Expression for sum of $n$ exponentials
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So I have this sum of exponentials and I would like to find an expression for it.
$$sum^n_i=1 e^mu(i-1) $$
Note that $i$ is not an imaginary indicator. I am aware there is a formula for summing a purely exponential sum, but I am not clear as to what happens to the $mu$.
sequences-and-series summation geometric-series
edited Jan 13 at 8:23
TheSimpliFire
12.4k62460
asked Jan 12 at 20:22
DioDio
998
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add a comment |
$begingroup$
So I have this sum of exponentials and I would like to find an expression for it.
$$sum^n_i=1 e^mu(i-1) $$
Note that $i$ is not an imaginary indicator. I am aware there is a formula for summing a purely exponential sum, but I am not clear as to what happens to the $mu$.
sequences-and-series summation geometric-series
edited Jan 13 at 8:23
TheSimpliFire
12.4k62460
asked Jan 12 at 20:22
DioDio
998
$endgroup$
add a comment |
$begingroup$
So I have this sum of exponentials and I would like to find an expression for it.
$$sum^n_i=1 e^mu(i-1) $$
Note that $i$ is not an imaginary indicator. I am aware there is a formula for summing a purely exponential sum, but I am not clear as to what happens to the $mu$.
sequences-and-series summation geometric-series
edited Jan 13 at 8:23
TheSimpliFire
12.4k62460
asked Jan 12 at 20:22
DioDio
998
$endgroup$
So I have this sum of exponentials and I would like to find an expression for it.
$$sum^n_i=1 e^mu(i-1) $$
Note that $i$ is not an imaginary indicator. I am aware there is a formula for summing a purely exponential sum, but I am not clear as to what happens to the $mu$.
sequences-and-series summation geometric-series
sequences-and-series summation geometric-series
edited Jan 13 at 8:23
TheSimpliFire
12.4k62460
asked Jan 12 at 20:22
DioDio
998
edited Jan 13 at 8:23
TheSimpliFire
12.4k62460
asked Jan 12 at 20:22
DioDio
998
edited Jan 13 at 8:23
TheSimpliFire
12.4k62460
edited Jan 13 at 8:23
TheSimpliFire
12.4k62460
edited Jan 13 at 8:23
TheSimpliFire
12.4k62460
12.4k62460
asked Jan 12 at 20:22
DioDio
998
asked Jan 12 at 20:22
DioDio
998
asked Jan 12 at 20:22
DioDio
998
998
add a comment |
add a comment |
4 Answers
4
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Define $a=e^mu$ when $mune 0$. Then you have $$sum^n_i=1 e^mu(i-1) =sum^n_i=1 a^i-1=1+a+cdots+a^n-1=a^n-1over a-1=e^mu n-1over e^mu -1 $$For $mu =0 $ we obtain$$sum^n_i=1 e^mu(i-1)=n$$
answered Jan 12 at 20:52
Mostafa AyazMostafa Ayaz
15.3k3939
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The sum $S$ can be rewritten as $$S= sum_i=0^n-1 e^mu i=sum_i=0^n-1 (e^mu)^i=frac1-e^mu n1-e^mu$$ since the geometric series $$sum_i=0^n-1 x^i=frac1-x^n1-x$$
answered Jan 12 at 20:29
aledenaleden
2,032511
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add a comment |
$begingroup$
One may recall that
$$
sum_i=1^nx^i-1=frac1-x^n1-x,qquad xneq1.
$$ What if you put $x=e^mu$?
answered Jan 12 at 20:29
Olivier OloaOlivier Oloa
108k17177294
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Hint:
This the sum of the $n$ first terms of the geometric series with ratio $mathrm e^mu$, since $;mathrm e^mu(i-1)=(mathrm e^mu)^i-1$.
answered Jan 12 at 20:29
BernardBernard
119k740113
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Define $a=e^mu$ when $mune 0$. Then you have $$sum^n_i=1 e^mu(i-1) =sum^n_i=1 a^i-1=1+a+cdots+a^n-1=a^n-1over a-1=e^mu n-1over e^mu -1 $$For $mu =0 $ we obtain$$sum^n_i=1 e^mu(i-1)=n$$
answered Jan 12 at 20:52
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
Define $a=e^mu$ when $mune 0$. Then you have $$sum^n_i=1 e^mu(i-1) =sum^n_i=1 a^i-1=1+a+cdots+a^n-1=a^n-1over a-1=e^mu n-1over e^mu -1 $$For $mu =0 $ we obtain$$sum^n_i=1 e^mu(i-1)=n$$
answered Jan 12 at 20:52
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
Define $a=e^mu$ when $mune 0$. Then you have $$sum^n_i=1 e^mu(i-1) =sum^n_i=1 a^i-1=1+a+cdots+a^n-1=a^n-1over a-1=e^mu n-1over e^mu -1 $$For $mu =0 $ we obtain$$sum^n_i=1 e^mu(i-1)=n$$
answered Jan 12 at 20:52
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
Define $a=e^mu$ when $mune 0$. Then you have $$sum^n_i=1 e^mu(i-1) =sum^n_i=1 a^i-1=1+a+cdots+a^n-1=a^n-1over a-1=e^mu n-1over e^mu -1 $$For $mu =0 $ we obtain$$sum^n_i=1 e^mu(i-1)=n$$
answered Jan 12 at 20:52
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 12 at 20:52
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 12 at 20:52
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 12 at 20:52
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
add a comment |
add a comment |
$begingroup$
The sum $S$ can be rewritten as $$S= sum_i=0^n-1 e^mu i=sum_i=0^n-1 (e^mu)^i=frac1-e^mu n1-e^mu$$ since the geometric series $$sum_i=0^n-1 x^i=frac1-x^n1-x$$
answered Jan 12 at 20:29
aledenaleden
2,032511
$endgroup$
add a comment |
$begingroup$
The sum $S$ can be rewritten as $$S= sum_i=0^n-1 e^mu i=sum_i=0^n-1 (e^mu)^i=frac1-e^mu n1-e^mu$$ since the geometric series $$sum_i=0^n-1 x^i=frac1-x^n1-x$$
answered Jan 12 at 20:29
aledenaleden
2,032511
$endgroup$
add a comment |
$begingroup$
The sum $S$ can be rewritten as $$S= sum_i=0^n-1 e^mu i=sum_i=0^n-1 (e^mu)^i=frac1-e^mu n1-e^mu$$ since the geometric series $$sum_i=0^n-1 x^i=frac1-x^n1-x$$
answered Jan 12 at 20:29
aledenaleden
2,032511
$endgroup$
The sum $S$ can be rewritten as $$S= sum_i=0^n-1 e^mu i=sum_i=0^n-1 (e^mu)^i=frac1-e^mu n1-e^mu$$ since the geometric series $$sum_i=0^n-1 x^i=frac1-x^n1-x$$
answered Jan 12 at 20:29
aledenaleden
2,032511
answered Jan 12 at 20:29
aledenaleden
2,032511
answered Jan 12 at 20:29
aledenaleden
2,032511
answered Jan 12 at 20:29
aledenaleden
2,032511
2,032511
add a comment |
add a comment |
$begingroup$
One may recall that
$$
sum_i=1^nx^i-1=frac1-x^n1-x,qquad xneq1.
$$ What if you put $x=e^mu$?
answered Jan 12 at 20:29
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
add a comment |
$begingroup$
One may recall that
$$
sum_i=1^nx^i-1=frac1-x^n1-x,qquad xneq1.
$$ What if you put $x=e^mu$?
answered Jan 12 at 20:29
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
add a comment |
$begingroup$
One may recall that
$$
sum_i=1^nx^i-1=frac1-x^n1-x,qquad xneq1.
$$ What if you put $x=e^mu$?
answered Jan 12 at 20:29
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
One may recall that
$$
sum_i=1^nx^i-1=frac1-x^n1-x,qquad xneq1.
$$ What if you put $x=e^mu$?
answered Jan 12 at 20:29
Olivier OloaOlivier Oloa
108k17177294
answered Jan 12 at 20:29
Olivier OloaOlivier Oloa
108k17177294
answered Jan 12 at 20:29
Olivier OloaOlivier Oloa
108k17177294
answered Jan 12 at 20:29
Olivier OloaOlivier Oloa
108k17177294
108k17177294
add a comment |
add a comment |
$begingroup$
Hint:
This the sum of the $n$ first terms of the geometric series with ratio $mathrm e^mu$, since $;mathrm e^mu(i-1)=(mathrm e^mu)^i-1$.
answered Jan 12 at 20:29
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
Hint:
This the sum of the $n$ first terms of the geometric series with ratio $mathrm e^mu$, since $;mathrm e^mu(i-1)=(mathrm e^mu)^i-1$.
answered Jan 12 at 20:29
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
Hint:
This the sum of the $n$ first terms of the geometric series with ratio $mathrm e^mu$, since $;mathrm e^mu(i-1)=(mathrm e^mu)^i-1$.
answered Jan 12 at 20:29
BernardBernard
119k740113
$endgroup$
Hint:
This the sum of the $n$ first terms of the geometric series with ratio $mathrm e^mu$, since $;mathrm e^mu(i-1)=(mathrm e^mu)^i-1$.
answered Jan 12 at 20:29
BernardBernard
119k740113
answered Jan 12 at 20:29
BernardBernard
119k740113
answered Jan 12 at 20:29
BernardBernard
119k740113
answered Jan 12 at 20:29
BernardBernard
119k740113
119k740113
add a comment |
add a comment |
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