Sum of a Positive Number and its Reciprocal from Calculus
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- Опубликовано: 5 июн 2024
- This is a short, animated visual proof demonstrating that sum of a positive real number and its reciprocal is always greater than or equal to 2.
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It turns out that this theorem is equivalent to the Arithmetic Mean-Geometric Mean inequality. The equivalence is implied by the following two proofs: • Visual Proof of AM-GM ...
• x + 1/x is greater tha...
And here are two alternate ways to prove this fact via the Pythagorean Theorem and triangle areas:
• x plus 1/x is greater ...
• x plus 1/x is greater ...
This animation is based on a proof by Roger B. Nelsen from the December 1994 issue of Mathematics Magazine, page 374 (doi.org/10.2307/2690999). You can find this here too: www.maa.org/sites/default/fil...
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bro that's crazy. I never looked at functions like this
I never was teached like this
That's a creative use of calculus ngl
another method using calculus:
compaire 1/x + x and 2 so we do
1/x + x - 2
= (x² - 2x + 1)/x
= (x - 1)²/x
(x-1)² always positive
and we have x > 0
That means (x-1)²/x ≥ 0
1/x + x - 2 ≥ 0
1/x + x ≥ 2
but I liked your method
I wished you'd give us a little more time at the end of the video to contemplate your question, but it helped to watch the video a second time.
There's a nice visual proof for the AM-GM inequality, using the curveness of ln (or e^x, doesn't really matter) like here.
wow, that's a way to prove it!
so cool!
The function f(x) = x + 1/x is an example of a function that is invariant under inversion, meaning it is unchanged when x is replaced with 1/x.
Another way of saying the same thing is that f(x) = x + 1/x is a solution to the functional equation f(x) = f(1/x). A function is invariant under inversion if and only if it is a solution to this functional equation.
It is a kind of symmetry similar to functions that satisfy f(-x)=f(x), called "even" functions, which are symmetrical about the y-axis, except here the function on the interval (0, 1] is symmetric with the function on [1, ∞).
There is a relationship between even functions and functions that are invariant under inversion. Hint: think about how log functions transform multiplication into addition.
f(log(x)) is an even function?
This took me way to long, but I think I got the relationship?
given an even function f(x)
f(log(x)) is guaranteed to be invariant under inversion
(took me multiple avenues/approaches on like guessing what the relationship could be, but landed me this, I wouldn’t be surprised if this isn’t the relationship)
Edit: hmm, log can just be any function g where g(ab) = g(a) + g(b) and g and g(1) = 0
(implies g(x) = -g(1/x))
so given h(x) = f(g(x)), then h(x) = h(1/x)
Hi, what text book would you recommend for studying that type of behaviours?
I was just learning about this function, RUclips algorithm works in mysterious ways
through your microphone
Such a beautiful proof.
Considering its given that x>0 couldnt you multiply both sides by x without any consequence?
Beautiful
I have never seen this, and now I can’t not see it!😊
Math is life ❤
Fun fact: the area bounded by the tangent line of f(x) = 1/x at any point x = a with the positive x and y axes is always 2.
Another way to prove this :
(x-1)^2 >= 0 (because a square is always positive)
x^2-2x+1 >= 0
x^2+1 >= 2x
Now divide by x, the inequality won’t change because x>0
x+1/x >= 2
The equality happens when x=1
Could also be proved using the property AM is greater than or equal to GM
COOL
Hi! I’m curious, What software do you use to edit/create these videos?
These are all done in manim, which is the python library created by 3blue1brown
yay jensen's
1
1 + 1/1 = 2, it's still true
@@user-LinaRexTone Yes, that is where equality holds.
▶️.
No, I don’t see it. It seems to break as soon as you input x=2 or x=3. Am I missing something? If so, then what?
When x=2: the function y=1/x passes through (2,1/2) while the line passes through (2,0). When x=3: curve goes through (3,1/3) while line goes through (3,-1). In both cases, the line passes through a lower point that the curve, which is to say that the line's y-value, 2-x, is less than the curve's y-value, 1/x. That is, 2-x
Yeah I think I can see it 😉
|N+1/n|≥2 for n in R
Let x > 0. f(x) = 1/x. This is a harded equation of a calculusi lmao
x = 1
So what ?
Whhaaat
this is the third proof you have of this
Fourth!
1
1 + 1/1 = 2, it's still true