Why is this?
04-07-2022, 02:12 AM
Post: #1
 lrdheat Senior Member Posts: 857 Joined: Feb 2014
Why is this?
Likely simple math, but I will ask anyway…I was messing around with graphing various functions. When I graphed x*e^-(x^2), I discovered that the max was equal to sin (45 degrees). Why?
04-07-2022, 03:37 AM
Post: #2
 Valentin Albillo Senior Member Posts: 1,100 Joined: Feb 2015
RE: Why is this?
(04-07-2022 02:12 AM)lrdheat Wrote:  Likely simple math, but I will ask anyway…I was messing around with graphing various functions. When I graphed x*e^-(x^2), I discovered that the max was equal to sin (45 degrees). Why?

Sin(45o) simply equals Sqrt(2)/2, which holds no mystery whatsoever.

V.

All My Articles & other Materials here:  Valentin Albillo's HP Collection

04-07-2022, 05:05 PM
Post: #3
 Leonid Member Posts: 54 Joined: Dec 2013
RE: Why is this?
(04-07-2022 02:12 AM)lrdheat Wrote:  Likely simple math, but I will ask anyway…I was messing around with graphing various functions. When I graphed x*e^-(x^2), I discovered that the max was equal to sin (45 degrees). Why?
By the way, global max $\sqrt[x]{x}$ is e
04-08-2022, 01:40 AM
Post: #4
 lrdheat Senior Member Posts: 857 Joined: Feb 2014
RE: Why is this?
Yes, and this is more intuitive to me. Is it purely coincidental that the max
of x*e^(-x^2) is (sqrt 2)/2, or is there a direct relationship to the sine function? I’m thinking that I am missing some basic math…
04-08-2022, 02:12 AM (This post was last modified: 04-08-2022 02:44 AM by Albert Chan.)
Post: #5
 Albert Chan Senior Member Posts: 2,679 Joined: Jul 2018
RE: Why is this?
f = x * exp(-x^2) is product of linear growth and exponential decay. A peak is expected.
Peak of f also at peak of ln(f) = ln(x) - x^2

(ln(f))' = 1/x - 2x = 0, we have x^2 = 1/2, or x = √(1/2)

---

g = x^(1/x) = exp(ln(x)/x)
ln(x) does not grow as fast as x, so again g is decaying after a peak.
Peak of g also at peak of ln(g) = ln(x)/x

(ln(g))' = (x*(1/x) - 1*ln(x)) / x^2 = (1 - ln(x)) / x^2 = 0, we have ln(x) = 1, or x = e
04-08-2022, 02:28 AM
Post: #6
 Valentin Albillo Senior Member Posts: 1,100 Joined: Feb 2015
RE: Why is this?
(04-08-2022 01:40 AM)lrdheat Wrote:  Yes, and this is more intuitive to me. Is it purely coincidental that the max
of x*e^(-x^2) is (sqrt 2)/2, or is there a direct relationship to the sine function? I’m thinking that I am missing some basic math…

- The derivative of y = x * e^(-x^2) is y' = e^(-x^2) * (1 - 2 * x^2)

- To find the extrema of y you must solve y' = 0.

- The exponential term is never 0 and the polynomial term 1 - 2 * x^2 is 0 for x = Sqrt(2) / 2, which is where y has a maximum.

- There's no relationship whatsoever to the sine function, just numerical coincidence with the numerical value of sin(45 deg), that's all.

V.

All My Articles & other Materials here:  Valentin Albillo's HP Collection

04-08-2022, 02:32 AM
Post: #7
 lrdheat Senior Member Posts: 857 Joined: Feb 2014
RE: Why is this?
Yes, it makes sense as far as the idea of an increase and then a decay in these functions. That the max of x*e^(-x^2) occurred at x=(sqrt 2)/2 suggested a trigonometric relationship to me, but how? It is beginning to seem coincidental now, and there is not a direct trigonometric relationship…
 « Next Oldest | Next Newest »

User(s) browsing this thread: 1 Guest(s)