Mixing
Getting an explicit formula from a recursive one.
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I have a recursive formula of a sequence and I would like to find an explicit version, if possible. Here it is:
$$a_k = a_{k-1} cdot k - k + 2$$
$$a_2 := 6$$
Is there any general approach I can take? Anything I tried hasn't worked yet (though I haven't been working on it for that long, so if you can just push me in the right direction that's great too). If I knew the formula, I could prove it by induction, but I don't. Thanks in advance.
sequences-and-series
asked Jan 21 at 15:20
MateInTwoMateInTwo
207
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add a comment |
$begingroup$
I have a recursive formula of a sequence and I would like to find an explicit version, if possible. Here it is:
$$a_k = a_{k-1} cdot k - k + 2$$
$$a_2 := 6$$
Is there any general approach I can take? Anything I tried hasn't worked yet (though I haven't been working on it for that long, so if you can just push me in the right direction that's great too). If I knew the formula, I could prove it by induction, but I don't. Thanks in advance.
sequences-and-series
asked Jan 21 at 15:20
MateInTwoMateInTwo
207
$endgroup$
2
$begingroup$
I think you mean to say "explicit" not "implicit".
$endgroup$
– saulspatz
Jan 21 at 15:24
$begingroup$
Of course, sorry
$endgroup$
– MateInTwo
Jan 21 at 15:39
add a comment |
$begingroup$
I have a recursive formula of a sequence and I would like to find an explicit version, if possible. Here it is:
$$a_k = a_{k-1} cdot k - k + 2$$
$$a_2 := 6$$
Is there any general approach I can take? Anything I tried hasn't worked yet (though I haven't been working on it for that long, so if you can just push me in the right direction that's great too). If I knew the formula, I could prove it by induction, but I don't. Thanks in advance.
sequences-and-series
asked Jan 21 at 15:20
MateInTwoMateInTwo
207
$endgroup$
I have a recursive formula of a sequence and I would like to find an explicit version, if possible. Here it is:
$$a_k = a_{k-1} cdot k - k + 2$$
$$a_2 := 6$$
Is there any general approach I can take? Anything I tried hasn't worked yet (though I haven't been working on it for that long, so if you can just push me in the right direction that's great too). If I knew the formula, I could prove it by induction, but I don't. Thanks in advance.
sequences-and-series
sequences-and-series
asked Jan 21 at 15:20
MateInTwoMateInTwo
207
asked Jan 21 at 15:20
MateInTwoMateInTwo
207
edited Jan 21 at 15:39
MateInTwo
asked Jan 21 at 15:20
MateInTwoMateInTwo
207
asked Jan 21 at 15:20
MateInTwoMateInTwo
207
asked Jan 21 at 15:20
MateInTwoMateInTwo
207
207
2
$begingroup$
I think you mean to say "explicit" not "implicit".
$endgroup$
– saulspatz
Jan 21 at 15:24
$begingroup$
Of course, sorry
$endgroup$
– MateInTwo
Jan 21 at 15:39
add a comment |
2
$begingroup$
I think you mean to say "explicit" not "implicit".
$endgroup$
– saulspatz
Jan 21 at 15:24
$begingroup$
Of course, sorry
$endgroup$
– MateInTwo
Jan 21 at 15:39
2
2
$begingroup$
I think you mean to say "explicit" not "implicit".
$endgroup$
– saulspatz
Jan 21 at 15:24
$begingroup$
I think you mean to say "explicit" not "implicit".
$endgroup$
– saulspatz
Jan 21 at 15:24
$begingroup$
Of course, sorry
$endgroup$
– MateInTwo
Jan 21 at 15:39
$begingroup$
Of course, sorry
$endgroup$
– MateInTwo
Jan 21 at 15:39
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Hint: Divide both sides by $k!$ and let $$b_k=frac{a_k}{k!}.$$
answered Jan 21 at 15:22
SongSong
16.7k11044
$endgroup$
$begingroup$
Oh wow, that was absolutely fantastic, thanks! So $a_k leq k!e +1$ is always true. That was really fun. How'd you get the idea to divide by k! ? Don't think I could've thought of that.
$endgroup$
– MateInTwo
Jan 22 at 18:51
add a comment |
$begingroup$
$a_k = ka_{k-1} - k + 2 = k((k-1)a_{k-2} - (k -1) + 2) -k + 2 = ... = k!a_{0} +2 - k + k(2 - (k-1)) + ... + k!(2 - (k-k))$
It may be possible to simplify the sum.
answered Jan 21 at 15:28
lightxbulblightxbulb
1,040311
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
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active
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$begingroup$
Hint: Divide both sides by $k!$ and let $$b_k=frac{a_k}{k!}.$$
answered Jan 21 at 15:22
SongSong
16.7k11044
$endgroup$
$begingroup$
Oh wow, that was absolutely fantastic, thanks! So $a_k leq k!e +1$ is always true. That was really fun. How'd you get the idea to divide by k! ? Don't think I could've thought of that.
$endgroup$
– MateInTwo
Jan 22 at 18:51
add a comment |
$begingroup$
Hint: Divide both sides by $k!$ and let $$b_k=frac{a_k}{k!}.$$
answered Jan 21 at 15:22
SongSong
16.7k11044
$endgroup$
$begingroup$
Oh wow, that was absolutely fantastic, thanks! So $a_k leq k!e +1$ is always true. That was really fun. How'd you get the idea to divide by k! ? Don't think I could've thought of that.
$endgroup$
– MateInTwo
Jan 22 at 18:51
add a comment |
$begingroup$
Hint: Divide both sides by $k!$ and let $$b_k=frac{a_k}{k!}.$$
answered Jan 21 at 15:22
SongSong
16.7k11044
$endgroup$
Hint: Divide both sides by $k!$ and let $$b_k=frac{a_k}{k!}.$$
answered Jan 21 at 15:22
SongSong
16.7k11044
answered Jan 21 at 15:22
SongSong
16.7k11044
answered Jan 21 at 15:22
SongSong
16.7k11044
answered Jan 21 at 15:22
SongSong
16.7k11044
16.7k11044
$begingroup$
Oh wow, that was absolutely fantastic, thanks! So $a_k leq k!e +1$ is always true. That was really fun. How'd you get the idea to divide by k! ? Don't think I could've thought of that.
$endgroup$
– MateInTwo
Jan 22 at 18:51
add a comment |
$begingroup$
Oh wow, that was absolutely fantastic, thanks! So $a_k leq k!e +1$ is always true. That was really fun. How'd you get the idea to divide by k! ? Don't think I could've thought of that.
$endgroup$
– MateInTwo
Jan 22 at 18:51
$begingroup$
Oh wow, that was absolutely fantastic, thanks! So $a_k leq k!e +1$ is always true. That was really fun. How'd you get the idea to divide by k! ? Don't think I could've thought of that.
$endgroup$
– MateInTwo
Jan 22 at 18:51
$begingroup$
Oh wow, that was absolutely fantastic, thanks! So $a_k leq k!e +1$ is always true. That was really fun. How'd you get the idea to divide by k! ? Don't think I could've thought of that.
$endgroup$
– MateInTwo
Jan 22 at 18:51
add a comment |
$begingroup$
$a_k = ka_{k-1} - k + 2 = k((k-1)a_{k-2} - (k -1) + 2) -k + 2 = ... = k!a_{0} +2 - k + k(2 - (k-1)) + ... + k!(2 - (k-k))$
It may be possible to simplify the sum.
answered Jan 21 at 15:28
lightxbulblightxbulb
1,040311
$endgroup$
add a comment |
$begingroup$
$a_k = ka_{k-1} - k + 2 = k((k-1)a_{k-2} - (k -1) + 2) -k + 2 = ... = k!a_{0} +2 - k + k(2 - (k-1)) + ... + k!(2 - (k-k))$
It may be possible to simplify the sum.
answered Jan 21 at 15:28
lightxbulblightxbulb
1,040311
$endgroup$
add a comment |
$begingroup$
$a_k = ka_{k-1} - k + 2 = k((k-1)a_{k-2} - (k -1) + 2) -k + 2 = ... = k!a_{0} +2 - k + k(2 - (k-1)) + ... + k!(2 - (k-k))$
It may be possible to simplify the sum.
answered Jan 21 at 15:28
lightxbulblightxbulb
1,040311
$endgroup$
$a_k = ka_{k-1} - k + 2 = k((k-1)a_{k-2} - (k -1) + 2) -k + 2 = ... = k!a_{0} +2 - k + k(2 - (k-1)) + ... + k!(2 - (k-k))$
It may be possible to simplify the sum.
answered Jan 21 at 15:28
lightxbulblightxbulb
1,040311
answered Jan 21 at 15:28
lightxbulblightxbulb
1,040311
answered Jan 21 at 15:28
lightxbulblightxbulb
1,040311
answered Jan 21 at 15:28
lightxbulblightxbulb
1,040311
1,040311
add a comment |
add a comment |
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2
$begingroup$
I think you mean to say "explicit" not "implicit".
$endgroup$
– saulspatz
Jan 21 at 15:24
$begingroup$
Of course, sorry
$endgroup$
– MateInTwo
Jan 21 at 15:39