 24 at 22:26
Jacky Chong
17.6k21128
answered Nov 24 at 22:26
Jacky Chong
17.6k21128
answered Nov 24 at 22:26
Jacky Chong
17.6k21128
answered Nov 24 at 22:26
Jacky Chong
17.6k21128
17.6k21128
(The case $y=1$ obviously has to be dealt with separately.)
– Clement C.
Nov 24 at 23:35
@ClementC. Yes. I will leave it as an exercise. =)
– Jacky Chong
Nov 24 at 23:43
^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
– Chase Ryan Taylor
Nov 24 at 23:44
@ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
– Clement C.
Nov 25 at 0:00
@ClementC. Which term is that?
– Chase Ryan Taylor
Nov 25 at 0:12
|
show 2 more comments
(The case $y=1$ obviously has to be dealt with separately.)
– Clement C.
Nov 24 at 23:35
@ClementC. Yes. I will leave it as an exercise. =)
– Jacky Chong
Nov 24 at 23:43
^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
– Chase Ryan Taylor
Nov 24 at 23:44
@ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
– Clement C.
Nov 25 at 0:00
@ClementC. Which term is that?
– Chase Ryan Taylor
Nov 25 at 0:12
(The case $y=1$ obviously has to be dealt with separately.)
– Clement C.
Nov 24 at 23:35
(The case $y=1$ obviously has to be dealt with separately.)
– Clement C.
Nov 24 at 23:35
@ClementC. Yes. I will leave it as an exercise. =)
– Jacky Chong
Nov 24 at 23:43
@ClementC. Yes. I will leave it as an exercise. =)
– Jacky Chong
Nov 24 at 23:43
^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
– Chase Ryan Taylor
Nov 24 at 23:44
^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
– Chase Ryan Taylor
Nov 24 at 23:44
@ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
– Clement C.
Nov 25 at 0:00
@ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
– Clement C.
Nov 25 at 0:00
@ClementC. Which term is that?
– Chase Ryan Taylor
Nov 25 at 0:12
@ClementC. Which term is that?
– Chase Ryan Taylor
Nov 25 at 0:12
|
show 2 more comments
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