Monotone Function Puns

u(x_1,x_2)=min{x_1,x_2}

v(x_1,x_2)=min{x_1,2(x_2)}

The problem says to use two bundles to show that v is not a monotone transformation of u. As you can guess, I have no idea how to do this.

Take some bundles:

u(1,1)=1, u(2,2)=2

v(1,1)=1, v(2,2)=2

While by eyeball it seems like the two functions aren't monotonic transformations of each other, it seems like any bundles you throw in preserve the same ordering. Both functions have income offer curves that are positively sloped straight lines from the origin. How could they not be monotonic transformations of each other?

Edit: looks like monotone transformations of the min function don't change the ratio that the consumer consumes in. In that sense, v is definitely not a monotone transformation of u. but how do I show that?

I'm having some problems proving the inverse is continuous. The hint in the book is to use the standard epsilon-delta definition of continuity. I believe the easiest route is a proof by contradiction, but with all of the quantifiers in the statement, I may be incorrectly negating the statement I am trying to prove. Also, I have at my disposal the intermediate value theorem, which most of my proof relies on. Below is the theorem:

\begin{theorem}Let $a < b$ be real numbers, and let $ f:[a, b] \to \mathbb{R} $ be a function which is both continuous and strictly monotone increasing. Then $f$ is a bijection from $[a, b]$ to $[f(a), f(b)]$, and the inverse $f^{-1}: [f(a), f(b)] \to [a, b]$ is also continuous and strictly monotone increasing.\end{theorem}
\begin{proof}Let $x_1, x_2 \in [a, b]$ be real numbers such that $f(x_1) = f(x_2)$. From the trichotomy of the real numbers, we have that exactly one of the following is true: $x_1 = x_2$, $x_1 < x_2$, or $x_1 > x_2$. Suppose $x_1 \not = x_2$. Then, by definition of strictly increasing monotone functions, we have that $f(x_1) \not = f(x_2)$. Thus, $x_1 = x_2$, and $f$ is injective.

Now let $y \in [f(a), f(b)]$ be a real number. Then, by the intermediate value theorem, there exists a real number $c \in [a, b]$ such that $f(c) = y$. Thus, $f$ is a surjection from $[a, b]$ to $[f(a), f(b)]$. Since $f$ is both injective and surjective, we can conclude that $f$ is a bijection from $[a, b]$ to $[f(a), f(b)]$.

To show that $f^{-1}$ is strictly monotone increasing, let $y_1, y_2 \in [f(a), f(b)]$ be real numbers such that $y_1 < y_2$. Then,by the intermediate value theorem, there exist $x_1, x_2 \in [a, b]$ such that $f(x_1) = y_1$ and $f(x_2) = y_2$. Since $f$ is strictly monotone increasing, we have $x_1 < x_2$. Using the definition of an inverse, we have\begin{align*}f^{-1}(y_1) &= f^{-1}(f(x_1)) \\&= x_1 \\&< x_2 \\&= f^{-1}(f(x_2)) \\&=f^{-1}(y_2) \text{,}\end{align*}showing that $f^{-1}$ is strictly monotone increasing.

Finally, we will show that $f^{-1}$ is continuous. Let $y_0 \in [f(a), f(b)]$ be a real number, and let $\epsilon > 0 $ be a real number. As before, there exists a real number $x_0 \in [a, b]$ such that $f(x_0) = y_0$. Likewise, for any real number $y \in [f(a), f(b)]$, the intermediate value theorem tells us that there exists a real n

Let me preface this with that I know there exists a "monotonic transformation", but this is not what I mean. In a monotonic transformation, there is no specific transformation defined -- the only thing that matters is that whatever the transformation is, it doesn't change the order of preference, and non-increasing or non-decreasing. A simple example: if you have f(x) = x, then a monotonic transformation would be g(x) = x+3, because any two numbers g(a) and g(b), where b>a, f(b) > f(a) and g(b) > g(a). So the order of preference doesn't change, and by looking at dg/dx, the function is always increasing.

What I'm asking about is a specific "transform" (I'm not sure if this would be an accurate term) that doesn't care about order, and in fact changes the order to fit a monotone function over a given interval. So in the same way that the Fourier transform, changes a function in the time domain to one in the frequency domain, I'm looking for a "transform" that changes a function to one that is always increasing over the same exact domain.

Here's how I'm imagining how something like this might work. If we look at a discrete function that has more than one maxima and minima, this transform would take all of those values, and reorder them such that the smallest value for a given interval comes first, and the largest value comes last. So, for a function over x with an interval of [0,4] with say points x = {0,1,2,3,4} and f(x) = {0,4,2,5,1}, then this transform would output for our values x = {0,1,2,3,4} and g(x) = {0,1,2,4,5}. In other words, our function has been "sorted". Our function has been transformed into a monotonic function that occupies the same interval and incorporates all of the same values; the x and f(x) has just been swapped around to "sort" f(x) in increasing values. Now of course, this is a discrete case to illustrate what I mean; however I'm wondering if something like this could be done on a continuous function, where the values of g(x) at x={0,1,2,3,4} might be different do to the "rearrangement" of output values. But, the "total" should remain the same. In other words, ∫f(x)dx = ∫g(x)dx for x-interval [a,b].

If this doesn't exist, is it even possible to develop such a transform? I see value in such a transformation where you know the continuous dynamics of a system, and wish to see the median, or perhaps a percentile of an independent parameter of the response of this system. As an example, maybe you know the continuous w

You are given a monotone function f: N= {1,2,...} to R so that f(1)=0,f(2)=1.

f also satisfies f(mn)=f(m)+f(n)

Prove that f(n)=log_2 (n)

I was looking for a monotone function from R to (0,1) which preserved rationals and I found one -- see below. But now I'm wondering if there are such functions which are infinitely differentiable on R as well.

My function which has an undefined second derivative at 0 is as follows.

f(x) = (1/2)(1 + (x / (1 + |x|)))

Here's a graph of f together with its first derivative.

I did find that there are bounded analytic functions from R into R which preserve the rationals, but the construction involved uniform limits of products of rational functions and the resulting function was far from monotone -- see the responses to this question for details.

If we have a function f that has the derivative (e^(2x) - e^2) * ln(x) and we try to show monotonicity of f, we can just check the derivative.

However, I‘m not quite sure why the following holds.

We know the derivative > 0 for all x in R>0 {1} . Also the derivative evaluated at 1 is = 0. Hence we get that f is strictly increasing on (0,inf)

However this I do not quite understand. We can split it to show that it strictly increasing on (0,1] and [1,inf)

But does the fact that f‘(1)= 0 not change strict monotonicity? Is it because there is only one such x?

Utility is the ability to satisfy certain wants and needs. It is considered an essential concept in economics because, it shows the satisfaction received from the consumption of a good or service, and it also considered to be a method of assigning a number to every consumption bundle.

Let f, g, h: ℝ → ℝ be functions such that any linear combination of them is monotonic. Prove that they are linearly dependent.

Is it possible (and if not, why?) to prove the mean value theorem for monotonous functions by applying a rotation matrix to the points of the function so that the line (f(b)-f(a))/(b-a) is horizontal in the rotated plane and then apply rolles theorem to say that there exists the desired derivative on the interval f’(x), x in (a,b)? I know that it could only works for monotonous functions because otherwise the rotated function might not be a function anymore (more than 1 value for a point in the domain). I can prove that the values of the function in the rotated space are on the same horizontal line and that the interval (a,b) conserves its order and non-emptiness, would there still be some other problem with this? For example a derivative would have to be defined in the new basis and that would crash?

Let f : (0; infinity) -> R, f(x) = xlna - alnx, where a is a real number in (0; infinity)

Determine the values of a for which f is monotonous on the interval (0; infinity). So the values for a in which the function is either increasing or decreasing for any positive x.

I first tried to take the derivative. f'(x) = lna - a/x. If we assume that f'(x)>0 (increasing function) then ln(a) > a/x and I don't know where to take it from here since there are two unknown variables and only one inequation.

Next I tried the old-school method of calculating [f(x)-f(y)]/(x-y) and seeing if it's positive or negative for any x, y part of (0; infinity). (Or you could think of them as x_1 and x_2)

I'll denote a positive number with (+) and a negative number with (-)

[f(x)-f(y)]/(x-y) = [xlna - alnx - ylna + alny]/(x-y) = [ln(a) * (x-y) + a(lny - lnx)]/(x-y) = [lna * (x-y) + ln(y/x)^(a) ]/(x-y)

Let's assume x>y. Thus x-y >0 so x-y = (+)

[lna * (+) + ln(y/x)^(a) ]/(+) = lna + ln(y/x)^(a) (it's not "equal" here but it will have the same sign)

We know x>y => y/x < 1 => ln(y/x) < 0

the whole thing becomes lna + (-)^(a)

I don't know where to take it from here either. ln(a) can be either positive or negative depending on whether a is bigger or smaller than 1. That wouldn't be a problem in of its own since I can just take two separate cases (that was the point of the exercise anyway) but a negative number raised to a positive power can be anything and I also need to determine whether lna will be bigger or smaller than ln(y/x)^(a) . Not only we have 2, but we have 3 unknowns and one inequation!

A real func is called monotonic iff, it is either non-increasing or, non-decreasing.

Consider the ordered set of numbers up to n: {0 ... n} and it's inclusion into the natural numbers N. This inclusion is a function. The function is monotonic: if a <= b in {0 ... n} then a <= b in N. But it is more than that, a <= b in {0 ... n} if and only if a <= b in N. Is there an order-theoretic name for this stronger property?

For context: I'm writing in a theorem prover (Isabelle) and reasoning about the conversion of words {0 .. 2^32 - 1} to natural numbers. I've written a little lemma and just want to give it the right name.

Hi, i am learning for my final exam of Calculus, we got a sample test, but I cannot answer this theoretical task. As far as I know sin(x) would be bounded and not monotonic, but it is periodic. Thank you <33

Example image. Note that the function is monotonic along both the x-axis and y-axis.

To be precise, I'm most interested in the case of two independent variables, like in the linked image.

Assume that I already have the function f(x,y).

To take this a step further: What if we operate strictly discretely (the independent variables and the dependent variable are always integers)?

Edit: Actually, I should add one further complication: I want to prove monotonicity across a specific range of values along each axis. Specifically (a,inf) along the x-axis, and (b,inf) along the y-axis.

Hi guys.

Are there any methods that only approximate monotonic function by using Neural network?

It is well known that a Neural network is able to approximate any functions. If the Neural network is shallow and all of its weights are positive, a function generated by the NN might be monotonic increasing function. In the case of deep neural network, however, it is hard to answer to this issue.

Any references or ideas are welcome.

Thanks.

I'm reading Rudin's Principles of Mathematical Analysis, and came across 4-20 "Let f be monotonic on (a, b). Then the set of points of (a,b) at which f is discontinuous is at most countable."

I feel like I follow the proof, but like... this proof is telling me that of all the functions who are discontinuous at the Cantor set, none are monotonic. I find this baffling.

Here's a pseudo-counterexample: Let f(0)=0, and dy/dx=0 wherever f(x) is continuous, but at any of the discontinuities of the Cantor set, shift f(x) on the right of the discontinuity so f(x+) = f(x-) + 0.1. Ie its a horizontal line, and you just break then shift the line up by 0.1 every time you hit a discontinuity. The discontinuities are uncountable, so it diverges, but... still monotonic, right?

And I'm even more confused, because it seems like Rudin's proof of 4.20 seems to hold for my case. Ie, because f is monotonic, at every discontinuity, there is a rational between f(x-) and f(x+), and thus the set of discontinuities maps to a subset of the rationals and is thus countable. But my function seems to have uncountably many discontinuities, and also seems to let us map discontinuities to a subset of the rationals, which is absurd...

Any thoughts?

Thanks!

If you have already waited 20 minutes for the bus, the bus can not arrive in less than 20 minutes because you can not "un-wait" the time you have already waited.

If you have 5 donuts and you eat some donuts - it is impossible to have more than 5 donuts when you are finished eating. This is because once you remove items from a set, the set can not have more items than the original number of items in the set.

If you walked 345 meters by 6 PM, the total number of meters you will have walked today can not be less than 345 meters. This is because you can not "un-walk" the meters you have already walked.

Do these concepts have proper names in math? E.g Commutative, reflexive, symmetric, etc?

How would you describe this property using mathematical terms - a set of objects that have certain properties, such that once a certain type of opperation is performed on objects in the set, the "cardinality" of set achieves a new infimum and supremum?

Do such terms in mathematics exist that can correspond to the examples I laid out?

Thanks!

I feel that Rudin just didn't motivate this, and it is causing me to become very confused. I don't know WHY it is there.

A - domain of the function

x1, x2 - elements of the domain

(∀x1, x2 ∈ A)(x1 ≤ x2 ⇔ f(x1) ≤ f(x2)).

(1/4)^(1-1/x) < 16*2^(2x - 3)

...

2^(-2+2/x ) < 2^(1+2x)

How would I apply the theorem to this inequality? What is happening here?

First - Previous

Captain Amanda Trent was seeing double.

The creature that had hold of her ship somehow pushed its voice directly into her brain. The process had not been pleasant. It also had not been particularly enlightening. Whatever it was trying to say, she couldn't understand.

"What the hell was that?" Tactical officer Weber was shaking and slapping his head, like he was trying to get water out of his ear.

"We didn't respond to its radio signal, maybe it thought we couldn't hear it." Comms officer Tran seemed relatively unfazed by the mental assault.

"Damage Report."

"EMP, but no explosion, so not a nuke. The only damage so far seems to be fried electronics." Tran was listening in on the damage control teams as they began their work. "No hull breaches, that thing has wrapped us up nice and tight with its tentacles, but either isn't strong enough to squeeze us apart, or doesn't want to."

"Why is the reactor offline?" The reactor was hardened against nuclear strike, which of course also meant against EMP. All of the ship's critical systems were. But right now the only systems that were working were those with local battery backup power.

"Unknown ma'am, engineering says they'll report when they figure it out."

Captain Trent was blind and useless sitting on the bridge. She kicked her mag boots together to activate them, then released her harness. "I'm going to the observation deck to see what the hell is going on out there, keep me updated."

The bridge was located near the center of the ship. Alliance design philosophy was to put the important stuff behind as much metal as possible. So it took the captain a few minutes of climbing and walking to reach her destination.

The observation deck had the largest window on the ship, but Amanda couldn't see much of the creature. She pressed herself against the glass to try to get an angle up and down the ship's outer hull. An enormous eye stared back at her. The creature shifted until the eye filled the entire window in front of the captain. For a few moments they simply stared at each other.

The silence was broken by a screeching sound, worse than fingernails on a blackboard. A deep groove was carving itself into the floor of the room, with no apparent mechanism behind the damage. Then another groove and another. It was carving bl

If I have a function that is strictly increasing, does this also imply that it is monotone increasing?

Show that if I= [a,b] and f:I -- > {R} is increasing on I, then f is continuous at a if and only f(a)= inf {f(x) : element (a,b)}

let f be an increasing function on I = [a,b]. If f is continuous on a, then for and epsilon > 0 there exist a delta >0 such that abs(f(x)-f(a)) <epsilon. It follows that f(a) = inf {f(x) : x element of (a,b)}.

Can someone please explain this part to me. Were does the part that starts with "such that" to the end come from. I don't understand.