Something Fun I Stumbled Across

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Thinkaholic
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Hi! I know all of you might know what I'm about to post, but I just discovered it for myself, and I want to share my enthusiasm.
Let
Formula f(x) = square root of x divided by (x - 1).

and
Equation F(x) = d/dx f(x) = (1/2(sqrt((x-1)/x) - sqrt(x/(x-1))))/(x-1), with F(x) on the left and the derivative of f(x) on the right.
(here, I'll be restricting the domain of f(x) to the positive real numbers.)
Here is a graph of the two, with f(x) in blue and F(x) in black:

Graph on a grid with x from -2 to 10 and y from -4 to 4: a blue curve decreasing from above the top edge near x=1 toward y=1 as x grows, and a black curve rising from below near x=1 toward y=0 as x grows.

1st question: Where does f(x) intersect with the line y=x?

you could write
Equation: square root of (x / (x - 1)) = x

squaring both sides of the equation, multiplying both sides by x-1, and subtracting x from both sides gives
Equation x^3 - x^2 - x = 0 written in plain black type.

Factoring x from the LHS and dividing both sides by x leaves you with
Equation x^2 - x - 1 = 0 written in plain black text on a white background.

This is the minimal polynomial for the golden ratio, or φ, and the minimal polynomial for -φ^-1, or -Φ. This means that the quadratic above has two solutions at φ and -Φ. -Φ cannot be the solution we are looking for, as, as stated above, I am only dealing with f(x) within the domain of the positive real numbers (positive x values only). So, the intersection of f(x) and y=x is at (φ,φ)!

Question 2: What is the value of
Mathematical limit expression: lim as x approaches infinity of f(x), rendered in standard notation.
?

Using L'Hospital's Rule, we obtain that
Equation chain: lim as x approaches infinity of f(x) = lim sqrt(x/(x-1)) = lim (d/dx sqrt(x))/(d/dx sqrt(x-1)) = lim sqrt((x-1)/x) = lim 1/f(x) = 1/lim f(x).

calling the limit as x approaches infinity of f(x) "L", then this becomes
Equation L = 1/L, with L on the left and the fraction 1 over L on the right.

and obviously L=1. So
Limit expression: lim as x approaches infinity of f(x).
.

Question 3: What is the derivative of f(x) at (φ,φ)?

If we take the derivative of f(x), plug in φ for x, and make sure to remember that φ-1=Φ and that φ^-1=Φ, we simplify:

Equation: (1/2)(sqrt(Φ/φ) - sqrt(φ/Φ)) / Φ = (1/2)(sqrt(1/φ^2) - sqrt(φ^2)) / (1/φ) = (1/2)φ(1/φ - φ) = (1/2)φ(φ - 1 - φ) = -(1/2)φ
Question 4 (finale): What is
Improper integral from phi to infinity of F(x) dx, where the lower limit is the golden ratio phi and the upper limit is infinity.
?
From the fundamental theorem of calculus

Equation: integral from phi to infinity of F(x) dx = lim as x approaches infinity of f(x) minus f(phi) = 1 - phi = -Phi.


So we could rewrite this as:

Equation: (1/2) φ times the integral from φ to infinity of [(sqrt((x-1)/x) - sqrt(x/(x-1))) / (x-1)] dx = -1
Hope I made no typos! Sorry if this is too long, but I want to share these interesting facts with y'all. Also, hopefully the type doesn't mess up, I used rendered LaTeX and pasted the images here. Also, the prefix is beginner, as most of the calculus stuff is taught in high school, but I really don't know what this is, so sorry if that is wrong.
 
Attachments
  • Graph on a grid with x from -2 to 10 and y from -4 to 4: a blue curve decreasing from above the top edge near x=1 toward y=1 as x grows, and a black curve rising from below near x=1 toward y=0 as x grows.
    upload_2018-5-9_19-1-12.png
    3.4 KB · Views: 534
  • Formula f(x) = square root of x divided by (x - 1).
    gif.gif
    560 bytes · Views: 837
  • Equation F(x) = d/dx f(x) = (1/2(sqrt((x-1)/x) - sqrt(x/(x-1))))/(x-1), with F(x) on the left and the derivative of f(x) on the right.
    gif.gif
    1.2 KB · Views: 734
  • Equation: square root of (x / (x - 1)) = x
    gif.gif
    423 bytes · Views: 516
  • Equation x^3 - x^2 - x = 0 written in plain black type.
    gif.gif
    351 bytes · Views: 508
  • Equation x^2 - x - 1 = 0 written in plain black text on a white background.
    gif.gif
    323 bytes · Views: 514
  • Mathematical limit expression: lim as x approaches infinity of f(x), rendered in standard notation.
    gif.gif
    467 bytes · Views: 508
  • Equation chain: lim as x approaches infinity of f(x) = lim sqrt(x/(x-1)) = lim (d/dx sqrt(x))/(d/dx sqrt(x-1)) = lim sqrt((x-1)/x) = lim 1/f(x) = 1/lim f(x).
    gif.gif
    2.6 KB · Views: 512
  • Equation L = 1/L, with L on the left and the fraction 1 over L on the right.
    gif.gif
    248 bytes · Views: 530
  • Limit expression: lim as x approaches infinity of f(x).
    gif.gif
    520 bytes · Views: 530
  • Equation: (1/2)(sqrt(Φ/φ) - sqrt(φ/Φ)) / Φ = (1/2)(sqrt(1/φ^2) - sqrt(φ^2)) / (1/φ) = (1/2)φ(1/φ - φ) = (1/2)φ(φ - 1 - φ) = -(1/2)φ
    gif.gif
    2.1 KB · Views: 527
  • Improper integral from phi to infinity of F(x) dx, where the lower limit is the golden ratio phi and the upper limit is infinity.
    gif.gif
    623 bytes · Views: 502
  • Equation: integral from phi to infinity of F(x) dx = lim as x approaches infinity of f(x) minus f(phi) = 1 - phi = -Phi.
    gif.gif
    1.4 KB · Views: 505
  • Equation: (1/2) φ times the integral from φ to infinity of [(sqrt((x-1)/x) - sqrt(x/(x-1))) / (x-1)] dx = -1
    gif.gif
    1.2 KB · Views: 518
Last edited:
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When I "restricted the domain of f to the postitve real numbers" instead of differentiating the f(x) I gave at first, I differentiated the square root of x divided by the square root of x-1. That way the domain of the function and its derivative is restricted to the positive real numbers, and that is why you may have obtained a different answer for the derivative because you used the chain rule on the f(x) I gave, which differentiated f(x) with respect to all values of x.