Find the Minimum | NO Calculus Allowed !
Вставка
- Опубліковано 9 чер 2024
- Check out Brilliant and get 20% off their annual premium subscription! =D brilliant.org/FlammableMaths
Support my channel and become a Patron! Please help keep the videos going!!! =) / mathable
Subscribe to @NPCooking69 for some tight ingredient bamboozeling! :D • The Underdog of the Po...
My Merch! =D papaflammy.myteespring.co/
Diffy Q Playlist: • Separable Differential...
Today a viewer suggested problem: find the minimum value of x+1/x without using calculus! I'll show two approaches today and I hope you'll enjoy the video :) Enjoy! =D
Help me create more free content! =)
/ mathable
Merch :v - papaflammy.myteespring.co/
www.amazon.com/shop/flammable...
shop.spreadshirt.de/papaflammy
Twitter: / flammablemaths
Instagram: / uncomfortably_cursed_m...
0:00 Intro
0:23 Method 1 Calculus lmao
3:13 Method 2
6:30 The Truly elegant and circular Method 3
You forgot the most important way:
Start with axiomatic set theory and derive all relevant theorem from scratch
The most beautiful way is to notice that the function is convex on the positives, which is easily proven. Thus it only has one minimum. Next (now comes the beautiful part) it is invariant under the transformation x->1/x. Thus, also the minimum has to be invariant (since there is only one). This finally leads to x=1.
x=-1 is also invariant so you would just have to find that there is also one maximum, and since the function has one discontinuity (x=0, where there is a singularity), you could show 1 is the minimum and -1 is max since x approaching 0+ the function goes to ∞ and same for x approaching ∞, thus 1 must be the minimum. maybe this isnt necessary and ive misunderstood your proof but nice anyway (i just think it was slightly incomplete)
@@alfiealfie35however if you consider the William jamama theorem you‘ll see that your mOm is fat
@@alfiealfie35 yes, in principle you also need to show that on the negatives there is no other minimum. But the function is only convex on the positives, so my proof only concerned the positives. To address the maximum on the negatives, you could argue that the function obeys f(-x) = -f(x), which directly implies that -1 has to be a (local) maximum. But notice that the function has no global minimum for x in the whole of R (since it goes to -infinity for x -> 0^-). So the restriction to the positives is somewhat natural.
That... is super elegant.
AM GM is the best until u study calculus
Pretty straightforward
@@luisisaurio I know but I'm saying that it is lengthy as compared to calculus
Tan u =x
F(x)=tan u + cot u
tan u = cosec 2u - cot 2u
cot u= cosec 2u +cot 2u
F(x)=2cosec 2u
=>2
Excellent.
based
ima do this next time I have to find a minimum and confuse everyone lol
Haha that's a lot of firepower for this problem, but still quite the interesting method. I'd be wary of invoking too many identities though, since you wanna ensure everything you're doing is justifiable without using calculus.
My first question is, over what domain? It's clear we can make the function arbitrarily small by approaching x=0 from the negative
Or approaching -∞
buh?
For all x in R there is no minimum near 0 because it isn't defined
No matter how small the x you will always find smaller point that is lower than that.
The same thing goes for maximum near 0
If that's clear, then it should also be clear what the intended question was. But yes, I suppose it should have been worded better.
The question isn’t asking you to find the minimum value, it’s asking for the minimum point. A “minimum point” has to have a gradient of 0 and has to be concave up. This is different from the “minimum value”, which is just the smallest value found along an interval.
It looks like my hands are tied huh😂
In Spanish notation, the parentheses are superfluous and usually omitted.
I did something somewhat similar to method 3, but not quite as rigorous:
Let x + 1/x = b and multiply by x.
Then you get a quadratic x^2 - bx + 1 = 0
This has a repeated root at stationary points (graphically, this is obvious), and we get b = 2 as the relevant minimum.
I've done a lot of inequality problems for fun with my colleague and I've found that they're always either AMGM, Jensen, or Cauchy-Schwarz
You can imagine an horizontal line with value c that intersects with the function such that you can use the quadratic equation to find all possible c which gives you at the end c²-4>=0
Because we want only real solution.
and because we can have two roots for x that means that the horizontal line intersects twice with the function so we choose
|c|=2 to get one root for x and knowing that |c| can't be less than 2 we know for sure that the minimum is at x = 1 when c = 2
What a coincidence!! Today only, I watched a visual proof of this.. They took a right angle triangle with two legs as x-1/x and 2 respectively. So the third side must be - x+1/x (By Pythagorean theorem). Now we know that hypotenuse must be greater than legs - so x+1/x ≥ 2
We have to find the minimum of that function without using calculus.
In the first step we use calculus to show that the function is always greater or equal to two...
MY SOLUTION (SPOILERS!)
I have the stinking suspicion that the answer is gonna be f(1) = 2
only considering the positive domain, of course
pf: f(1/x) = f(x) which means I can just focus my attention on x >= 1
\begin{SKIPPABLE}
split f(x) into x and 1/x. Notice that for all x >= 1, 0 < x = 1, x < f(x) = 2.
\end{SKIPPABLE}
Show that f(x) is increasing along the interval [1, +inf)
Assume we have some x >= 1 and some positive offset s > 0. Then the goal is to show that f(x) < f(x+s).
We have the following chain:
f(x)
= x + 1/x
= (x+s)(x + 1/x)/(x+s)
= (x^2 + 1 + sx + s/x)/(x+s)
= (x^2 + 1 + sx)/(x+s) + (s/(x+s))(1/x)
Let f(x) = x + 1/x | x € R\{0}
Notice that
f(1/x) = 1/x + x = f(x)
From this we get two cnadidates for extremum points, x=+-1 , sinc for example for all x>1, f(x) > f(1), then necessarily for all 0 2 = f(1), so our guess will be that 1 is a minimum point
Let x>0
We will try to find a solution to the inequality
f(x) > f(1)
x + 1/x > 2
x² + 1 > 2x
x² - 2x + 1 > 0
(x-1)² > 0
This inequality holds for all x=/1 in the domain
==>
x = 1 is a minimum of the function, solved using algebra only
You could also use graph where take f(x) = max(x,1/x) which would give you the minimum point of the graph at (1,2)
That only works by coincidence. In general if you want to minimise a sum of functions f(x) + g(x), you cannot simply apply this logic.
Spanish notation would make parentheses redundant when factorials are applied to large expressions. ¡Clearly another W from the superior orthography!
|¡i!|=√(π cschπ) is just a headache.
First, x + 1/x -> -infinty as x -> 0 from the left, so I'm going to assume we are minimizing over positive x. Notably, when x > 0, y = x + 1/x > 0 as well. Doing some rearranging, we get that x^2 - yx + 1 = 0. We want to minimize y, so we choose the smallest positive y such that there exists a positive x which satisfies the equality. Since it is a quadratic, we know there exists a real x which satisfies it whenever the discriminant is non-negative. Here, the discriminant is y^2 - 4, and the smallest positive y that keeps it non-negative is the one that makes it zero, i.e. y = 2. Plugging this back into the quadratic, we can see that x =1 > 0, so this is a valid choice and thus it is the minimum.
Haven’t watched the video yet but it’s literally: AM-GM: x+1/x >= 2sqrt(x*1/x). So x + 1/x >= 2 meaning the it’s smaller value is 2
If x is positive, then noting that f(x) = f(1/x) implies an extrema at x = 1/x --> x =1 --> f(1) = 2. endpoints imply a minimum since f(0+) = f(+infty) = +infty.
Well, the endpoints aren't sufficient for the minimum. You also need to know there's only one extremum. And that's not implied by f(x) = f(1/x) alone (indeed you can construct functions with as many extrema as you like satisfying that functional equation).
Setting that aside, how do you even know x = 1 is an extremum? For a continuous function, say. And by "extremum" I mean that it's the local maximum or minimum in some sufficiently small interval around it. I don't disagree, hell, I actually completely agree that this should be the case, I'm just unsure how you'd show it to be so.
Edit: I know how to do it for a differentiable function with continuous derivative. I'm willing to conjecture that this is also a necessary condition, as if it doesn't hold, you can potentially let the function vary like x^2sin(1/x) near x=1 (much like the "topologist's sine curve") and fill in the gap at x=1 to establish continuity, also maintaining differentiability, with a derivative everywhere. And the derivative is indeed 0 at x=1. But it's not a minimum, as you can always find points smaller in any interval ; not a maximum for the same reason. Hell, similar reasoning shows it can't even be an inflection point. It's a weird critical point, and a somewhat pathological counterexample admittedly.
Hi Papa Flammy, I've been wondering for quite some time about how I can find a global maximum and minimum of a function? I realize that taking the derivative test only guarentees local extrema points and not global ones. Is it sufficient to say that, if we have f: I-> R, if I is closed then it must have a minimum and maximum by the EVT, therefore if we have an increasing function on some part of the domain, or a decreasing function on some part of the domain, if we then substitute the outermost part of the domain where f is decreasing/increasing, and then compare it to the local extreme points which we get by differentiating our f, see which is bigger and conclude on that? Is this enough? Also, say f: I-> R, I is an open interval, then we do the same approach however we take the limit and now say they would be our supremum and infimum? I have searched wide and far through the internet but I cannot quite find a good solution about finding global min and maxes of functions.
Also, have you considered doing videos on complex integration? Things like contour integration I believe need a lot more "Bigger picture videos" then there are on the internet at the moment, I can't quite find a good intuition about understanding how they work.
Thank you for your time and hope you have a great day!
There is no shortcut for finding the global minimum. You have to find and compare all local minima and check the boundaries.
So, lets just assume that minimum exist and it equals m. So there exist x =/= 0 such as:
x + 1/x = m
we can multiply it by x and just solve for x. We get rather nasty x1 and x2, but we know that extremum of that will be at arithmetic mean of x1 and x2. Luckily the mean is rather nice:
mean(x1, x2) = m/2
So we know that the point we are looking for has following form: (m/2; m). So it must be true that:
f(m/2) = m
m/2 + 1/(m/2) = m
which has two nice solution:
m = -2 and m = 2
But we know we are looking for a point that has form (m/2, m), so we get two points:
(-1, -2) and (1, 2)
Problem is I am not really sure how to proof which point is minimum and which point is maximum without calculus.
last time i was this early to a papa flammy video, andrew was being fed in the basement
Assuming real numbers, A + B is always greater or equal to 2sqrt(AB) because (sqrt (A) - sqrt (B))^2 must either be positive or zero; in other words:
(sqrt (A) - sqrt (B))^2 not less than 0;
Multiply out and add 2 sqrt(A)sqrt(B) to both sides
We get: A + B not less than 2 sqrt(A)sqrt (B)
In this case the minimum is 2sqrt(x* 1/x) = 2 which is our answer.
Well, It doesn't prove it's a minimum. You've just found some value, that doesn't exceed the function. You could've argued the same thing for 1, for that matter.
method 3 is mvp for me... though I would approach in a bit different way.
1) Suppose we have positive a and x, such that a=1/x+x
2) Multiply it by x and gather all terms on one side: x^2-a*x+1=0
3) we got quadratic equation but we already know the x (by assumption 1), so we know there is at least one root. So the determinant must be nonnegative D = a^2-4*1*1 = a^2-4 >= 0
4) and so a>=2
In the end we got: if a=1/x+x then a>=2. I like this method more, since we find "2" and not consider it from the begining
Muirhead:
(1, -1) majorizes (0,0) -> sum_sym xy^-1 >= sum_sym 1 = 2
Let y = 1, then sum_sym xy^-1 = x + x^-1
-> x+x^-1>=2 and equality when x = 1.
This proof is only valid when x is a nonnegative teal number, but if x could be a negative real number, then the minium value would approach -inf.
idk about minimum but choose negative delta when delta is smol, or very big, either is fine
f(x) = 2cosh(lnx) for x>0 and cosh has a minimum of 1, so f(x) > = 2 and = 2 when x = 1
negative infinity
over which interval
Complete the square on
sqrt(x)^2 + (1/sqrt(x))^2
= (sqrt(x) + 1/sqrt(x) )^2 -2
= (sqrt(x) - 1/sqrt(x) )^2 + 2
My rule of thumb is "the minimum value of x+f(x) is reached when x=f(x)",not sure if this works for all f(x), and may need AM GM for multiple terms
Doesn't work at all. Consider f(x) = 1/x^2.
Advice: Before declaring a rule of thumb, maybe try it out on a non-trivial example. Better yet, justify it.
@@fahrenheit2101 good to know
rip
Just plot the function. It will be obvious as x approaches +-1 the function changes directions. 1 is a local minimum. -1 is a local maximum.
Normally you use calculus to plot functions unless you want to do it by putting lots of different points in a coordinate system and then drawing the graph along the points
@@antenym8947 You can still plot a function with simple intuition. You only need calculus for the stationary points.
Here we know the graph lies entirely in the 1st and 3rd quadrants, and in particular, between the y-axis and the line y=x, and this is all easy to justify. Then, you can consider limits, and you're effectively done. But from there, you either need a lot of specific points, or some more clever work to actually get the minimum.
Minimum at x approaches -infinity.
When you realize positivity of x^2 in real world
No possible answer as the limit in 0- is -∞.
Depends on the meaning of the question. Find a local minimum works.
Or you could infimise the function, and get -∞ as your answer.
By using Calculus, it won't come. It doesn't make sense... If you consider negative numbers also (except zero)
To understand what I am saying,
Take f(x) = x + 1/x
So, f'(x) = 1 - 1/x^2
Thereby, the critical points are x = 1,-1
f''(x) = 2/x^3
This gives us, local maxima at x = -1 and local minima at x = 1 with the maximum value being -1 and minimum value being 1....
only it's 2 and -2, and what doesn't make sense? it's local minima and maxima, not global
It means there's a stationary point at x=1 which is smaller than all other f values in a δ neighborhood.
Why didnt a get a notification?
Easy-peasy, plot the graph and prove that there is no global minimum and one local using the definition. No calculus needed
Correct me if I’m wrong, but Method 2 doesn’t proof 2 is the minimum, just that it’s a lower bound.
It works because f(1) = 2. Obviously, since f(x) ≥ 2 for all x > 0, the minimum can't be less than 2, and since f(1) = 2, anything higher wouldn't be a lower bound. So the minimum must be exactly 2.
@@blastjer Yes but that relies on insights from method 1 using calculus.
@@woody442 Not necessarily. If you just happen to try the AM-GM inequality, you get the same result, no calculus needed. Of course, you're right that he knew the minimum because he had already found it from calculus, and that prompted the use of AM-GM, but logically method 2 is a perfectly valid proof on its own.
@@blastjer but you don’t know f‘(x)=0 -> x=1, -1.
So what values of x would you try to know the minimum is 2?
@@woody442 You don't need to know that f'(x) = 0 => x = 1, -1. AM-GM gets the inequality f(x) >= 2, and you can tell that f(1) = 2 by inspection. You don't need to try particular values of x.
It's easier. The X part gets further from 0 as the absolute value of x increases. The 1/X part gets closer to zero as the absolute value of X increases. They are pulling in separate directions. So, to find the extrema, just set X = 1/X and solve. The answers are 1 and -1. To find which is the max and which is the min, just plug 1 and -1 into the original equation and solve.
That's not even close to a convincing argument. It happens to work, but would not generalize.
Indeed the same logic applies to x + 1/x^2 , but the method does not apply.
@@fahrenheit2101 1/x^2 pulls in the same direction as x when x is negative and even.
@@InventiveHarvest We were never discussing negative x, but if you're still not convinced, 1/x^3 fails just as hard.
@@InventiveHarvest Also, nothing's "pulling". Pure intuition can only get you so far. Your approach as stated is not valid. There may be credence to it, but it would need to be expanded on significantly to make sense of it.
Are you the guy who made that video 54 years ago
0:03 Rosana Rodríguez-López approves. Maybe.
So this is the 911th video.
Coolio
with calculus you'll also find that the function is minimised at at x=1 and the min is f(1)=2 after taking the second derivative
Maid outfit for 400K?
You should use calculus, that s all