Mixing
Expression for sum of $n$ exponentials
$begingroup$
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
$endgroup$
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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$begingroup$
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
15.6k3939
$endgroup$
add a comment |
$begingroup$
The sum $S$ can be rewritten as $$S= sum_{i=0}^{n-1} e^{mu i}=sum_{i=0}^{n-1} (e^{mu})^i=frac{1-e^{mu n}}{1-e^{mu}}$$ since the geometric series $$sum_{i=0}^{n-1} x^{i}=frac{1-x^n}{1-x}$$
answered Jan 12 at 20:29
aledenaleden
2,327511
$endgroup$
add a comment |
$begingroup$
One may recall that
$$
sum_{i=1}^nx^{i-1}=frac{1-x^n}{1-x},qquad xneq1.
$$ What if you put $x=e^mu$?
answered Jan 12 at 20:29
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
add a comment |
$begingroup$
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
121k740116
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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
15.6k3939
$endgroup$
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
15.6k3939
$endgroup$
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
15.6k3939
$endgroup$
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
15.6k3939
answered Jan 12 at 20:52
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 20:52
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 20:52
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
$begingroup$
The sum $S$ can be rewritten as $$S= sum_{i=0}^{n-1} e^{mu i}=sum_{i=0}^{n-1} (e^{mu})^i=frac{1-e^{mu n}}{1-e^{mu}}$$ since the geometric series $$sum_{i=0}^{n-1} x^{i}=frac{1-x^n}{1-x}$$
answered Jan 12 at 20:29
aledenaleden
2,327511
$endgroup$
add a comment |
$begingroup$
The sum $S$ can be rewritten as $$S= sum_{i=0}^{n-1} e^{mu i}=sum_{i=0}^{n-1} (e^{mu})^i=frac{1-e^{mu n}}{1-e^{mu}}$$ since the geometric series $$sum_{i=0}^{n-1} x^{i}=frac{1-x^n}{1-x}$$
answered Jan 12 at 20:29
aledenaleden
2,327511
$endgroup$
add a comment |
$begingroup$
The sum $S$ can be rewritten as $$S= sum_{i=0}^{n-1} e^{mu i}=sum_{i=0}^{n-1} (e^{mu})^i=frac{1-e^{mu n}}{1-e^{mu}}$$ since the geometric series $$sum_{i=0}^{n-1} x^{i}=frac{1-x^n}{1-x}$$
answered Jan 12 at 20:29
aledenaleden
2,327511
$endgroup$
The sum $S$ can be rewritten as $$S= sum_{i=0}^{n-1} e^{mu i}=sum_{i=0}^{n-1} (e^{mu})^i=frac{1-e^{mu n}}{1-e^{mu}}$$ since the geometric series $$sum_{i=0}^{n-1} x^{i}=frac{1-x^n}{1-x}$$
answered Jan 12 at 20:29
aledenaleden
2,327511
answered Jan 12 at 20:29
aledenaleden
2,327511
answered Jan 12 at 20:29
aledenaleden
2,327511
answered Jan 12 at 20:29
aledenaleden
2,327511
2,327511
add a comment |
add a comment |
$begingroup$
One may recall that
$$
sum_{i=1}^nx^{i-1}=frac{1-x^n}{1-x},qquad xneq1.
$$ 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
$$
sum_{i=1}^nx^{i-1}=frac{1-x^n}{1-x},qquad xneq1.
$$ 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
$$
sum_{i=1}^nx^{i-1}=frac{1-x^n}{1-x},qquad xneq1.
$$ What if you put $x=e^mu$?
answered Jan 12 at 20:29
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
One may recall that
$$
sum_{i=1}^nx^{i-1}=frac{1-x^n}{1-x},qquad xneq1.
$$ 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
121k740116
$endgroup$
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
121k740116
$endgroup$
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
121k740116
$endgroup$
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
121k740116
answered Jan 12 at 20:29
BernardBernard
121k740116
answered Jan 12 at 20:29
BernardBernard
121k740116
answered Jan 12 at 20:29
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
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