anti-logarithm (base $10$)
$begingroup$
Let's say that I start with the $log_{10}^7$, which is $0.84509804$, and I want to calculate $10^{0.84509804}$. $0.84509804$ is close to $431/510$. Without access to anything electronic, it seems that I would have to find the $510^{th}$ root of $10^{431}$. I believe that would be an iterative process requiring calculating $X^{510}$ for each iteration. Is that how the table values were calculated prior to 1950? It seems unbelievably tedious.
I apologize for my lack of proper notation.
logarithms
edited Jan 11 at 5:14
Lee
330111
asked Jan 11 at 4:45
Bob7430Bob7430
11
$endgroup$
add a comment |
$begingroup$
Let's say that I start with the $log_{10}^7$, which is $0.84509804$, and I want to calculate $10^{0.84509804}$. $0.84509804$ is close to $431/510$. Without access to anything electronic, it seems that I would have to find the $510^{th}$ root of $10^{431}$. I believe that would be an iterative process requiring calculating $X^{510}$ for each iteration. Is that how the table values were calculated prior to 1950? It seems unbelievably tedious.
I apologize for my lack of proper notation.
logarithms
edited Jan 11 at 5:14
Lee
330111
asked Jan 11 at 4:45
Bob7430Bob7430
11
$endgroup$
2
$begingroup$
nice question. You're right that the iterative process would be way too slow. Our ancestors knew many tricks for approximating roots, which our computers are still using under the hood; for example, Newton's method.
$endgroup$
– hunter
Jan 11 at 4:49
$begingroup$
Yes, Newton's method. It's been more than 30 years since I took Numerical Analysis, but I should have remembered that. Thanks.
$endgroup$
– Bob7430
Jan 11 at 5:06
1
$begingroup$
There was a lot of tedium in calculating tables. In years past "calculator" was a profession, not an electronic device. It was a full time job. There is a lot of regularity in tables which can relieve some of the work. I don't know what approaches were used. I can imagine that if you were going to do a table of base 10 logs, you might start with 1.1. Having done that you could get 1.21 by doubling, 1.331 by tripling and other values with small corrections. Then find the ones in between by interpolation.
$endgroup$
– Ross Millikan
Jan 11 at 5:10
add a comment |
$begingroup$
Let's say that I start with the $log_{10}^7$, which is $0.84509804$, and I want to calculate $10^{0.84509804}$. $0.84509804$ is close to $431/510$. Without access to anything electronic, it seems that I would have to find the $510^{th}$ root of $10^{431}$. I believe that would be an iterative process requiring calculating $X^{510}$ for each iteration. Is that how the table values were calculated prior to 1950? It seems unbelievably tedious.
I apologize for my lack of proper notation.
logarithms
edited Jan 11 at 5:14
Lee
330111
asked Jan 11 at 4:45
Bob7430Bob7430
11
$endgroup$
Let's say that I start with the $log_{10}^7$, which is $0.84509804$, and I want to calculate $10^{0.84509804}$. $0.84509804$ is close to $431/510$. Without access to anything electronic, it seems that I would have to find the $510^{th}$ root of $10^{431}$. I believe that would be an iterative process requiring calculating $X^{510}$ for each iteration. Is that how the table values were calculated prior to 1950? It seems unbelievably tedious.
I apologize for my lack of proper notation.
logarithms
logarithms
edited Jan 11 at 5:14
Lee
330111
asked Jan 11 at 4:45
Bob7430Bob7430
11
edited Jan 11 at 5:14
Lee
330111
asked Jan 11 at 4:45
Bob7430Bob7430
11
edited Jan 11 at 5:14
Lee
330111
edited Jan 11 at 5:14
Lee
330111
edited Jan 11 at 5:14
Lee
330111
330111
asked Jan 11 at 4:45
Bob7430Bob7430
11
asked Jan 11 at 4:45
Bob7430Bob7430
11
asked Jan 11 at 4:45
Bob7430Bob7430
11
11
2
$begingroup$
nice question. You're right that the iterative process would be way too slow. Our ancestors knew many tricks for approximating roots, which our computers are still using under the hood; for example, Newton's method.
$endgroup$
– hunter
Jan 11 at 4:49
$begingroup$
Yes, Newton's method. It's been more than 30 years since I took Numerical Analysis, but I should have remembered that. Thanks.
$endgroup$
– Bob7430
Jan 11 at 5:06
1
$begingroup$
There was a lot of tedium in calculating tables. In years past "calculator" was a profession, not an electronic device. It was a full time job. There is a lot of regularity in tables which can relieve some of the work. I don't know what approaches were used. I can imagine that if you were going to do a table of base 10 logs, you might start with 1.1. Having done that you could get 1.21 by doubling, 1.331 by tripling and other values with small corrections. Then find the ones in between by interpolation.
$endgroup$
– Ross Millikan
Jan 11 at 5:10
add a comment |
2
$begingroup$
nice question. You're right that the iterative process would be way too slow. Our ancestors knew many tricks for approximating roots, which our computers are still using under the hood; for example, Newton's method.
$endgroup$
– hunter
Jan 11 at 4:49
$begingroup$
Yes, Newton's method. It's been more than 30 years since I took Numerical Analysis, but I should have remembered that. Thanks.
$endgroup$
– Bob7430
Jan 11 at 5:06
1
$begingroup$
There was a lot of tedium in calculating tables. In years past "calculator" was a profession, not an electronic device. It was a full time job. There is a lot of regularity in tables which can relieve some of the work. I don't know what approaches were used. I can imagine that if you were going to do a table of base 10 logs, you might start with 1.1. Having done that you could get 1.21 by doubling, 1.331 by tripling and other values with small corrections. Then find the ones in between by interpolation.
$endgroup$
– Ross Millikan
Jan 11 at 5:10
2
2
$begingroup$
nice question. You're right that the iterative process would be way too slow. Our ancestors knew many tricks for approximating roots, which our computers are still using under the hood; for example, Newton's method.
$endgroup$
– hunter
Jan 11 at 4:49
$begingroup$
nice question. You're right that the iterative process would be way too slow. Our ancestors knew many tricks for approximating roots, which our computers are still using under the hood; for example, Newton's method.
$endgroup$
– hunter
Jan 11 at 4:49
$begingroup$
Yes, Newton's method. It's been more than 30 years since I took Numerical Analysis, but I should have remembered that. Thanks.
$endgroup$
– Bob7430
Jan 11 at 5:06
$begingroup$
Yes, Newton's method. It's been more than 30 years since I took Numerical Analysis, but I should have remembered that. Thanks.
$endgroup$
– Bob7430
Jan 11 at 5:06
1
1
$begingroup$
There was a lot of tedium in calculating tables. In years past "calculator" was a profession, not an electronic device. It was a full time job. There is a lot of regularity in tables which can relieve some of the work. I don't know what approaches were used. I can imagine that if you were going to do a table of base 10 logs, you might start with 1.1. Having done that you could get 1.21 by doubling, 1.331 by tripling and other values with small corrections. Then find the ones in between by interpolation.
$endgroup$
– Ross Millikan
Jan 11 at 5:10
$begingroup$
There was a lot of tedium in calculating tables. In years past "calculator" was a profession, not an electronic device. It was a full time job. There is a lot of regularity in tables which can relieve some of the work. I don't know what approaches were used. I can imagine that if you were going to do a table of base 10 logs, you might start with 1.1. Having done that you could get 1.21 by doubling, 1.331 by tripling and other values with small corrections. Then find the ones in between by interpolation.
$endgroup$
– Ross Millikan
Jan 11 at 5:10
add a comment |
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$begingroup$
nice question. You're right that the iterative process would be way too slow. Our ancestors knew many tricks for approximating roots, which our computers are still using under the hood; for example, Newton's method.
$endgroup$
– hunter
Jan 11 at 4:49
$begingroup$
Yes, Newton's method. It's been more than 30 years since I took Numerical Analysis, but I should have remembered that. Thanks.
$endgroup$
– Bob7430
Jan 11 at 5:06
1
$begingroup$
There was a lot of tedium in calculating tables. In years past "calculator" was a profession, not an electronic device. It was a full time job. There is a lot of regularity in tables which can relieve some of the work. I don't know what approaches were used. I can imagine that if you were going to do a table of base 10 logs, you might start with 1.1. Having done that you could get 1.21 by doubling, 1.331 by tripling and other values with small corrections. Then find the ones in between by interpolation.
$endgroup$
– Ross Millikan
Jan 11 at 5:10