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Why does the approximation for exponents $(a+b)^c approx a^c-bc (a+1)^cb$ work?
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I was working with some code involving exponents in an environment where exponents can only be calculated if the base of the exponent is an integer. I needed a good fast way to approximate this without causing overflow issues. I accidentally stumbled upon an incredible approximation method and I'm not sure why it works.
Suppose you have an exponent in the form of $x^y$ where $x$ is not an integer and you want to approximate the value using only exponents which have integers for their base-values.
Break $x$ into two parts, an integer part, and an additive. For example $3.7to 3 + 0.7$.
Therefor, $xto(a+b)$ where $a$ is an integer part.
The approximation formula is:
$$(a+b)^c approx a^c-bc (a+1)^cb$$
Or in my original form:
$$(a+b)^c approx ((a+b)-b)^c(1-b) ((a+b)+(1-b))^cb$$
It's remarkably close to the right solution seemingly every time. Granted I've only been able to check about 100 cases, but I'm fascinated.
For example:
$$37.5^28 ≈ 37^14cdot38^14$$
And sure enough, if we divide both parts, the ratio is 1.002 which is very close.
Edit: Thanks to RayDansh pointing out in the comments, this is accurate IFF $a+b$ is big. In fact, the larger $a$ gets the more accurate this approximation seems to get.
Can anyone shed some light as to why this approximation method I've stumbled upon works?
exponential-function exponentiation approximation
edited Aug 15 at 3:29
AccidentalFourierTransform
1,301627
asked Aug 14 at 23:41
Albert Renshaw
6791625
 |Â
show 3 more comments
up vote
6
down vote
favorite
I was working with some code involving exponents in an environment where exponents can only be calculated if the base of the exponent is an integer. I needed a good fast way to approximate this without causing overflow issues. I accidentally stumbled upon an incredible approximation method and I'm not sure why it works.
Suppose you have an exponent in the form of $x^y$ where $x$ is not an integer and you want to approximate the value using only exponents which have integers for their base-values.
Break $x$ into two parts, an integer part, and an additive. For example $3.7to 3 + 0.7$.
Therefor, $xto(a+b)$ where $a$ is an integer part.
The approximation formula is:
$$(a+b)^c approx a^c-bc (a+1)^cb$$
Or in my original form:
$$(a+b)^c approx ((a+b)-b)^c(1-b) ((a+b)+(1-b))^cb$$
It's remarkably close to the right solution seemingly every time. Granted I've only been able to check about 100 cases, but I'm fascinated.
For example:
$$37.5^28 ≈ 37^14cdot38^14$$
And sure enough, if we divide both parts, the ratio is 1.002 which is very close.
Edit: Thanks to RayDansh pointing out in the comments, this is accurate IFF $a+b$ is big. In fact, the larger $a$ gets the more accurate this approximation seems to get.
Can anyone shed some light as to why this approximation method I've stumbled upon works?
exponential-function exponentiation approximation
edited Aug 15 at 3:29
AccidentalFourierTransform
1,301627
asked Aug 14 at 23:41
Albert Renshaw
6791625
Well, here's your intuition. The exponent terms you've discovered are such that whichever of $a$ and $a+1$ are closer to $a+b$, those will be weighted more than the other term. And both happen to be remarkably close to $a+b$. Hence your discovery.
– Rushabh Mehta
Aug 14 at 23:48
2
What about $1.5^100≈1^50*2^50$? Quite inaccurate.
– RayDansh
Aug 14 at 23:48
Just trying to show how the approximation loses accuracy as $x$ decreases and $y$ increases.
– RayDansh
Aug 14 at 23:49
1
@RayDansh I think the point is that it improves for large $a$.
– Ian
Aug 14 at 23:49
@RayDansh Right you are; I was only testing with large numbers. Seems to get 100% accurate asaapproached infinity. How odd
– Albert Renshaw
Aug 14 at 23:50
 |Â
show 3 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I was working with some code involving exponents in an environment where exponents can only be calculated if the base of the exponent is an integer. I needed a good fast way to approximate this without causing overflow issues. I accidentally stumbled upon an incredible approximation method and I'm not sure why it works.
Suppose you have an exponent in the form of $x^y$ where $x$ is not an integer and you want to approximate the value using only exponents which have integers for their base-values.
Break $x$ into two parts, an integer part, and an additive. For example $3.7to 3 + 0.7$.
Therefor, $xto(a+b)$ where $a$ is an integer part.
The approximation formula is:
$$(a+b)^c approx a^c-bc (a+1)^cb$$
Or in my original form:
$$(a+b)^c approx ((a+b)-b)^c(1-b) ((a+b)+(1-b))^cb$$
It's remarkably close to the right solution seemingly every time. Granted I've only been able to check about 100 cases, but I'm fascinated.
For example:
$$37.5^28 ≈ 37^14cdot38^14$$
And sure enough, if we divide both parts, the ratio is 1.002 which is very close.
Edit: Thanks to RayDansh pointing out in the comments, this is accurate IFF $a+b$ is big. In fact, the larger $a$ gets the more accurate this approximation seems to get.
Can anyone shed some light as to why this approximation method I've stumbled upon works?
exponential-function exponentiation approximation
edited Aug 15 at 3:29
AccidentalFourierTransform
1,301627
asked Aug 14 at 23:41
Albert Renshaw
6791625
I was working with some code involving exponents in an environment where exponents can only be calculated if the base of the exponent is an integer. I needed a good fast way to approximate this without causing overflow issues. I accidentally stumbled upon an incredible approximation method and I'm not sure why it works.
Suppose you have an exponent in the form of $x^y$ where $x$ is not an integer and you want to approximate the value using only exponents which have integers for their base-values.
Break $x$ into two parts, an integer part, and an additive. For example $3.7to 3 + 0.7$.
Therefor, $xto(a+b)$ where $a$ is an integer part.
The approximation formula is:
$$(a+b)^c approx a^c-bc (a+1)^cb$$
Or in my original form:
$$(a+b)^c approx ((a+b)-b)^c(1-b) ((a+b)+(1-b))^cb$$
It's remarkably close to the right solution seemingly every time. Granted I've only been able to check about 100 cases, but I'm fascinated.
For example:
$$37.5^28 ≈ 37^14cdot38^14$$
And sure enough, if we divide both parts, the ratio is 1.002 which is very close.
Edit: Thanks to RayDansh pointing out in the comments, this is accurate IFF $a+b$ is big. In fact, the larger $a$ gets the more accurate this approximation seems to get.
Can anyone shed some light as to why this approximation method I've stumbled upon works?
exponential-function exponentiation approximation
edited Aug 15 at 3:29
AccidentalFourierTransform
1,301627
asked Aug 14 at 23:41
Albert Renshaw
6791625
edited Aug 15 at 3:29
AccidentalFourierTransform
1,301627
edited Aug 15 at 3:29
AccidentalFourierTransform
1,301627
edited Aug 15 at 3:29
AccidentalFourierTransform
1,301627
1,301627
asked Aug 14 at 23:41
Albert Renshaw
6791625
asked Aug 14 at 23:41
Albert Renshaw
6791625
asked Aug 14 at 23:41
Albert Renshaw
6791625
6791625
Well, here's your intuition. The exponent terms you've discovered are such that whichever of $a$ and $a+1$ are closer to $a+b$, those will be weighted more than the other term. And both happen to be remarkably close to $a+b$. Hence your discovery.
– Rushabh Mehta
Aug 14 at 23:48
2
What about $1.5^100≈1^50*2^50$? Quite inaccurate.
– RayDansh
Aug 14 at 23:48
Just trying to show how the approximation loses accuracy as $x$ decreases and $y$ increases.
– RayDansh
Aug 14 at 23:49
1
@RayDansh I think the point is that it improves for large $a$.
– Ian
Aug 14 at 23:49
@RayDansh Right you are; I was only testing with large numbers. Seems to get 100% accurate asaapproached infinity. How odd
– Albert Renshaw
Aug 14 at 23:50
 |Â
show 3 more comments
Well, here's your intuition. The exponent terms you've discovered are such that whichever of $a$ and $a+1$ are closer to $a+b$, those will be weighted more than the other term. And both happen to be remarkably close to $a+b$. Hence your discovery.
– Rushabh Mehta
Aug 14 at 23:48
2
What about $1.5^100≈1^50*2^50$? Quite inaccurate.
– RayDansh
Aug 14 at 23:48
Just trying to show how the approximation loses accuracy as $x$ decreases and $y$ increases.
– RayDansh
Aug 14 at 23:49
1
@RayDansh I think the point is that it improves for large $a$.
– Ian
Aug 14 at 23:49
@RayDansh Right you are; I was only testing with large numbers. Seems to get 100% accurate asaapproached infinity. How odd
– Albert Renshaw
Aug 14 at 23:50
Well, here's your intuition. The exponent terms you've discovered are such that whichever of $a$ and $a+1$ are closer to $a+b$, those will be weighted more than the other term. And both happen to be remarkably close to $a+b$. Hence your discovery.
– Rushabh Mehta
Aug 14 at 23:48
Well, here's your intuition. The exponent terms you've discovered are such that whichever of $a$ and $a+1$ are closer to $a+b$, those will be weighted more than the other term. And both happen to be remarkably close to $a+b$. Hence your discovery.
– Rushabh Mehta
Aug 14 at 23:48
2
2
What about $1.5^100≈1^50*2^50$? Quite inaccurate.
– RayDansh
Aug 14 at 23:48
What about $1.5^100≈1^50*2^50$? Quite inaccurate.
– RayDansh
Aug 14 at 23:48
Just trying to show how the approximation loses accuracy as $x$ decreases and $y$ increases.
– RayDansh
Aug 14 at 23:49
Just trying to show how the approximation loses accuracy as $x$ decreases and $y$ increases.
– RayDansh
Aug 14 at 23:49
1
1
@RayDansh I think the point is that it improves for large $a$.
– Ian
Aug 14 at 23:49
@RayDansh I think the point is that it improves for large $a$.
– Ian
Aug 14 at 23:49
@RayDansh Right you are; I was only testing with large numbers. Seems to get 100% accurate as
a approached infinity. How odd– Albert Renshaw
Aug 14 at 23:50
@RayDansh Right you are; I was only testing with large numbers. Seems to get 100% accurate as
a approached infinity. How odd– Albert Renshaw
Aug 14 at 23:50
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
18
down vote
accepted
Write your approximation
$$a^c-bc cdot (a+1)^cb=a^cleft(1+frac 1aright)^bc$$
and your approximation is $$left(1+frac 1aright)^bapprox 1+frac ba$$
Which is the first two terms of the binomial expansion. It will be reasonably accurate when $frac ba ll 1$ The next term is $frac b(b-1)2a^2$
answered Aug 15 at 0:01
Ross Millikan
279k22188355
2
Doesn't get clearer than this +1
– user159517
Aug 15 at 0:31
add a comment |Â
up vote
6
down vote
As a general rule, if you see a lot of products and exponents it may clear things up to take logs. In your case,
$$(a+b)^c ≈ a^c-bc cdot (a+1)^cb$$
becomes
$$c log(a+b) approx c(1-b) log(a) + cb log(a+1)$$
or just
$$log(a+b) approx (1-b) log(a) + b log(a+1).$$
This is equivalent to doing a linear interpolation of $log x$ between the points $a$ and $a+1$. This will be pretty accurate when $a$ is large because $log x$ will be close to linear between $a$ and $a+1$.
edited Aug 15 at 0:01
Holo
4,3072629
answered Aug 14 at 23:59
Jair Taylor
8,48932144
add a comment |Â
up vote
1
down vote
The answer is obvious. If we expand both sides of the formula by the binomial theorem, we get for the first two: $(a+b)^c=a^c+cba^c-1+ ...$ for both forms of the equation. Now, it is immediately apparent that this approximation will improve as a gets larger because the discarded terms become less significant. That is, for larger a, $a^n>>a^n-1$.
answered Aug 14 at 23:57
Dr Peter McGowan
4737
Of course it improves, but why is it significantly better than just keeping $a^c$?
– Ian
Aug 14 at 23:58
Look at the rest of the binomial expansion terms that are omitted above
– Dr Peter McGowan
Aug 15 at 0:00
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
18
down vote
accepted
Write your approximation
$$a^c-bc cdot (a+1)^cb=a^cleft(1+frac 1aright)^bc$$
and your approximation is $$left(1+frac 1aright)^bapprox 1+frac ba$$
Which is the first two terms of the binomial expansion. It will be reasonably accurate when $frac ba ll 1$ The next term is $frac b(b-1)2a^2$
answered Aug 15 at 0:01
Ross Millikan
279k22188355
2
Doesn't get clearer than this +1
– user159517
Aug 15 at 0:31
add a comment |Â
up vote
18
down vote
accepted
Write your approximation
$$a^c-bc cdot (a+1)^cb=a^cleft(1+frac 1aright)^bc$$
and your approximation is $$left(1+frac 1aright)^bapprox 1+frac ba$$
Which is the first two terms of the binomial expansion. It will be reasonably accurate when $frac ba ll 1$ The next term is $frac b(b-1)2a^2$
answered Aug 15 at 0:01
Ross Millikan
279k22188355
2
Doesn't get clearer than this +1
– user159517
Aug 15 at 0:31
add a comment |Â
up vote
18
down vote
accepted
up vote
18
down vote
accepted
Write your approximation
$$a^c-bc cdot (a+1)^cb=a^cleft(1+frac 1aright)^bc$$
and your approximation is $$left(1+frac 1aright)^bapprox 1+frac ba$$
Which is the first two terms of the binomial expansion. It will be reasonably accurate when $frac ba ll 1$ The next term is $frac b(b-1)2a^2$
answered Aug 15 at 0:01
Ross Millikan
279k22188355
Write your approximation
$$a^c-bc cdot (a+1)^cb=a^cleft(1+frac 1aright)^bc$$
and your approximation is $$left(1+frac 1aright)^bapprox 1+frac ba$$
Which is the first two terms of the binomial expansion. It will be reasonably accurate when $frac ba ll 1$ The next term is $frac b(b-1)2a^2$
answered Aug 15 at 0:01
Ross Millikan
279k22188355
answered Aug 15 at 0:01
Ross Millikan
279k22188355
answered Aug 15 at 0:01
Ross Millikan
279k22188355
answered Aug 15 at 0:01
Ross Millikan
279k22188355
279k22188355
2
Doesn't get clearer than this +1
– user159517
Aug 15 at 0:31
add a comment |Â
2
Doesn't get clearer than this +1
– user159517
Aug 15 at 0:31
2
2
Doesn't get clearer than this +1
– user159517
Aug 15 at 0:31
Doesn't get clearer than this +1
– user159517
Aug 15 at 0:31
add a comment |Â
up vote
6
down vote
As a general rule, if you see a lot of products and exponents it may clear things up to take logs. In your case,
$$(a+b)^c ≈ a^c-bc cdot (a+1)^cb$$
becomes
$$c log(a+b) approx c(1-b) log(a) + cb log(a+1)$$
or just
$$log(a+b) approx (1-b) log(a) + b log(a+1).$$
This is equivalent to doing a linear interpolation of $log x$ between the points $a$ and $a+1$. This will be pretty accurate when $a$ is large because $log x$ will be close to linear between $a$ and $a+1$.
edited Aug 15 at 0:01
Holo
4,3072629
answered Aug 14 at 23:59
Jair Taylor
8,48932144
add a comment |Â
up vote
6
down vote
As a general rule, if you see a lot of products and exponents it may clear things up to take logs. In your case,
$$(a+b)^c ≈ a^c-bc cdot (a+1)^cb$$
becomes
$$c log(a+b) approx c(1-b) log(a) + cb log(a+1)$$
or just
$$log(a+b) approx (1-b) log(a) + b log(a+1).$$
This is equivalent to doing a linear interpolation of $log x$ between the points $a$ and $a+1$. This will be pretty accurate when $a$ is large because $log x$ will be close to linear between $a$ and $a+1$.
edited Aug 15 at 0:01
Holo
4,3072629
answered Aug 14 at 23:59
Jair Taylor
8,48932144
add a comment |Â
up vote
6
down vote
up vote
6
down vote
As a general rule, if you see a lot of products and exponents it may clear things up to take logs. In your case,
$$(a+b)^c ≈ a^c-bc cdot (a+1)^cb$$
becomes
$$c log(a+b) approx c(1-b) log(a) + cb log(a+1)$$
or just
$$log(a+b) approx (1-b) log(a) + b log(a+1).$$
This is equivalent to doing a linear interpolation of $log x$ between the points $a$ and $a+1$. This will be pretty accurate when $a$ is large because $log x$ will be close to linear between $a$ and $a+1$.
edited Aug 15 at 0:01
Holo
4,3072629
answered Aug 14 at 23:59
Jair Taylor
8,48932144
As a general rule, if you see a lot of products and exponents it may clear things up to take logs. In your case,
$$(a+b)^c ≈ a^c-bc cdot (a+1)^cb$$
becomes
$$c log(a+b) approx c(1-b) log(a) + cb log(a+1)$$
or just
$$log(a+b) approx (1-b) log(a) + b log(a+1).$$
This is equivalent to doing a linear interpolation of $log x$ between the points $a$ and $a+1$. This will be pretty accurate when $a$ is large because $log x$ will be close to linear between $a$ and $a+1$.
edited Aug 15 at 0:01
Holo
4,3072629
answered Aug 14 at 23:59
Jair Taylor
8,48932144
edited Aug 15 at 0:01
Holo
4,3072629
edited Aug 15 at 0:01
Holo
4,3072629
edited Aug 15 at 0:01
Holo
4,3072629
4,3072629
answered Aug 14 at 23:59
Jair Taylor
8,48932144
answered Aug 14 at 23:59
Jair Taylor
8,48932144
answered Aug 14 at 23:59
Jair Taylor
8,48932144
8,48932144
add a comment |Â
add a comment |Â
up vote
1
down vote
The answer is obvious. If we expand both sides of the formula by the binomial theorem, we get for the first two: $(a+b)^c=a^c+cba^c-1+ ...$ for both forms of the equation. Now, it is immediately apparent that this approximation will improve as a gets larger because the discarded terms become less significant. That is, for larger a, $a^n>>a^n-1$.
answered Aug 14 at 23:57
Dr Peter McGowan
4737
Of course it improves, but why is it significantly better than just keeping $a^c$?
– Ian
Aug 14 at 23:58
Look at the rest of the binomial expansion terms that are omitted above
– Dr Peter McGowan
Aug 15 at 0:00
add a comment |Â
up vote
1
down vote
The answer is obvious. If we expand both sides of the formula by the binomial theorem, we get for the first two: $(a+b)^c=a^c+cba^c-1+ ...$ for both forms of the equation. Now, it is immediately apparent that this approximation will improve as a gets larger because the discarded terms become less significant. That is, for larger a, $a^n>>a^n-1$.
answered Aug 14 at 23:57
Dr Peter McGowan
4737
Of course it improves, but why is it significantly better than just keeping $a^c$?
– Ian
Aug 14 at 23:58
Look at the rest of the binomial expansion terms that are omitted above
– Dr Peter McGowan
Aug 15 at 0:00
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The answer is obvious. If we expand both sides of the formula by the binomial theorem, we get for the first two: $(a+b)^c=a^c+cba^c-1+ ...$ for both forms of the equation. Now, it is immediately apparent that this approximation will improve as a gets larger because the discarded terms become less significant. That is, for larger a, $a^n>>a^n-1$.
answered Aug 14 at 23:57
Dr Peter McGowan
4737
The answer is obvious. If we expand both sides of the formula by the binomial theorem, we get for the first two: $(a+b)^c=a^c+cba^c-1+ ...$ for both forms of the equation. Now, it is immediately apparent that this approximation will improve as a gets larger because the discarded terms become less significant. That is, for larger a, $a^n>>a^n-1$.
answered Aug 14 at 23:57
Dr Peter McGowan
4737
edited Aug 15 at 0:02
answered Aug 14 at 23:57
Dr Peter McGowan
4737
answered Aug 14 at 23:57
Dr Peter McGowan
4737
answered Aug 14 at 23:57
Dr Peter McGowan
4737
4737
Of course it improves, but why is it significantly better than just keeping $a^c$?
– Ian
Aug 14 at 23:58
Look at the rest of the binomial expansion terms that are omitted above
– Dr Peter McGowan
Aug 15 at 0:00
add a comment |Â
Of course it improves, but why is it significantly better than just keeping $a^c$?
– Ian
Aug 14 at 23:58
Look at the rest of the binomial expansion terms that are omitted above
– Dr Peter McGowan
Aug 15 at 0:00
Of course it improves, but why is it significantly better than just keeping $a^c$?
– Ian
Aug 14 at 23:58
Of course it improves, but why is it significantly better than just keeping $a^c$?
– Ian
Aug 14 at 23:58
Look at the rest of the binomial expansion terms that are omitted above
– Dr Peter McGowan
Aug 15 at 0:00
Look at the rest of the binomial expansion terms that are omitted above
– Dr Peter McGowan
Aug 15 at 0:00
add a comment |Â
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Well, here's your intuition. The exponent terms you've discovered are such that whichever of $a$ and $a+1$ are closer to $a+b$, those will be weighted more than the other term. And both happen to be remarkably close to $a+b$. Hence your discovery.
– Rushabh Mehta
Aug 14 at 23:48
2
What about $1.5^100≈1^50*2^50$? Quite inaccurate.
– RayDansh
Aug 14 at 23:48
Just trying to show how the approximation loses accuracy as $x$ decreases and $y$ increases.
– RayDansh
Aug 14 at 23:49
1
@RayDansh I think the point is that it improves for large $a$.
– Ian
Aug 14 at 23:49
@RayDansh Right you are; I was only testing with large numbers. Seems to get 100% accurate as
aapproached infinity. How odd– Albert Renshaw
Aug 14 at 23:50