I have found a formula for dividing numbers in easy steps
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I found an easy method for division and it depends on some factors.
I wanted to find an answer for $1000/101$ with easy steps. My starting point is here. I formulated this method by 2 hours of hard work. It is an infinite series, but taking 4 or 6 units of the series we can arrive at an answer easily.
To find $100/11$
We know $100/10 = 10$, then $100/11 = 10 - 1 + 0.1 - 0.01$ (each term is the preceding term divided by $10$; division by $10$ is easy) $= 9 + 0.09$ (simplify by taking two members of the series, so easy steps) $= 9.09$
To find $100/12$
We know $100/11 = 9.09$, then $100/12 = 9.09 - 0.909 + 0.0909 - 0.00909$ (each member of the series is found by dividing by $10$ with the preceding member) giving $8.181+.08181$ (calculations are easy) or $8.29$
I took only $4$ members of the series, and if we take $6$ members of the series, we get a better result.
To find $1000/101$
We know $1000/100 = 10$, then $1000/101 = 10 - 0.1 + 0.001 - .0001
= 9.9 + 0.0009 = 9.9009$
To find $1000/102$
We know, $1000/101 = 9.9$ approximately, then $1000/102 = 9.9 - 0.099 + 0.00099- 0.0000099 = 9.801 + 0.0009801 = 9.801$ (approx)
To find $100/3$
We know $100/2 = 50$, then $100/3 = 50 - 25 + 12.5 - 6.25 + 3.125 - 1.5625 = 25 + 6.25 + 1.5625 = 32.8125$ or approximately $33$.
Since they are a small number to get a perfect answer (we need to take more members of the series; for big numbers $4$ members of the series is sufficient)
Has somebody found this before me? Where should I submit the infinite series that I have found for further evaluation?
approximation
edited Sep 9 at 18:38
José Carlos Santos
123k17101186
asked Sep 9 at 6:33
rock-onn
16614
add a comment |Â
up vote
31
down vote
favorite
I found an easy method for division and it depends on some factors.
I wanted to find an answer for $1000/101$ with easy steps. My starting point is here. I formulated this method by 2 hours of hard work. It is an infinite series, but taking 4 or 6 units of the series we can arrive at an answer easily.
To find $100/11$
We know $100/10 = 10$, then $100/11 = 10 - 1 + 0.1 - 0.01$ (each term is the preceding term divided by $10$; division by $10$ is easy) $= 9 + 0.09$ (simplify by taking two members of the series, so easy steps) $= 9.09$
To find $100/12$
We know $100/11 = 9.09$, then $100/12 = 9.09 - 0.909 + 0.0909 - 0.00909$ (each member of the series is found by dividing by $10$ with the preceding member) giving $8.181+.08181$ (calculations are easy) or $8.29$
I took only $4$ members of the series, and if we take $6$ members of the series, we get a better result.
To find $1000/101$
We know $1000/100 = 10$, then $1000/101 = 10 - 0.1 + 0.001 - .0001
= 9.9 + 0.0009 = 9.9009$
To find $1000/102$
We know, $1000/101 = 9.9$ approximately, then $1000/102 = 9.9 - 0.099 + 0.00099- 0.0000099 = 9.801 + 0.0009801 = 9.801$ (approx)
To find $100/3$
We know $100/2 = 50$, then $100/3 = 50 - 25 + 12.5 - 6.25 + 3.125 - 1.5625 = 25 + 6.25 + 1.5625 = 32.8125$ or approximately $33$.
Since they are a small number to get a perfect answer (we need to take more members of the series; for big numbers $4$ members of the series is sufficient)
Has somebody found this before me? Where should I submit the infinite series that I have found for further evaluation?
approximation
edited Sep 9 at 18:38
José Carlos Santos
123k17101186
asked Sep 9 at 6:33
rock-onn
16614
1
This is a good one; it should go into the next edition of Dead Reckoning. You may want to get a copy of that book; it is full of such approximations.
– richard1941
Sep 12 at 5:49
add a comment |Â
up vote
31
down vote
favorite
up vote
31
down vote
favorite
I found an easy method for division and it depends on some factors.
I wanted to find an answer for $1000/101$ with easy steps. My starting point is here. I formulated this method by 2 hours of hard work. It is an infinite series, but taking 4 or 6 units of the series we can arrive at an answer easily.
To find $100/11$
We know $100/10 = 10$, then $100/11 = 10 - 1 + 0.1 - 0.01$ (each term is the preceding term divided by $10$; division by $10$ is easy) $= 9 + 0.09$ (simplify by taking two members of the series, so easy steps) $= 9.09$
To find $100/12$
We know $100/11 = 9.09$, then $100/12 = 9.09 - 0.909 + 0.0909 - 0.00909$ (each member of the series is found by dividing by $10$ with the preceding member) giving $8.181+.08181$ (calculations are easy) or $8.29$
I took only $4$ members of the series, and if we take $6$ members of the series, we get a better result.
To find $1000/101$
We know $1000/100 = 10$, then $1000/101 = 10 - 0.1 + 0.001 - .0001
= 9.9 + 0.0009 = 9.9009$
To find $1000/102$
We know, $1000/101 = 9.9$ approximately, then $1000/102 = 9.9 - 0.099 + 0.00099- 0.0000099 = 9.801 + 0.0009801 = 9.801$ (approx)
To find $100/3$
We know $100/2 = 50$, then $100/3 = 50 - 25 + 12.5 - 6.25 + 3.125 - 1.5625 = 25 + 6.25 + 1.5625 = 32.8125$ or approximately $33$.
Since they are a small number to get a perfect answer (we need to take more members of the series; for big numbers $4$ members of the series is sufficient)
Has somebody found this before me? Where should I submit the infinite series that I have found for further evaluation?
approximation
edited Sep 9 at 18:38
José Carlos Santos
123k17101186
asked Sep 9 at 6:33
rock-onn
16614
I found an easy method for division and it depends on some factors.
I wanted to find an answer for $1000/101$ with easy steps. My starting point is here. I formulated this method by 2 hours of hard work. It is an infinite series, but taking 4 or 6 units of the series we can arrive at an answer easily.
To find $100/11$
We know $100/10 = 10$, then $100/11 = 10 - 1 + 0.1 - 0.01$ (each term is the preceding term divided by $10$; division by $10$ is easy) $= 9 + 0.09$ (simplify by taking two members of the series, so easy steps) $= 9.09$
To find $100/12$
We know $100/11 = 9.09$, then $100/12 = 9.09 - 0.909 + 0.0909 - 0.00909$ (each member of the series is found by dividing by $10$ with the preceding member) giving $8.181+.08181$ (calculations are easy) or $8.29$
I took only $4$ members of the series, and if we take $6$ members of the series, we get a better result.
To find $1000/101$
We know $1000/100 = 10$, then $1000/101 = 10 - 0.1 + 0.001 - .0001
= 9.9 + 0.0009 = 9.9009$
To find $1000/102$
We know, $1000/101 = 9.9$ approximately, then $1000/102 = 9.9 - 0.099 + 0.00099- 0.0000099 = 9.801 + 0.0009801 = 9.801$ (approx)
To find $100/3$
We know $100/2 = 50$, then $100/3 = 50 - 25 + 12.5 - 6.25 + 3.125 - 1.5625 = 25 + 6.25 + 1.5625 = 32.8125$ or approximately $33$.
Since they are a small number to get a perfect answer (we need to take more members of the series; for big numbers $4$ members of the series is sufficient)
Has somebody found this before me? Where should I submit the infinite series that I have found for further evaluation?
approximation
approximation
edited Sep 9 at 18:38
José Carlos Santos
123k17101186
asked Sep 9 at 6:33
rock-onn
16614
edited Sep 9 at 18:38
José Carlos Santos
123k17101186
asked Sep 9 at 6:33
rock-onn
16614
edited Sep 9 at 18:38
José Carlos Santos
123k17101186
edited Sep 9 at 18:38
José Carlos Santos
123k17101186
edited Sep 9 at 18:38
José Carlos Santos
123k17101186
123k17101186
asked Sep 9 at 6:33
rock-onn
16614
asked Sep 9 at 6:33
rock-onn
16614
asked Sep 9 at 6:33
rock-onn
16614
16614
1
This is a good one; it should go into the next edition of Dead Reckoning. You may want to get a copy of that book; it is full of such approximations.
– richard1941
Sep 12 at 5:49
add a comment |Â
1
This is a good one; it should go into the next edition of Dead Reckoning. You may want to get a copy of that book; it is full of such approximations.
– richard1941
Sep 12 at 5:49
1
1
This is a good one; it should go into the next edition of Dead Reckoning. You may want to get a copy of that book; it is full of such approximations.
– richard1941
Sep 12 at 5:49
This is a good one; it should go into the next edition of Dead Reckoning. You may want to get a copy of that book; it is full of such approximations.
– richard1941
Sep 12 at 5:49
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
55
down vote
accepted
What you are using are so called geometric series:
$$frac11+n=frac1nfrac11-(-1/n)=frac1nsum_k=0^inftyleft(-frac1nright)^k$$
thus
$$fracx1+n=frac xnsum_k=0^inftyleft(-frac1nright)^k=fracxn-fracxn^2+fracxn^3-dots.$$
Of course the fact that you found the result by observation is quite impressive.
answered Sep 9 at 6:54
b00n heT
8,90211833
5
This also shows that the OP made a mistake in 1000/102, where he divided by 100 instead of 101.
– Carsten S
Sep 9 at 15:02
We can divide by 100 not with 101 when we divide by 100 itself.
– rock-onn
Sep 9 at 16:22
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
55
down vote
accepted
What you are using are so called geometric series:
$$frac11+n=frac1nfrac11-(-1/n)=frac1nsum_k=0^inftyleft(-frac1nright)^k$$
thus
$$fracx1+n=frac xnsum_k=0^inftyleft(-frac1nright)^k=fracxn-fracxn^2+fracxn^3-dots.$$
Of course the fact that you found the result by observation is quite impressive.
answered Sep 9 at 6:54
b00n heT
8,90211833
5
This also shows that the OP made a mistake in 1000/102, where he divided by 100 instead of 101.
– Carsten S
Sep 9 at 15:02
We can divide by 100 not with 101 when we divide by 100 itself.
– rock-onn
Sep 9 at 16:22
add a comment |Â
up vote
55
down vote
accepted
What you are using are so called geometric series:
$$frac11+n=frac1nfrac11-(-1/n)=frac1nsum_k=0^inftyleft(-frac1nright)^k$$
thus
$$fracx1+n=frac xnsum_k=0^inftyleft(-frac1nright)^k=fracxn-fracxn^2+fracxn^3-dots.$$
Of course the fact that you found the result by observation is quite impressive.
answered Sep 9 at 6:54
b00n heT
8,90211833
5
This also shows that the OP made a mistake in 1000/102, where he divided by 100 instead of 101.
– Carsten S
Sep 9 at 15:02
We can divide by 100 not with 101 when we divide by 100 itself.
– rock-onn
Sep 9 at 16:22
add a comment |Â
up vote
55
down vote
accepted
up vote
55
down vote
accepted
What you are using are so called geometric series:
$$frac11+n=frac1nfrac11-(-1/n)=frac1nsum_k=0^inftyleft(-frac1nright)^k$$
thus
$$fracx1+n=frac xnsum_k=0^inftyleft(-frac1nright)^k=fracxn-fracxn^2+fracxn^3-dots.$$
Of course the fact that you found the result by observation is quite impressive.
answered Sep 9 at 6:54
b00n heT
8,90211833
What you are using are so called geometric series:
$$frac11+n=frac1nfrac11-(-1/n)=frac1nsum_k=0^inftyleft(-frac1nright)^k$$
thus
$$fracx1+n=frac xnsum_k=0^inftyleft(-frac1nright)^k=fracxn-fracxn^2+fracxn^3-dots.$$
Of course the fact that you found the result by observation is quite impressive.
answered Sep 9 at 6:54
b00n heT
8,90211833
answered Sep 9 at 6:54
b00n heT
8,90211833
answered Sep 9 at 6:54
b00n heT
8,90211833
answered Sep 9 at 6:54
b00n heT
8,90211833
8,90211833
5
This also shows that the OP made a mistake in 1000/102, where he divided by 100 instead of 101.
– Carsten S
Sep 9 at 15:02
We can divide by 100 not with 101 when we divide by 100 itself.
– rock-onn
Sep 9 at 16:22
add a comment |Â
5
This also shows that the OP made a mistake in 1000/102, where he divided by 100 instead of 101.
– Carsten S
Sep 9 at 15:02
We can divide by 100 not with 101 when we divide by 100 itself.
– rock-onn
Sep 9 at 16:22
5
5
This also shows that the OP made a mistake in 1000/102, where he divided by 100 instead of 101.
– Carsten S
Sep 9 at 15:02
This also shows that the OP made a mistake in 1000/102, where he divided by 100 instead of 101.
– Carsten S
Sep 9 at 15:02
We can divide by 100 not with 101 when we divide by 100 itself.
– rock-onn
Sep 9 at 16:22
We can divide by 100 not with 101 when we divide by 100 itself.
– rock-onn
Sep 9 at 16:22
add a comment |Â
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1
This is a good one; it should go into the next edition of Dead Reckoning. You may want to get a copy of that book; it is full of such approximations.
– richard1941
Sep 12 at 5:49