Guest Posted February 13, 2010 Report Share Posted February 13, 2010 I was thinking about removing the integer restriction on factorials I'll call it a "fractorial" , use the factorial symbol of "!" and think of it like x! = x(x-1)(x-2)…(x-k) where x is any real number greater than 1 and the last term (x-k) is the last positive term in the series. So 4! = 24 1.5! = 0.75 2.5! = 1.875 1) Find 4.1! 2) Find 4.333… ! 3) Find a rational solution for w! = 0.11 4) Find a rational solution for x! = 0.897 5a) Find a rational solution for y! = 0.703125 5b) Find a rational solution between 1 and 2 for y! = 0.703125, or prove that it does not exist 6) Find a rational solution for z! = 2.252197265625 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2010 Report Share Posted February 13, 2010 I'm sorry to rain on your parade, but factorials are not actually confined to integers. In fact, the gamma function yields the factorial value of all real numbers. that includes negatives. for example, (-.5)!= sqrt(pi) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2010 Report Share Posted February 13, 2010 Aha -- very interesting magician -- thanks for the info. I think I've learned more math since joining this forum than I did in school. (I hope my ignorance doesn't spoil anyone's fun in trying to figure out the puzzle.) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 14, 2010 Report Share Posted February 14, 2010 1)2.93601 2)14.9794238... 3)1.1 4)1.57098... 5)(1+sqrt(3.8125))/2 6)2.5625 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 14, 2010 Report Share Posted February 14, 2010 1)2.93601 2)14.9794238... 3)1.1 4)1.57098... 5)(1+sqrt(3.8125))/2 6)2.5625 I got the same answers you got for 1,2,3, and 6. I came up with something different for 4 and 5. 4) I don't think there can be a rational solution in the 1 to 2 range. 5) Your solution looks "irrational." Looks like the rest didn't find this puzzle to be very interesting. Quote Link to comment Share on other sites More sharing options...
0 plainglazed Posted February 15, 2010 Report Share Posted February 15, 2010 I for one found this puzzle very interesting....the same irrational result as magician for 5. for the quadratic expression for possibilities between 1 and 2 but my math isnt strong enough to prove irrational or otherwise for polynomials of higher order. muy bueno! (yes, I enjoyed your other as well) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 15, 2010 Report Share Posted February 15, 2010 I for one found this puzzle very interesting....the same irrational result as magician for 5. for the quadratic expression for possibilities between 1 and 2 but my math isnt strong enough to prove irrational or otherwise for polynomials of higher order. muy bueno! (yes, I enjoyed your other as well) Muey bueno! Well, I think that you did prove it was irrational if you made a quadratic equation and the part under the square root (b2 -4ac) came out to an integer that wasn't a perfect square. (I had x2 +8x -45 = 0 for that one, which includes sqrt[64-4*1*(-45)] = sqrt(224) which is irrational.) I'm not sure if you took the same approach I did. On #4, for example, I changed it to a fraction = 897/1000 (I'd reduce this if possible -- but this one can't) If the x is between 1 and 2, I'd have x! = as the product of two fractions with equal denominators. I could see that the denominators couldn't be rational because those denominators would have to be sqrt(1000) which is irrational. But the cube root of 1000 is 10, so if there is a rational solution, there'd have to be three factors. So I'd make the equation (a/10)[(a+10)/10][(a+20)/10] = 897/1000. Then I'd see what the last digit of "a" could be so that "a" cubed ends in a 7. It finishes quickly from there. I found #6 to be interesting, because the denominator has several roots, which made for more trials. I suppose a person could play around with a spreadsheet and get lucky on some of them. Glad you liked it Quote Link to comment Share on other sites More sharing options...
0 plainglazed Posted February 15, 2010 Report Share Posted February 15, 2010 Ok, delving a little deeper... x(x-1)(x-2)=.897 ==> x3-3x2+2x-.897=0 ==> x=2.3 and x(x-1)(x-2)=.703125 ==> x3-3x2+2x-.703125=0 ==> x=2.5 could there be more? Quote Link to comment Share on other sites More sharing options...
0 plainglazed Posted February 15, 2010 Report Share Posted February 15, 2010 realized typo after window to edit expired - x=2.25 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 15, 2010 Report Share Posted February 15, 2010 realized typo after window to edit expired - x=2.25 And wondered the same thing on a couple of them -- could there be more? I had a denominator of 64 on # 5, after reducing = 26. I reasoned that for a rational solution, the only possible number of factors would have to be a factor of 6 greater than 1, so there might be a rational root below 2, one below 3 and one below 6. It seems like if you opened it to irrational solutions you'd have an infinite amount - at least one between every two integers. And if you open it to complex numbers I suppose there would be many more? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 17, 2010 Report Share Posted February 17, 2010 I f you do not use only integers as factors or factorial, then you have to throw out the rule that these factors are no longer limited by 1 to N(the number itself). The only way a number greater than 1 can have a factor of less than 1 is that if the other factor in that set of factors is greater than the number itself. Once you do this, then each pair of factors for the number 1.5 using decimals, the product of this number is always 1.5, and every time you use this, the sets of factors will be (2,0.75)(3,0.5)(4,0.375)(5,0.3)(6,0.25)(8,0.1825)(10,0.15) and so on, this would make the product of all the factor 1.5 to the power of how many sets you carry it out to, if you carry it out to infinity, then the factor of any number greater than 1 would be infinity. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 17, 2010 Report Share Posted February 17, 2010 I'm not sure which question you are answering. I f you do not use only integers as factors or factorial, then you have to throw out the rule that these factors are no longer limited by 1 to N(the number itself). See definition in original post. The number itself is the largest factor. The only way a number greater than 1 can have a factor of less than 1 is that if the other factor in that set of factors is greater than the number itself. True if the number has only two factors. Once you do this, then each pair of factors for the number 1.5 using decimals, the product of this number is always 1.5, and every time you use this, the sets of factors will be (2,0.75)(3,0.5)(4,0.375)(5,0.3)(6,0.25)(8,0.1825)(10,0.15) and so on, this would make the product of all the factor 1.5 to the power of how many sets you carry it out to, if you carry it out to infinity, then the factor of any number greater than 1 would be infinity. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 18, 2010 Report Share Posted February 18, 2010 (edited) I was interested in Magicians note in #2, with which I fully concur. The Gamma function, defined for positive x as the integral from 0 to infinity of yxe-y with respect to y, means that, when x is integral, then Gamma(x+1) = x!, though the definition, as noted by Magician, holds valid for non-integral x. For any positive x, a useful approximation to x!, considered to be valid for large values for x, is Stirlings approximation, which is the product of two terms:- (1) The square root of 2 pi x, (2) (x/e)x. This approximation can be improved immensely by multiplying these two terms by the following:- (3) The exponent, to base e, of the term 1/12x 1/360x3 + 1/1260x5 1/1680x7 + 1/1188x9. Even for x = 1 this yields an over-estimate of a mere 0.05%, whilst for x = 3, the error is lessened to less than a millionth of a percent. If one wishes to evaluate x!, or for that matter Gamma(x), for small positive values for x, a convenient method is to evaluate say Gamma(x+20) and then divide down using the relation Gamma(x) = Gamma(x+1) / x. Of course, another approach is to subscribe to Mathematica and get Wolfram to do the work for you. Edited February 18, 2010 by jerbil Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 23, 2010 Report Share Posted February 23, 2010 I was interested in Magician's note in #2, with which I fully concur. The Gamma function, defined for positive x as the integral from 0 to infinity of yxe-y with respect to y, means that, when x is integral, then Gamma(x+1) = x!, though the definition, as noted by Magician, holds valid for non-integral x. For any positive x, a useful approximation to x!, considered to be valid for "large values for x," is Stirling's approximation, which is the product of two terms:- (1) The square root of 2 pi x, (2) (x/e)x. This approximation can be improved immensely by multiplying these two terms by the following:- (3) The exponent, to base e, of the term 1/12x – 1/360x3 + 1/1260x5 – 1/1680x7 + 1/1188x9. Even for x = 1 this yields an over-estimate of a mere 0.05%, whilst for x = 3, the error is lessened to less than a millionth of a percent. If one wishes to evaluate x!, or for that matter Gamma(x), for small positive values for x, a convenient method is to evaluate say Gamma(x+20) and then divide down using the relation Gamma(x) = Gamma(x+1) / x. Of course, another approach is to subscribe to Mathematica and get Wolfram to do the work for you. Hadn't seen you on here for a while, jerbil. You posted a lot that's new to me. Thanks for the info -- looks interesting. Quote Link to comment Share on other sites More sharing options...
Question
Guest
I was thinking about removing the integer restriction on factorials
I'll call it a "fractorial" , use the factorial symbol of "!" and think of it like
x! = x(x-1)(x-2)…(x-k)
where x is any real number greater than 1
and the last term (x-k) is the last positive term in the series.
So
4! = 24
1.5! = 0.75
2.5! = 1.875
1) Find 4.1!
2) Find 4.333… !
3) Find a rational solution for w! = 0.11
4) Find a rational solution for x! = 0.897
5a) Find a rational solution for y! = 0.703125
5b) Find a rational solution between 1 and 2 for y! = 0.703125,
or prove that it does not exist
6) Find a rational solution for z! = 2.252197265625
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