Wikipedia:Reference desk/Archives/Mathematics/2015 April 13

= April 13 =

Algebra
I am trying to get 1-e^(-1/L)=0.95 into the form e^(1/L)=20. The = should be greater than or equal to but I couldn't get that sign on here for some reason. I have tried doing the algebra but the closest I could get to was e^(-1/L)=0.05, where = means less than or equal to. I am not sure how to turn the -1/L into 1/L or change e into a. Positive value without changing the sign. Any hints? Clover345 (talk) 22:20, 13 April 2015 (UTC)


 * Is this what you were trying to write ?
 * I can convert this inequality:

1 - e(-1/L) ≥ 0.95
 * into this one:

e(-1/L) ≤ 0.05
 * but how do I get to here:

e(1/L) ≤ 20
 * As far as hints: "What happens when you change an exponent from positive to negative ?" (If you don't know, get a calculator and type in 23 and 2 -3, then compare the results.)  StuRat (talk) 22:35, 13 April 2015 (UTC)


 * I can't find a particularly suitable link that a student might benefit from. Perhaps guiding, but not directly applicable (because it only deals with integer exponents), is Exponentiation § Negative exponents. —Quondum 23:12, 13 April 2015 (UTC)


 * Looks good to me. (The rule doesn't only apply to integer exponents.) StuRat (talk) 23:26, 13 April 2015 (UTC)


 * Also relevant is Inequality (mathematics). -- ToE 01:28, 14 April 2015 (UTC)


 * BTW, whenever you can't get a proper "greater than or equals" sign (or "less than or equals" sign), it's fine to use ">=" or "<=". StuRat (talk) 23:53, 13 April 2015 (UTC)


 * Incidentally, that last inequality should be reversed:
 * e1/L ≥ 20
 * —Quondum 00:19, 14 April 2015 (UTC)

Thanks everyone. I know that the negative indece would get me 1-1/e^(1/2) >=0.95. but that doesn't get me any closer to the expression I want. The bit that really confuses me is where that 20 comes from. Clover345 (talk) 01:17, 14 April 2015 (UTC)
 * Do you understand the following properties of inequalities:
 * If x < y then -x > -y.
 * If x < y then 1/x > 1/y when x & y are both positive or both negative.
 * And the following properties of exponents:
 * x-1 = 1/x
 * (xa)b = xa⋅b
 * which can be combined to form
 * x-a = 1/xa
 * -- ToE 01:48, 14 April 2015 (UTC)
 * I see that you have used all but the second inequality property listed above, and that so far you have:
 * 1 - e(-1/L) ≥ 0.95
 * -e(-1/L) ≥ -0.05
 * e(-1/L) ≤ 0.05
 * So what happens when you apply the unused property? (That is, how do you solve 1/x ≤ 0.05 when you know that x > 0?  And why do you know x > 0?) Also, your most recent edit appears to have introduced a typo, substituting "2" for "L", which makes the inequality false. -- ToE 07:59, 14 April 2015 (UTC)

So I now have 20<=L but how does this turn into e^(1/L)>=20? Clover345 (talk) 15:53, 14 April 2015 (UTC)
 * Umm, you have e(-1/L) ≤ 0.05, and you already said that you know this is equivalent to 1/(e(1/L)) ≤ 0.05, so how did you come up with 20<=L (which is incorrect) and not the answer you are seeking? (Hint: To get what you want just apply the second inequality property from above.) -- ToE 16:46, 14 April 2015 (UTC)

I got it from the question in your hint. This is really puzzling me. This is basic algebra and I'm confused. Clover345 (talk) 19:29, 14 April 2015 (UTC)
 * In ToE's hint, x is not L, it's $$e^{1/L}$$. -- Meni Rosenfeld (talk) 08:03, 15 April 2015 (UTC)

Rosenblaty transform
In a Rosenblatt transform, if you have g(x,y)=x-yz where z is a constant, can you just drop z so it becomes mu-x+sigma-x*U-x - mu-y+sigma-y*UY? Clover345 (talk) 22:47, 13 April 2015 (UTC)


 * I'm not sure I know the term. Are you talking about multivariate transformations like the ones mentioned in this and this? Looking at the redlinks we have for both of these terms, we could certainly use an article on the topic. (We do, however, have an article on Murray Rosenblatt.) -- Impsswoon (talk) 10:29, 14 April 2015 (UTC)

Here is a paper on the topic. http://archi2.ace.nitech.ac.jp/yzhao2/peper-hm/MM-SS01.pdf Clover345 (talk) 18:47, 14 April 2015 (UTC) Equations 29-32 are relevant. Clover345 (talk) 18:26, 15 April 2015 (UTC)