Math People, Refresh My Memory On An Exponent

[RutgersRaRa]

Backdrop: I'm 50 pages into Trading Bases, a book by Joe Peta, who got his MBA at Stanford, went to Wall Street and became a trader, then ended up churning baseball data into a new (and lucrative) art form. Part of his process relied upon an earlier methodology by a gentleman (whose name escapes me) who is credited with developing the "baseball pythagorean theorem," which is a statistical analysis of runs scored and runs allowed to determine a correlation between runs scored and runs allowed, and a team's winning percentage. The earlier researcher came up with the following formula: (# of runs scored)^2 divided by [(# of runs scored)^2 + (# of runs allowed)^2] = Team's Winning Percentage. Through regression analysis and other processes, Mr. Peta determined that the correlation is more accurate when the factors aren't raised to the 2nd power (squared), but raised to a power of 1.83.

My question is, what is the process of raising a number to the 1.83 power? I used to know this and no longer do. Skillet, 4Real, anyone, can you help a brutha out? TIA.

 

[RU85inFla]

You need a calculator that has a y raised to the x power button. In excel you would use =a1^1.83 where cell a1 has the value you want to raise yo the 1.83 power. You are multiplying a given number by itself the number of times in the exponent.

 

[RutgersRaRa]

You need a calculator that has a y raised to the x power button. In excel you would use =a1^1.83 where cell a1 has the value you want to raise yo the 1.83 power. You are multiplying a given number by itself the number of times in the exponent.

I'm talking about pen-and-paper, pre-calculator process.

 

[RU85inFla]

then we need someone with an 8th grader.

 

[RutgersRaRa]

M

then we need someone with an 8th grader.

You're showing your age!! All 8th graders use calculators.

 

[DJ Spanky]

M

You're showing your age!! All 8th graders use calculators.

You're showing your age: all 8th graders use their phones!

 

[BeKnighted]

The Pythagorean Theorem was Bill James.

Many years ago, I would have used a log table, since that was the easiest way to do it - you find the log of a1, multiply by the exponent, then go back to the log table to find the result.

 

[GoodOl'Rutgers]

Beware: RaRa is just making a list of the fans who know what exponents are to further his identity theft database.

 

[RutgersRaRa]

Beware: RaRa is just making a list of the fans who know what exponents are to further his identity theft database.

I have just raised you to the third exponent. You are now virtually unreachable, even by my data techniques. Others, such as RUTexan, are still in play.

As to the original problem and the answers relating to "log tables" and other aspects of this thought-provoking question i came up with because I was bored and love you people, any way of doing this without log tables, log jams, or logarithms? I'm talking to you, Skillet--stop hiding behind the time-zone difference and get in this thread.

 

[RutgersRaRa]

The Pythagorean Theorem was Bill James.

Many years ago, I would have used a log table, since that was the easiest way to do it - you find the log of a1, multiply by the exponent, then go back to the log table to find the result.

This sounds like cheating, and if there's one thing I will have no part of, it's something along these lines.

 

[BeKnighted]

Many years ago, I would have used a log table, since that was the easiest way to do it - you find the log of a1, multiply by the exponent, then go back to the log table to find the result.

This sounds like cheating, and if there's one thing I will have no part of, it's something along these lines.

The other approach was to do it on a slide rule, but I guess you'd consider that cheating, too. The manual calculation takes a very long time.

 

[RutgersRaRa]

The other approach was to do it on a slide rule, but I guess you'd consider that cheating, too. The manual calculation takes a very long time.

I'm not interested in the calculation as much as the way to write the problem out. Maybe it's easier to conceptualize the issue of non-whole exponents by asking, what does 10^1.5 equal, and how do we write it out?

 

[RU848789]

I'm talking about pen-and-paper, pre-calculator process.

It's actually quite complicated with pen and paper, so just use the friggin' calculator, lol, since anything raised to a decimal power, solved by hand will either involve log tables, natural logs, or power series (or a slide rule, lol), depending on how you attack the problem. The basic issue is something like 10 to the 1.83 power really is 10^2 /10^0.17, which equals 67.6. The first part is easy (10^2=100), but the second part is hard (10^0.17=1.47). Or you could look at it as 10^1.83 equals 10^1 (which is 10) times 10^0.83 (6.76), if you want to multiply. Either way, if you really want to calculate it all out, take a look at the example in the link, which has multiple options.

http://math.stackexchange.com/questions/21381/how-to-calculate-a-decimal-power-of-a-number

 

[RutgersRaRa]

It's actually quite complicated with pen and paper, so just use the friggin' calculator, lol, since anything raised to a decimal power, solved by hand will either involve log tables, natural logs, or power series (or a slide rule, lol), depending on how you attack the problem. The basic issue is something like 10 to the 1.83 power really is 10^2 /10^0.17, which equals 67.6. The first part is easy (10^2=100), but the second part is hard (10^0.17=1.47). Or you could look at it as 10^1.83 equals 10^1 (which is 10) times 10^0.83 (6.76), if you want to multiply. Either way, if you really want to calculate it all out, take a look at the example in the link, which has multiple options.

http://math.stackexchange.com/questions/21381/how-to-calculate-a-decimal-power-of-a-number

You're bringing back my memory here, Numbers. So far, so good. I just don't have any memory of using tables back in the day, and solved everything longhand. At least that's how I remember it.

 

[RU4Real]

My guess is that if OP had bought a Honda Accord like he was supposed to, instead of blowing all his money on some fancy German car, he'd be able to do maths.

 

[DJ Spanky]

My guess is that if OP had bought a Honda Accord like he was supposed to, instead of blowing all his money on some fancy German car, he'd be able to do maths.

He still wouldn't have figured out how to use the interwebz properly though.

 

[Scarlet_Scourge]

Your Math friend: http://www.mathsisfun.com/index.htm

 

[RutgersRaRa]

My guess is that if OP had bought a Honda Accord like he was supposed to, instead of blowing all his money on some fancy German car, he'd be able to do maths.

Before buyng, I did the math on my phone, and it turns out that 1.8 deutsche marks = 1 hectare, and as everyone knows 1 hectare = 3.2 yen, so it was cheaper to buy German after converting hectares to dollars because 1.8 is less than 3.2. Get with the program, people. Get with the program or be left behind.

 

[anvilofstars]

It's been a long time since I took my math classes so I am sure there are better ways (now I only do it for fun sometimes just to see if I can do it) and after a quick google search I think the below is probably your best bet.

This is a case where you have to approximate I think. a^b = e^(a ln b). (b is 1.83 here).

Taylor approximation here of ln b is .68. e^.68a is your answer. ln (1 + x) = x -x^2/2 + x^3/3 - x^4/4 etc. The more of these you use the more accurate your answer will be.

e^.68a can be approximated using taylor approximation. 1 + x + x^2/2! + x^3/3! and so forth.

 

[SouthJerseyRU]

Anyone else find it ironic that someone with perhaps the most famous equation of all time their user pic is asking this question?

Others have answered about how to do the the calculations, for those interested on the baseball part of this topic, Baseball Prospectus has the 1st, 2nd, and 3rd order win-loss record on their site which are based on similar equations. http://www.baseballprospectus.com/standings/ .

Wikipedia has a good article about the use of Pythagorean Expectation in baseball, as well as variations on the formula for other sports.
https://en.wikipedia.org/wiki/Pythagorean_expectation

 

[PhilaPhans]

Logs and natural logs, people!

 

[RutgersRaRa]

Anyone else find it ironic that someone with perhaps the most famous equation of all time their user pic is asking this question?

C'mon now, it's not ironic, it's consistent. Besides, it's written on a cork, and i therefore cannot be held responsible for everything I type.

 

[RutgersRaRa]

It's been a long time since I took my math classes so I am sure there are better ways (now I only do it for fun sometimes just to see if I can do it) and after a quick google search I think the below is probably your best bet.

This is a case where you have to approximate I think. a^b = e^(a ln b). (b is 1.83 here).

Taylor approximation here of ln b is .68. e^.68a is your answer. ln (1 + x) = x -x^2/2 + x^3/3 - x^4/4 etc. The more of these you use the more accurate your answer will be.

e^.68a can be approximated using taylor approximation. 1 + x + x^2/2! + x^3/3! and so forth.

Based on the above, how would you simplify the number, 10^1.5, and what does it equal? It's smaller than 100 since 10^2 = 100, and the exponent in the latter is larger.

P.S. I don't remember using factorials in equations involving exponents, but damn I love me some factorials! Kinda bringing back some memories ITT.

 

[RU848789]

Based on the above, how would you simplify the number, 10^1.5, and what does it equal? It's smaller than 100 since 10^2 = 100, and the exponent in the latter is larger.

P.S. I don't remember using factorials in equations involving exponents, but damn I love me some factorials! Kinda bringing back some memories ITT.

Some decimals are easy to do, when they are simple fractions. 1.5 = 3/2, so it's really (10^3)^1/2, or the square root of 10^3 or the square root of 1000, which is pretty easy to do by trial and error, by hand. And it's even easier to do with a calculator, lol (approximately 31.227766).