Solve the recursion given by $f(n) = 2f(n-1) + frac{(-1)^n}{n!}$ using generating functions.
$begingroup$
Assume that $f(0) = 0, f(1) = -1$.
$g(x)$ is $f(n)$'s generating function, I got to the expression:
$ g(x) = frac{e^{-x}-1}{1-2x} $
But am now stuck, since I can't find a power series for the function in simple terms. Any ideas?
combinatorics discrete-mathematics power-series generating-functions recursion
edited Jan 24 at 17:12
OmG
2,527822
asked Jan 24 at 16:44
Amit LevyAmit Levy
747
$endgroup$
add a comment |
$begingroup$
Assume that $f(0) = 0, f(1) = -1$.
$g(x)$ is $f(n)$'s generating function, I got to the expression:
$ g(x) = frac{e^{-x}-1}{1-2x} $
But am now stuck, since I can't find a power series for the function in simple terms. Any ideas?
combinatorics discrete-mathematics power-series generating-functions recursion
edited Jan 24 at 17:12
OmG
2,527822
asked Jan 24 at 16:44
Amit LevyAmit Levy
747
$endgroup$
add a comment |
$begingroup$
Assume that $f(0) = 0, f(1) = -1$.
$g(x)$ is $f(n)$'s generating function, I got to the expression:
$ g(x) = frac{e^{-x}-1}{1-2x} $
But am now stuck, since I can't find a power series for the function in simple terms. Any ideas?
combinatorics discrete-mathematics power-series generating-functions recursion
edited Jan 24 at 17:12
OmG
2,527822
asked Jan 24 at 16:44
Amit LevyAmit Levy
747
$endgroup$
Assume that $f(0) = 0, f(1) = -1$.
$g(x)$ is $f(n)$'s generating function, I got to the expression:
$ g(x) = frac{e^{-x}-1}{1-2x} $
But am now stuck, since I can't find a power series for the function in simple terms. Any ideas?
combinatorics discrete-mathematics power-series generating-functions recursion
combinatorics discrete-mathematics power-series generating-functions recursion
edited Jan 24 at 17:12
OmG
2,527822
asked Jan 24 at 16:44
Amit LevyAmit Levy
747
edited Jan 24 at 17:12
OmG
2,527822
asked Jan 24 at 16:44
Amit LevyAmit Levy
747
edited Jan 24 at 17:12
OmG
2,527822
edited Jan 24 at 17:12
OmG
2,527822
edited Jan 24 at 17:12
OmG
2,527822
2,527822
asked Jan 24 at 16:44
Amit LevyAmit Levy
747
asked Jan 24 at 16:44
Amit LevyAmit Levy
747
asked Jan 24 at 16:44
Amit LevyAmit Levy
747
747
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think you should have $e^{-x}$ in your formula, not $e^x$.
Now you have $g(x) = (e^{-x}-1) cdot frac{1}{1-2x}$. Can you find power series for those both separately? Do you know how to multiply power series?
answered Jan 24 at 17:00
kccukccu
10.6k11229
$endgroup$
$begingroup$
The exponent was just a spelling mistake which I now fixed. I do know how to multiply power series, but then I get a sum as the solution, which doesn't feel simplified enough. Is that usually considered a solution to a recursion?
$endgroup$
– Amit Levy
Jan 24 at 17:02
$begingroup$
Sometimes there may be a way to simplify the sum, but sometimes not.
$endgroup$
– kccu
Jan 24 at 17:12
$begingroup$
In any case, you should still be able to say what happens to $f(n)$ asymptotically, which may be of interest.
$endgroup$
– kccu
Jan 24 at 17:14
add a comment |
$begingroup$
A simpler solution might be expand the recursion equation here.
$$f(n) = sum_{i = 1}^{n} 2^{n-i} frac{(-1)^i}{i!} = 2^nsum_{i = 1}^{n} 2^{-i} frac{(-1)^i}{i!}$$
answered Jan 24 at 17:11
OmGOmG
2,527822
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
I think you should have $e^{-x}$ in your formula, not $e^x$.
Now you have $g(x) = (e^{-x}-1) cdot frac{1}{1-2x}$. Can you find power series for those both separately? Do you know how to multiply power series?
answered Jan 24 at 17:00
kccukccu
10.6k11229
$endgroup$
$begingroup$
The exponent was just a spelling mistake which I now fixed. I do know how to multiply power series, but then I get a sum as the solution, which doesn't feel simplified enough. Is that usually considered a solution to a recursion?
$endgroup$
– Amit Levy
Jan 24 at 17:02
$begingroup$
Sometimes there may be a way to simplify the sum, but sometimes not.
$endgroup$
– kccu
Jan 24 at 17:12
$begingroup$
In any case, you should still be able to say what happens to $f(n)$ asymptotically, which may be of interest.
$endgroup$
– kccu
Jan 24 at 17:14
add a comment |
$begingroup$
I think you should have $e^{-x}$ in your formula, not $e^x$.
Now you have $g(x) = (e^{-x}-1) cdot frac{1}{1-2x}$. Can you find power series for those both separately? Do you know how to multiply power series?
answered Jan 24 at 17:00
kccukccu
10.6k11229
$endgroup$
$begingroup$
The exponent was just a spelling mistake which I now fixed. I do know how to multiply power series, but then I get a sum as the solution, which doesn't feel simplified enough. Is that usually considered a solution to a recursion?
$endgroup$
– Amit Levy
Jan 24 at 17:02
$begingroup$
Sometimes there may be a way to simplify the sum, but sometimes not.
$endgroup$
– kccu
Jan 24 at 17:12
$begingroup$
In any case, you should still be able to say what happens to $f(n)$ asymptotically, which may be of interest.
$endgroup$
– kccu
Jan 24 at 17:14
add a comment |
$begingroup$
I think you should have $e^{-x}$ in your formula, not $e^x$.
Now you have $g(x) = (e^{-x}-1) cdot frac{1}{1-2x}$. Can you find power series for those both separately? Do you know how to multiply power series?
answered Jan 24 at 17:00
kccukccu
10.6k11229
$endgroup$
I think you should have $e^{-x}$ in your formula, not $e^x$.
Now you have $g(x) = (e^{-x}-1) cdot frac{1}{1-2x}$. Can you find power series for those both separately? Do you know how to multiply power series?
answered Jan 24 at 17:00
kccukccu
10.6k11229
answered Jan 24 at 17:00
kccukccu
10.6k11229
answered Jan 24 at 17:00
kccukccu
10.6k11229
answered Jan 24 at 17:00
kccukccu
10.6k11229
10.6k11229
$begingroup$
The exponent was just a spelling mistake which I now fixed. I do know how to multiply power series, but then I get a sum as the solution, which doesn't feel simplified enough. Is that usually considered a solution to a recursion?
$endgroup$
– Amit Levy
Jan 24 at 17:02
$begingroup$
Sometimes there may be a way to simplify the sum, but sometimes not.
$endgroup$
– kccu
Jan 24 at 17:12
$begingroup$
In any case, you should still be able to say what happens to $f(n)$ asymptotically, which may be of interest.
$endgroup$
– kccu
Jan 24 at 17:14
add a comment |
$begingroup$
The exponent was just a spelling mistake which I now fixed. I do know how to multiply power series, but then I get a sum as the solution, which doesn't feel simplified enough. Is that usually considered a solution to a recursion?
$endgroup$
– Amit Levy
Jan 24 at 17:02
$begingroup$
Sometimes there may be a way to simplify the sum, but sometimes not.
$endgroup$
– kccu
Jan 24 at 17:12
$begingroup$
In any case, you should still be able to say what happens to $f(n)$ asymptotically, which may be of interest.
$endgroup$
– kccu
Jan 24 at 17:14
$begingroup$
The exponent was just a spelling mistake which I now fixed. I do know how to multiply power series, but then I get a sum as the solution, which doesn't feel simplified enough. Is that usually considered a solution to a recursion?
$endgroup$
– Amit Levy
Jan 24 at 17:02
$begingroup$
The exponent was just a spelling mistake which I now fixed. I do know how to multiply power series, but then I get a sum as the solution, which doesn't feel simplified enough. Is that usually considered a solution to a recursion?
$endgroup$
– Amit Levy
Jan 24 at 17:02
$begingroup$
Sometimes there may be a way to simplify the sum, but sometimes not.
$endgroup$
– kccu
Jan 24 at 17:12
$begingroup$
Sometimes there may be a way to simplify the sum, but sometimes not.
$endgroup$
– kccu
Jan 24 at 17:12
$begingroup$
In any case, you should still be able to say what happens to $f(n)$ asymptotically, which may be of interest.
$endgroup$
– kccu
Jan 24 at 17:14
$begingroup$
In any case, you should still be able to say what happens to $f(n)$ asymptotically, which may be of interest.
$endgroup$
– kccu
Jan 24 at 17:14
add a comment |
$begingroup$
A simpler solution might be expand the recursion equation here.
$$f(n) = sum_{i = 1}^{n} 2^{n-i} frac{(-1)^i}{i!} = 2^nsum_{i = 1}^{n} 2^{-i} frac{(-1)^i}{i!}$$
answered Jan 24 at 17:11
OmGOmG
2,527822
$endgroup$
add a comment |
$begingroup$
A simpler solution might be expand the recursion equation here.
$$f(n) = sum_{i = 1}^{n} 2^{n-i} frac{(-1)^i}{i!} = 2^nsum_{i = 1}^{n} 2^{-i} frac{(-1)^i}{i!}$$
answered Jan 24 at 17:11
OmGOmG
2,527822
$endgroup$
add a comment |
$begingroup$
A simpler solution might be expand the recursion equation here.
$$f(n) = sum_{i = 1}^{n} 2^{n-i} frac{(-1)^i}{i!} = 2^nsum_{i = 1}^{n} 2^{-i} frac{(-1)^i}{i!}$$
answered Jan 24 at 17:11
OmGOmG
2,527822
$endgroup$
A simpler solution might be expand the recursion equation here.
$$f(n) = sum_{i = 1}^{n} 2^{n-i} frac{(-1)^i}{i!} = 2^nsum_{i = 1}^{n} 2^{-i} frac{(-1)^i}{i!}$$
answered Jan 24 at 17:11
OmGOmG
2,527822
answered Jan 24 at 17:11
OmGOmG
2,527822
answered Jan 24 at 17:11
OmGOmG
2,527822
answered Jan 24 at 17:11
OmGOmG
2,527822
2,527822
add a comment |
add a comment |
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