30 Jul 12
1 edit
Originally posted by sonhouseTake the usual formula for K to F:
I got close just doing the formula, (((K-273.15)*1.8)-32)=F
How do you come up with a formula where the absolute value of K=F?
(K-273.15)*1.8+32=FWe want to find the point at which the temp value is the same for both scales, so substitute:
(c-273.15)*1.8+32=c
Solve for c:
1.8c-491.67+32=c
-459.67=-0.8c
574.5875=c
30 Jul 12
Originally posted by SwissGambit[/b]I am talking about the absolute value of K Vs absolute value of F, not C.
Take the usual formula for K to F:(K-273.15)*1.8+32=FWe want to find the point at which the temp value is the same for both scales, so substitute:(c-273.15)*1.8+32=c=c
Solve for c:
1.8c-491.67+32=c
-459.67=-0.8c
[b]574.5875
It is ~164, that is 164 degrees K = ~-164 degrees F, just wanted to see if there was an exact solution and how you derive a formula that would give that exact solution.
30 Jul 12
4 edits
Originally posted by sonhouseAhh, now I know what you want. I didn't understand exactly what you were asking for in the opening post.
I am talking about the absolute value of K Vs absolute value of F, not C.
It is ~164, that is 164 degrees K = ~-164 degrees F, just wanted to see if there was an exact solution and how you derive a formula that would give that exact solution.
The variable "c" was not meant to be Celsius, but a generic variable. Sorry about the confusion. I will use "x" this time. 🙂
The steps are the same as before. Just start with -x on the right side this time:
(x-273.15)*1.8+32=-x
1.8x -491.67+32=-x
-459.67=-2.8x
x=164.1678571
Edit: But if you speak in terms of absolute value, then both of my answers are solutions. 🙂
30 Jul 12
1 edit
Originally posted by SwissGambitI used the windows PC calculator and got this:
Ahh, now I know what you want. I didn't understand exactly what you were asking for in the opening post.
The variable "c" was not meant to be Celsius, but a generic variable. Sorry about the confusion. I will use "x" this time. 🙂
The steps are the same as before. Just start with -x on the right side this time:[quote](x-273.15)*1.8+32=-x
1.8x ...[text shortened]... t if you speak in terms of absolute value, then both of my answers are solutions. 🙂
164.1678571 428571 428571 428571.....
I separated the digits by the 1's to show the repeating series.
I wonder if there is a way, perhaps another number base, that would make that an exact solution, somewhat akin to the difference between calling it 4.333333333....Vs
4 1/3 (exact solution)
So in base 10, there is no exact solution just an infinite series.
30 Jul 12
1 edit
Originally posted by sonhouseIf there is a repeating decimal, it can be expressed as a fraction. You don't need another number base [not sure why that would help anyway 😛].
I used the windows PC calculator and got this:
164.1678571 428571 428571 428571.....
I separated the digits by the 1's to show the repeating series.
I wonder if there is a way, perhaps another number base, that would make that an exact solution, somewhat akin to the difference between calling it 4.333333333....Vs
4 1/3 (exact solution)
So in base 10, there is no exact solution just an infinite series.
45967/280 does the trick in this case.
31 Jul 12
2 edits
Originally posted by SwissGambitWell my calculator says 45967/280 covers the 164.1678571 part exactly but it misses out on the actual repeaters, the 428571's that repeat forever.
If there is a repeating decimal, it can be expressed as a fraction. You don't need another number base [not sure why that would help anyway 😛].
45967/280 does the trick in this case.
How did you suss out that fraction anyway? You have fraction sniffing software?
31 Jul 12
1 edit
Originally posted by sonhouseMy windows calc program gives
Well my calculator says 45967/280 covers the 164.1678571 part exactly but it misses out on the actual repeaters, the 428571's that repeat forever.
How did you suss out that fraction anyway? You have fraction sniffing software?
164.1678571 428571 428571 428571 4286
I am not sure why you are not seeing the repeating digits. You should.
I used an old algebra trick to get the fraction. Software didn't come into play until it was time to reduce the fraction. 🙂
N = 164.1678571 428571 428571
create another equation by multiplying both sides by some power of 10, such that the repeating digits line up with the first equation:
1000000N = 164167857.1428571 428571
Now subtract the first equation from the second:
999999N = 164167692.975
get rid of the decimals:
999999000N = 164167692975
solve for N and you have your fraction:
164167692975/999999000
now all that remains is to reduce it.
45967/280
31 Jul 12
Originally posted by SwissGambitbut with that method don't you need to include all the infinite number of repeats? That would seem to me to need an infinite number of zeros in the first part, 1000000000000000000000000000..........N to get the whole thing.
My windows calc program gives
164.1678571 428571 428571 428571 4286
I am not sure why you are not seeing the repeating digits. You should.
I used an old algebra trick to get the fraction. Software didn't come into play until it was time to reduce the fraction. 🙂
N = 164.1678571 428571 428571
create another equation by multiplying both sid ...[text shortened]... your fraction:
164167692975/999999000
now all that remains is to reduce it.
45967/280
31 Jul 12
2 edits
Originally posted by sonhouseI did include the infinite repeats. They were there when I subtracted one equation from the other, but they were lined up, and thus disappeared due to the subtraction. It's a pretty sweet trick.
but with that method don't you need to include all the infinite number of repeats? That would seem to me to need an infinite number of zeros in the first part, 1000000000000000000000000000..........N to get the whole thing.
Here's a simpler example to illustrate.
Make up some number with repeating digits:
14.512 757575757575...
Created 2nd equation and subtract the first:
100N = 1451.2 757575757575....
-(N = 14.512 757575757575....)
-------------------------
99N = 1436.763 (the repeating 75's died in the subtraction!)
99000N=1436763
N=1436763/99000
N=478921/33000 (unlucky - this couldn't be reduced much, but there it is. Plug it into a calculator and you should get all the repeating digits).
31 Jul 12
Originally posted by SwissGambitSo the answer would be the same no matter how many repeating digits were included? It would also be the same if you only used the first set of digits and ignored the rest?
I did include the infinite repeats. They were there when I subtracted one equation from the other, but they were lined up, and thus disappeared due to the subtraction. It's a pretty sweet trick.
Here's a simpler example to illustrate.
Make up some number with repeating digits:
14.512 757575757575...
Created 2nd equation and subtract th ...[text shortened]... ch, but there it is. Plug it into a calculator and you should get all the repeating digits).
31 Jul 12
Originally posted by sonhouseYou always include all repeating digits. You just write down enough to make sure they are lined up for the subtraction.
So the answer would be the same no matter how many repeating digits were included? It would also be the same if you only used the first set of digits and ignored the rest?