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Talks
Spring 2015

# Efficient Minimax Optimal Strategies for Universal Prediction

Tuesday, March 17th, 2015 2:50 pm3:15 pm

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Location:

Calvin Lab Auditorium

In game-theoretic formulations of prediction problems, a strategy makes a decision, observes an outcome and pays a loss. The aim is to minimize the regret, which is the amount by which the total loss incurred exceeds the total loss of the best decision in hindsight. This talk will focus on the minimax optimal strategy, which minimizes the regret, in three settings: prediction with log loss (a formulation of sequential probability density estimation that is closely related to sequential compression, coding, gambling and investment problems), sequential least squares (where decisions and outcomes lie in a subset of a Hilbert space, and loss is squared distance), and fixed-design linear regression (where the aim is to predict real-valued labels as well as the best linear function). We obtain the minimax optimal strategy for these problems under appropriate conditions, and show that it can be efficiently computed.

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# 1.6.6E: Inverse Functions

section 1.6 exercises

Assume that the function f is a one-to-one function.

1. If (f(6)=7), find (f^{-1} (7))

2. If (f(3)=2), find (f^{-1} (2))

3. If (f^{-1} (-4)=-8), find (f(-8))

4. If (f^{-1} (-2)=-1), find (f(-1))

5. If (f(5)=2), find ((f(5))^{-1})

6. If (f(1)=4), find ((f(1))^{-1})

7. Using the graph of (f(x)) shown

a. Find (f(0))

b. Solve (f(x)=0)

c. Find (f^{-1} (0))

d. Solve (f^{-1} (x)=0)

8. Using the graph shown

a. Find (g(1))

b. Solve (g(x)=1)

c. Find (g^{-1} (1))

d. Solve (g^{-1} (x)=1)

9. Use the table below to find the indicated quantities.

 (x) 0 1 2 3 4 5 6 7 8 9 (f(x)) 8 0 7 4 2 6 5 3 9 1

a. Find (f(1))

b. Solve (f(x)=3)

c. Find (f^{-1}(0))

d. Solve (f^{-1}(x)=7)

10. Use the table below to fill in the missing values.

 (t) 0 1 2 3 4 5 6 7 8 (h(t)) 6 0 1 7 2 3 5 4 9

a. Find (h(6))

b. Solve (h(t)=0)

c. Find (h^{-1} (5))

d. Solve (h^{-1} (t)=1)

For each table below, create a table for (f^{-1} (x).)

11.

 (x) 3 6 9 13 14 (f(x)) 1 4 7 12 16

For each function below, find (f^{-1} (x))

13. (f(x)=x+3)

14. (f(x)=x+5)

15. (f(x)= 2 - x)

16. (f(x)=3-x)

17. (f(x)=11x+7)

18. (f(x)=9+10x)

For each function, find a domain on which (f) is one-to-one and non-decreasing, then find the inverse of (f) restricted to that domain.

19. (f(x)=(x +7)^{2})

20. (f(x)=(x-6)^{2})

21. (f(x)=x^{2} -5)

22. (f(x)=x^{2} +1)

23. If (f(x)=x^{3} -5) and (g(x)=sqrt[{3}]{x+5}), find

a. (f(g(x)))

b. (g(f(x)))

c. What does this tell us about the relationship between (f(x)) and (g(x))?

24. If (f(x)=dfrac{x}{2+x}) and (g(x)=dfrac{2x}{1-x}), find

a. What does this tell us about the relationship between (f(x)) and (g(x))?

1. 6

3. -4

5. 1/2

7a. 3
b. 2
c. 2
d. 2

11.

 (x) 1 4 7 12 16 (f^{-1}(x)) 3 6 9 13 14

13. (f^{-1}(x) = x -3)

15. (f^{-1}(x) = -x + 2)

17. (f^{-1}(x) = dfrac{x - 7}{11})

19. Restricted domain (x ge -7), (f^{-1}(x) = sqrt{x} - 7)

21. Restricted domain (x ge 0), (f^{-1}(x) = sqrt{x + 5})

23a. (f(g(x)) = (sqrt[3]{x + 5})^3 - 5 = x)
b. (g(f(x)) = sqrt[3]{x^3 - 5 + 5} = x)
c. This means that they are inverse functions (of each other)

## Stewart Calculus 7e Solutions Chapter 6 Inverse Functions Exercise 6.8

Stewart Calculus 7e Solutions Chapter 6 Inverse Functions Exercise 6.8

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## Contents

with Jacobian determinant:

As an important result, the inverse function theorem has been given numerous proofs. The proof most commonly seen in textbooks relies on the contraction mapping principle, also known as the Banach fixed-point theorem (which can also be used as the key step in the proof of existence and uniqueness of solutions to ordinary differential equations). [2] [3]

Since the fixed point theorem applies in infinite-dimensional (Banach space) settings, this proof generalizes immediately to the infinite-dimensional version of the inverse function theorem [4] (see Generalizations below).

An alternate proof in finite dimensions hinges on the extreme value theorem for functions on a compact set. [5]

Yet another proof uses Newton's method, which has the advantage of providing an effective version of the theorem: bounds on the derivative of the function imply an estimate of the size of the neighborhood on which the function is invertible. [6]

### Manifolds Edit

The inverse function theorem can be rephrased in terms of differentiable maps between differentiable manifolds. In this context the theorem states that for a differentiable map F : M → N (of class C 1 > ), if the differential of F ,

is a diffeomorphism. Note that this implies that the connected components of M and N containing p and F(p) have the same dimension, as is already directly implied from the assumption that dFp is an isomorphism. If the derivative of F is an isomorphism at all points p in M then the map F is a local diffeomorphism.

### Banach manifolds Edit

These two directions of generalization can be combined in the inverse function theorem for Banach manifolds. [10]

### Constant rank theorem Edit

The inverse function theorem (and the implicit function theorem) can be seen as a special case of the constant rank theorem, which states that a smooth map with constant rank near a point can be put in a particular normal form near that point. [11] Specifically, if F : M → N has constant rank near a point p ∈ M , then there are open neighborhoods U of p and V of F ( p ) and there are diffeomorphisms u : T p M → U M o U!> and v : T F ( p ) N → V N o V!> such that F ( U ) ⊆ V and such that the derivative d F p : T p M → T F ( p ) N :T_

M o T_N!> is equal to v − 1 ∘ F ∘ u circ Fcirc u!> . That is, F "looks like" its derivative near p . Semicontinuity of the rank function implies that there is an open dense subset of the domain of F on which the derivative has constant rank. Thus the constant rank theorem applies to a generic point of the domain.

When the derivative of F is injective (resp. surjective) at a point p , it is also injective (resp. surjective) in a neighborhood of p , and hence the rank of F is constant on that neighborhood, and the constant rank theorem applies.

### Holomorphic functions Edit

If a holomorphic function F is defined from an open set U of C n ^!> into C n ^!> , and the Jacobian matrix of complex derivatives is invertible at a point p , then F is an invertible function near p . This follows immediately from the real multivariable version of the theorem. One can also show that the inverse function is again holomorphic. [12]

### Polynomial functions Edit

If it would be true, the Jacobian conjecture would be a variant of the inverse function theorem for polynomials. It states that if a vector-valued polynomial function has a Jacobian determinant that is an invertible polynomial (that is a nonzero constant), then it has an inverse that is also a polynomial function. It is unknown whether this is true or false, even in the case of two variables. This is a major open problem in the theory of polynomials.

## Inverse-gamma distribution

In probability theory and statistics, the inverse gamma distribution is a two-parameter family of continuous probability distributions on the positive real line, which is the distribution of the reciprocal of a variable distributed according to the gamma distribution. Perhaps the chief use of the inverse gamma distribution is in Bayesian statistics, where the distribution arises as the marginal posterior distribution for the unknown variance of a normal distribution, if an uninformative prior is used, and as an analytically tractable conjugate prior, if an informative prior is required.

α + ln  ( β Γ ( α ) ) − ( 1 + α ) ψ ( α )

However, it is common among Bayesians to consider an alternative parametrization of the normal distribution in terms of the precision, defined as the reciprocal of the variance, which allows the gamma distribution to be used directly as a conjugate prior. Other Bayesians prefer to parametrize the inverse gamma distribution differently, as a scaled inverse chi-squared distribution.

## 1.6.6E: Inverse Functions

What are inverse functions?

Take a function f: draw its graph in the usual way interchange x and y axes, and you have the graph of the inverse function f − 1 .
y = f ( x ) means x = f − 1 ( y ) .

This can be accomplished with a drawing on a piece of paper by turning the paper over, orienting so that the old first quadrant appears in the upper right corner, and looking through the paper at the old graph.

Do not confuse the inverse with the reciprocal function they are completely different concepts. Careless people may use notation that is the same for both. This is a bad thing to do since it promotes confusion.

Notice that if you substitute x = f − 1 ( y ) into y = f ( x ) you get x = f − 1 ( f ( x ) ) . This last equation can be used as an alternate definition of the inverse function to f .

There is a problem with defining inverse functions. A function can have only one ordered pair for each argument, while the same value can occur many times. This means that interchanging arguments and values which is what we do in creating an inverse, will create a non-function, unless the original function takes on each value exactly once.

When a function takes a value more than once, we have to do extra work to define an inverse function for it. Namely, we have to select one of its values to be the new argument and throw away the others. This can be done in many different ways when f is not single valued, so that there is always some arbitrariness in the definition of f − 1 when f is not single valued.

The clearest example for this is the function x 2 . It takes on each positive value twice. Both 4 and -4 have the same square. The standard thing to do for this function is to define its inverse, x 1 / 2 , to be the positive square root, ignoring the negative one. (The negative square root is then denoted by − x 1 / 2 .) This definition has two virtues: one is that positive numbers are more positive than negative ones. The other is, that with this definition (and not had we chosen the negative root as inverse) the square root of a product is the product of the square roots of its factors.

In general, you can make the choice of what you want to call the inverse of f by looking at the graph of f , selecting a domain on which f is single-valued, and making that the range of f − 1 .

Exercises 1.8 For what values can you define the inverse of the function cos  ( sin  x ) . (Hint: set f = cos  ( sin  x ) look at its inverse and figure out the answer.) Solution

## 1.4 Inverse Functions

An inverse function reverses the operation done by a particular function. In other words, whatever a function does, the inverse function undoes it. In this section, we define an inverse function formally and state the necessary conditions for an inverse function to exist. We examine how to find an inverse function and study the relationship between the graph of a function and the graph of its inverse. Then we apply these ideas to define and discuss properties of the inverse trigonometric functions.

### Definition

Note that f −1 f −1 is read as “f inverse.” Here, the −1 −1 is not used as an exponent and f −1 ( x ) ≠ 1 / f ( x ) . f −1 ( x ) ≠ 1 / f ( x ) . Figure 1.37 shows the relationship between the domain and range of f and the domain and range of f −1 . f −1 .

### Example 1.28

#### Determining Whether a Function Is One-to-One

For each of the following functions, use the horizontal line test to determine whether it is one-to-one.

### Example 1.29

#### Solution

Follow the steps outlined in the strategy.

Step 2. Rewrite as y = 1 3 x + 4 3 y = 1 3 x + 4 3 and let y = f −1 ( x ) . y = f −1 ( x ) .

Therefore, f −1 ( x ) = 1 3 x + 4 3 . f −1 ( x ) = 1 3 x + 4 3 .

You can verify that f −1 ( f ( x ) ) = x f −1 ( f ( x ) ) = x by writing

### Example 1.31

#### Restricting the Domain

Consider the function f ( x ) = ( x + 1 ) 2 . f ( x ) = ( x + 1 ) 2 .

### Inverse Trigonometric Functions

The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function (Figure 1.34). The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval [ − π 2 , π 2 ] . [ − π 2 , π 2 ] . By doing so, we define the inverse sine function on the domain [ −1 , 1 ] [ −1 , 1 ] such that for any x x in the interval [ −1 , 1 ] , [ −1 , 1 ] , the inverse sine function tells us which angle θ θ in the interval [ − π 2 , π 2 ] [ − π 2 , π 2 ] satisfies sin θ = x . sin θ = x . Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric functions , which are functions that tell us which angle in a certain interval has a specified trigonometric value.

### Definition

To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line y = x y = x (Figure 1.41).

### Media

Go to the following site for more comparisons of functions and their inverses.

We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions sin ( sin −1 ( 2 2 ) ) sin ( sin −1 ( 2 2 ) ) and sin −1 ( sin ( π ) ) . sin −1 ( sin ( π ) ) . For the first one, we simplify as follows:

For the second one, we have

Similarly, for the cosine function,

Similar properties hold for the other trigonometric functions and their inverses.

### Example 1.32

#### Evaluating Expressions Involving Inverse Trigonometric Functions

Evaluate each of the following expressions.

### Student Project

#### The Maximum Value of a Function

In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain, even if we don’t know its exact value at a given instant. For instance, if we have a function describing the strength of a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have a function that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails. Safe design often depends on knowing maximum values.

This project describes a simple example of a function with a maximum value that depends on two equation coefficients. We will see that maximum values can depend on several factors other than the independent variable x.

### Section 1.4 Exercises

For the following exercises, use the horizontal line test to determine whether each of the given graphs is one-to-one.

## No Inverse?

Let us see graphically what is going on here:

To be able to have an inverse we need unique values.

Just think . if there are two or more x-values for one y-value, how do we know which one to choose when going back?

 General Function No Inverse

Imagine we came from x1 to a particular y value, where do we go back to? x1 or x2?

In that case we can't have an inverse.

But if we can have exactly one x for every y we can have an inverse.

It is called a "one-to-one correspondence" or Bijective, like this

 Bijective Function Has an Inverse

A function has to be "Bijective" to have an inverse.

So a bijective function follows stricter rules than a general function, which allows us to have an inverse.

## Examples

For more practice with the concepts covered in this tutorial, visit the Inverse Functions Problems page at the link below. The solutions to the problems will be posted after these chapters are covered in your calculus course.

To test your knowledge of inverse function problems, try taking the general inverse functions test on the iLrn website or the advanced inverse functions test at the link below.

Please forward any questions, comments, or problems you have experienced with this website to Alex Karassev.

I know that the mod function is defined by floor function (a - n floor(na). But I'm still unable to create an inverse function with one inverse function for all values. I made a piecewise function that works for those values, but I feel like there should be a better method. Does anybody know an inverse rule for either floor or modulus?

The 'mod' function? There is no such thing.

Just do it as for any other function:

y=4x+3, hence x= (y-3)/4, so what is division by 4 in mod 7 arithmetic?

The first thing you should do is calculate the actual values of f:
f(0)= 4(0)+ 3= 3. f(1)= 4(1)+ 3= 7= 0 mod 7. f(2)= 4(2)+ 3= 11= 4 mod 7.
f(3)= 4(3)+ 3= 15= 1 mod y. f(4)= 4(4)+ 3= 19= 5 mod 7. f(5)= 4(5)+ 3= 23= 2 mod 7. f(6)= 4(6)+ 3= 27= 6 mod 7

So you want f -1 (0)= 1, f -1 (1)= 3, f -1 (2)= 5, f -1 (3)= 0, f -1 (4)= 2, f -1 (5)= 4, f -1 (6)= 6.

Yes, that's exactly what your function gives. Good work! That is, by the way, a single "function" and I have no problem with it. However, you might remember that -6 is 1 in mod 7. Since the first formula was given "mod 7" why not do the same with this:
f -1 (x)= 2x+ 1 (mod 7) or f -1 (x)= 2x- 6 (mod 7)
They are exactly the same.

The 'mod' function? There is no such thing.

My apologies. mod "operator" (or modulo operator) I was just asking if there was an inverse operator that I could use so I could switch the x and y variables and solve for x instead of switching the values around and making an entirely new function.

## 1.6.6E: Inverse Functions

In the last example from the previous section we looked at the two functions (fleft( x ight) = 3x - 2) and (gleft( x ight) = frac <3>+ frac<2><3>) and saw that

[left( ight)left( x ight) = left( ight)left( x ight) = x]

and as noted in that section this means that there is a nice relationship between these two functions. Let’s see just what that relationship is. Consider the following evaluations.

In the first case we plugged (x = - 1) into (fleft( x ight)) and got a value of (-5). We then turned around and plugged (x = - 5) into (gleft( x ight)) and got a value of -1, the number that we started off with.

In the second case we did something similar. Here we plugged (x = 2) into (gleft( x ight)) and got a value of(frac<4><3>), we turned around and plugged this into (fleft( x ight)) and got a value of 2, which is again the number that we started with.

Note that we really are doing some function composition here. The first case is really,

[left( ight)left( < - 1> ight) = gleft[ ight)> ight] = gleft[ < - 5> ight] = - 1]

and the second case is really,

Note as well that these both agree with the formula for the compositions that we found in the previous section. We get back out of the function evaluation the number that we originally plugged into the composition.

So, just what is going on here? In some way we can think of these two functions as undoing what the other did to a number. In the first case we plugged (x = - 1) into (fleft( x ight)) and then plugged the result from this function evaluation back into (gleft( x ight)) and in some way (gleft( x ight)) undid what (fleft( x ight)) had done to (x = - 1) and gave us back the original (x) that we started with.

Function pairs that exhibit this behavior are called inverse functions. Before formally defining inverse functions and the notation that we’re going to use for them we need to get a definition out of the way.

A function is called one-to-one if no two values of (x) produce the same (y). Mathematically this is the same as saying,

So, a function is one-to-one if whenever we plug different values into the function we get different function values.

Sometimes it is easier to understand this definition if we see a function that isn’t one-to-one. Let’s take a look at a function that isn’t one-to-one. The function (fleft( x ight) = ) is not one-to-one because both (fleft( < - 2> ight) = 4) and (fleft( 2 ight) = 4). In other words, there are two different values of (x) that produce the same value of (y). Note that we can turn (fleft( x ight) = ) into a one-to-one function if we restrict ourselves to (0 le x < infty ). This can sometimes be done with functions.

Showing that a function is one-to-one is often tedious and/or difficult. For the most part we are going to assume that the functions that we’re going to be dealing with in this course are either one-to-one or we have restricted the domain of the function to get it to be a one-to-one function.

Now, let’s formally define just what inverse functions are. Given two one-to-one functions (fleft( x ight)) and (gleft( x ight)) if

then we say that (fleft( x ight)) and (gleft( x ight)) are inverses of each other. More specifically we will say that (gleft( x ight)) is the inverse of (fleft( x ight)) and denote it by

[gleft( x ight) = >left( x ight)]

Likewise, we could also say that (fleft( x ight)) is the inverse of (gleft( x ight)) and denote it by

[fleft( x ight) = >left( x ight)]

The notation that we use really depends upon the problem. In most cases either is acceptable.

For the two functions that we started off this section with we could write either of the following two sets of notation.

[eginfleft( x ight) & = 3x - 2hspace<0.25in>hspace <0.25in>& >left( x ight) & = frac <3>+ frac<2><3> & & & gleft( x ight) & = frac <3>+ frac<2><3>hspace<0.25in>hspace<0.25in>& >left( x ight) & = 3x - 2end]

Now, be careful with the notation for inverses. The “-1” is NOT an exponent despite the fact that it sure does look like one! When dealing with inverse functions we’ve got to remember that

This is one of the more common mistakes that students make when first studying inverse functions.

The process for finding the inverse of a function is a fairly simple one although there are a couple of steps that can on occasion be somewhat messy. Here is the process

#### Finding the Inverse of a Function

Given the function (fleft( x ight)) we want to find the inverse function, (>left( x ight)).

1. First, replace (fleft( x ight)) with (y). This is done to make the rest of the process easier.
2. Replace every (x) with a (y) and replace every (y) with an (x).
3. Solve the equation from Step 2 for (y). This is the step where mistakes are most often made so be careful with this step.
4. Replace (y) with (>left( x ight)). In other words, we’ve managed to find the inverse at this point!
5. Verify your work by checking that [left( >> ight)left( x ight) = x] and [left( <> circ f> ight)left( x ight) = x] are both true. This work can sometimes be messy making it easy to make mistakes so again be careful.

That’s the process. Most of the steps are not all that bad but as mentioned in the process there are a couple of steps that we really need to be careful with since it is easy to make mistakes in those steps.

In the verification step we technically really do need to check that both (left( >> ight)left( x ight) = x) and (left( <> circ f> ight)left( x ight) = x) are true. For all the functions that we are going to be looking at in this course if one is true then the other will also be true. However, there are functions (they are beyond the scope of this course however) for which it is possible for only one of these to be true. This is brought up because in all the problems here we will be just checking one of them. We just need to always remember that technically we should check both.

Now, we already know what the inverse to this function is as we’ve already done some work with it. However, it would be nice to actually start with this since we know what we should get. This will work as a nice verification of the process.

So, let’s get started. We’ll first replace (fleft( x ight)) with (y).

Next, replace all (x)’s with (y)and all (y)’s with (x).

Finally replace (y) with (>left( x ight)).

Now, we need to verify the results. We already took care of this in the previous section, however, we really should follow the process so we’ll do that here. It doesn’t matter which of the two that we check we just need to check one of them. This time we’ll check that (left( >> ight)left( x ight) = x) is true.

[eginleft( >> ight)left( x ight) & = fleft[ <>left( x ight)> ight] & = fleft[ <3>+ frac<2><3>> ight] & = 3left( <3>+ frac<2><3>> ight) - 2 & = x + 2 - 2 & = xend]

The fact that we’re using (gleft( x ight)) instead of (fleft( x ight)) doesn’t change how the process works. Here are the first few steps.

Now, to solve for (y)we will need to first square both sides and then proceed as normal.

Finally let’s verify and this time we’ll use the other one just so we can say that we’ve gotten both down somewhere in an example.

So, we did the work correctly and we do indeed have the inverse.

The next example can be a little messy so be careful with the work here.

The first couple of steps are pretty much the same as the previous examples so here they are,

Now, be careful with the solution step. With this kind of problem it is very easy to make a mistake here.

[eginxleft( <2y - 5> ight) & = y + 4 2xy - 5x & = y + 4 2xy - y & = 4 + 5x left( <2x - 1> ight)y & = 4 + 5x y & = frac<<4 + 5x>><<2x - 1>>end]

So, if we’ve done all of our work correctly the inverse should be,

Finally, we’ll need to do the verification. This is also a fairly messy process and it doesn’t really matter which one we work with.

Okay, this is a mess. Let’s simplify things up a little bit by multiplying the numerator and denominator by (2x - 1).

Wow. That was a lot of work, but it all worked out in the end. We did all of our work correctly and we do in fact have the inverse.

There is one final topic that we need to address quickly before we leave this section. There is an interesting relationship between the graph of a function and the graph of its inverse.

Here is the graph of the function and inverse from the first two examples.

In both cases we can see that the graph of the inverse is a reflection of the actual function about the line (y = x). This will always be the case with the graphs of a function and its inverse.

1. Replace f(x) by y .
2. Switch the roles of colorx and colory . In other words, interchange x and y in the equation.
3. Solve for y in terms of x .
4. Replace y by >left( x ight) to get the inverse function
5. Sometimes, it is helpful to use the domain and range of the original function to identify the correct inverse function out of two possibilities. This happens when you get a “plus or minus” case in the end.

#### Examples of How to Find the Inverse Function of a Quadratic Function

Example 1: Find the inverse function of fleft( x ight) = + 2 , if it exists. State its domain and range.

The first thing I realize is that this quadratic function doesn’t have a restriction on its domain. I am sure that when I graph this, I can draw a horizontal line that will intersect it more than once. Therefore the inverse is not a function. I will not even bother applying the key steps above to find its inverse.

The diagram shows that it fails the Horizontal Line Test, thus the inverse is not a function. I will stop here.

Example 2: Find the inverse function of fleft( x ight) = + 2,,,x ge 0 , if it exists. State its domain and range.

This same quadratic function, as seen in Example 1, has a restriction on its domain which is x ge 0 . After plotting the function in xy- axis, I can see that the graph is a parabola cut in half for all x values equal to or greater than zero. This should pass the Horizontal Line Test which tells me that I can actually find its inverse function by following the suggested steps.

In its graph below, I clearly defined the domain and range because I will need this information to help me identify the correct inverse function in the end.

Remember that the domain and range of the inverse function come from the range, and domain of the original function, respectively. It’s called the swapping of domain and range.

Even without solving for the inverse function just yet, I can easily identify its domain and range using the information from the graph of the original function: domain is x ≥ 2 and range is y ≥ 0.

Do you see how I interchange the domain and range of the original function to get the domain and range of its inverse?

Now, let’s go ahead and algebraically solve for its inverse.

Graphing the original function with its inverse in the same coordinate axis…

Example 3: Find the inverse function of fleft( x ight) = - - 1,,,x le 0 , if it exists. State its domain and range.

This problem is very similar to Example 2. The range starts at colory=-1 , and it can go down as low as possible.

Now, these are the steps on how to solve for the inverse.

Applying square root operation results in getting two equations because of the positive and negative cases. To pick the correct inverse function out of the two, I suggest that you find the domain and range of each possible answer. Now, the correct inverse function should have a domain coming from the range of the original function and a range coming from the domain of the same function.

The following are the graphs of the original function and its inverse on the same coordinate axis.

Example 4: Find the inverse of the function below, if it exists. State its domain and range.

I would graph this function first and clearly identify the domain and range. Notice that the restriction in the domain cuts the parabola into two equal halves. I will deal with the left half of this parabola. Clearly, this has an inverse function because it passes the Horizontal Line Test.

Proceed with the steps in solving for the inverse function. In fact, there are two ways how to work this out.

where a , b , and c can contain variables.

This is expected since we are solving for a function, not exact values.

The key step here is to pick the appropriate inverse function in the end because we will have the plus (+) and minus (−) cases. We can do that by finding the domain and range of each and compare that to the domain and range of the original function. Remember that we swap the domain and range of the original function to get the domain and range of its inverse.

• The method of completing the squares allows us to isolate the variable in a quadratic trinomial. As you will see in the steps, the quadratic trinomial is converted into linear binomial raised to the power of 2 . Obviously, we can apply the square root operation to get rid of the exponent 2 , therefore leaving us with an easy equation to solve.

If you observe, the graphs of the function and its inverse are actually symmetrical along the line y = x (see dashed line). They are like mirror images of each other.

I hope that you gain some level of appreciation on how to find the inverse of a quadratic function. Although it can be a bit tedious, as you can see, overall it is not that bad. I recommend that you check out the related lessons on how to find inverses of other kinds of functions.


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# D’Alembert’s principle and conservation of energy principles

## Homework Statement

A load of 10 kg is pulled up a lubricated slideway, set at an angle of 30° to the horizontal, such that the load is accelerated from rest to a velocity of 1 m/s whilst travelling up the plane through a distance of 1 m. The frictional resistance to this motion is 10 N and g = 9.81 m/s2.

Using a) d’Alembert’s principle, then b) conservation of energy principles, find:

i) the work done in moving the load as described
ii) the maximum input power provided by the pulling device.

## Homework Equations

honestly no idea.

## The Attempt at a Solution

i have done my best to research d'Alembert's princeiples but i cant get a single hint anywhere.

Related Introductory Physics Homework Help News on Phys.org
PhanthomJay
Homework Helper
Gold Member

## Homework Statement

A load of 10 kg is pulled up a lubricated slideway, set at an angle of 30° to the horizontal, such that the load is accelerated from rest to a velocity of 1 m/s whilst travelling up the plane through a distance of 1 m. The frictional resistance to this motion is 10 N and g = 9.81 m/s2.

Using a) d’Alembert’s principle, then b) conservation of energy principles, find:

i) the work done in moving the load as described
ii) the maximum input power provided by the pulling device.

## Homework Equations

honestly no idea.

## The Attempt at a Solution

i have done my best to research d'Alembert's princeiples but i cant get a single hint anywhere.
Well things can get rather complex using the full description of his principle, as you may have found out by googling it on the Wiki site.
D'Alembert's principle is good to use in certain cases, this not being one of them.
Essentially, however, his principle takes Newton's 2nd law, $F_{net} = ma$, and rearranges it to $F_{net} - ma = 0$ . Here, the system can be said to be in a state of dynamic equilibrium, where the '-ma' term is called the ficticious inertial force acting opposite to the real net force. In equilibrium, the net work done by all forces, including the inertial force, is 0. You'll have to calculate the acceleration using the kinematic equations. In calculating the work done by the pulling force, you'll have to subtract out the work done by gravity and friction. I don't like it. Try starting first using conservation of energy to see what you get.


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## An observation about the new Chrome OS Smart-Lock and Instant Tethering features.

I own a Samsung Galaxy S9 smart-phone, and an Asus Flip C213 Chromebook. And, two relatively new features which Google rolled out are:

• Smart-Lock: The ability to unlock the Chromebook, using the presence of the phone, and
• Instant Tethering: It has always been possible to activate the Mobile Hot-Spot feature of the phone, assuming that a user has a plan that includes tethering, and then to connect the Chromebook (or other device) to it, in the form of a mobile, Wi-Fi Access Point. But, with Instant Tethering, the availability of the phone as a tether is supposed to be more quickly visible from the Chromebook, and theoretically, accessible with a single click.

What some people have reported is, that this feature does not always work 100%, even though the procedure was followed, which my readers can find in many other places on the Web, to set up the feature. I recently experienced as well that, on my first try, these two features were not working at all, when the Chrome OS version on my Chromebook was ’80.x’. Yet, even during the interval of my trials, an update to the Chrome OS version had presented itself, to version ’81.y’. And since the update, the features seem to work 50% of the time.

There was an additional step which can be taken, but should not be 100% necessary in this case, and which I took, which is outlined in this article:

https://www.howtogeek.com/fyi/chrome-os-instant-tethering-comes-to-more-android-phones-heres-how-to-do-it/

I will explain below, Why I changed the flag under:

chrome://flags/#instant-tethering

From ‘Default’, to ‘Enabled’. A reboot was required…

One reason these features may still not work 100% for me, could be the possibility of the phone going into ‘Deep Sleep’…

## About a minor (Home-Screen) nuisance I’ve experienced on Android deviceS.

I have owned several Android-based devices, and some of those were purchased from Samsung, those being:

• A Galaxy Tab S, First Generation,
• (An earlier Smart-Phone),
• A Galaxy S6 Smart-Phone,
• A Galaxy S9 Smart-Phone.

A feature which all these devices have, is the Touchwiz Home-Screen (program). This is the default of what the devices display, when not displaying a specific app, when not displaying the app drawer, and when not displaying ‘Bixby’ (most recently). An unfortunate behaviour of the devices is, that Touchwiz will sometimes crash. In my experience, when it does, this is no big deal, because it restarts automatically, and after a few minutes, even my Notification-Bar Entries will reappear. If certain apps fail to make their notifications reappear by themselves, then launching those apps from the application groups will make their notifications reappear.

I tend to rate each Android device, according to how rarely its Home-Screen will crash in this way. According to that, my Google Pixel C Tablet fared better because its home-screen has never crashed on me. My S9 Phone fared almost as well, in that Touchwiz seldom crashed. But now I think I’ve identified a situation which will frequently cause Touchwiz to crash on the S9 Phone.

Firstly, as I’m writing this, the firmware on that phone is at its latest version, that being the October 1 patch, of 2019, of Android 9.

I discovered that I can trigger this situation, as I was experimenting with the Super-Slow-Mo camera recording mode, in which the camera can record up to 0.4 seconds of video at 960FPS, at a resolution of 1280×720. When the camera does this, it generates a 20MB video, after that has been compressed via a standard H.264 CODEC into an .MP4 container-file. I have the default set, to record all camera footage to the external Micro SD Card. Having recorded the super-slow-mo video once, triggered this behaviour.

There is a simple way to interpret what has caused this, that does not seem to lay any blame on Samsung: When the camera is recording video that fast, it’s generating data faster than the external SD Card can store it. Therefore, the data takes up RAM, until some later point in time, when the O/S has transferred the data to the SD Card, by writing it out. This moment was reached several seconds later.

Here’s where the news gets a bit worse. I can download This 3rd-party app, that’s designed to test what speed of external SD Card I have. The reason I need to do this is the fact that I never seem to remember exactly what type of SD Card I purchased, for use with one specific device.

According to this app, my external SD Card can be written to sequentially at ~12MBytes/Sec. That makes it a Class 10 card. Yet, 20MB of data are to be stored in 0.4 seconds. In fact, simply running the benchmarking app caused a second Touchwiz crash, which was just as inconsequential as the first, that I was trying to investigate. What this seems to suggest is, that virtually no SD Card that I can buy, can really be fast enough to be written to at the speed with which the camera app can generate its data. The camera app will need to cache its footage in RAM, before that footage has been written to the SD Card.

Further, the footage is certainly being stored in RAM in an uncompressed form of data (384 raw frames), while what’s to be written to the SD Card is finally compressed. (:1)

And yet, either of these two apps will cause the Touchwiz crash. Hmm… I think that for the moment, I’ll just hold my horses, and record a maximum of 0.2 seconds of Super-Slow-Mo. Thankfully, this is a parameter that I can choose, with the little icon in the upper-right-hand corner of the view, before shooting.

(Updated 11/17/2019, 12h10 … )

## Comparing two Bose headphones, both of which use active technology.

In this posting I’m going to do something I rarely do, which is, something like a product review. I have purchased the following two headphones within the past few months:

The first set of headphones has an analog 3.5mm stereo input cable, which has a dual-purpose Mike / Headphone Jack, and comes either compatible with Samsung, or with Apple phones, while the second uses Bluetooth to connect to either brand of phone. I should add that the phone I use with either set of headphones is a Samsung Galaxy S9, which supports Bluetooth 5.

The first set of headphones requires a single, AAA alkaline battery to work properly. And this not only fuels its active noise cancelling, but also an equalizer chip that has become standard with many similar middle-price-range headphones. The second has a built-in rechargeable Lithium-Ion Battery, which is rumoured to be good for 10-15 hours of play-time, which I have not yet tested. Like the first, the second has an equalizer chip, but no active noise cancellation.

I think that right off the bat I should point out, that I don’t approve of this use of an equalizer chip, effectively, to compensate for the sound oddities of the internal voice-coils. I think that more properly, the voice-coils should be designed to deliver the best frequency response possible, by themselves. But the reality in the year 2019 is, that many headphones come with an internal equalizer chip instead.

What I’ve found is that the first set of headphones, while having excellent noise cancellation, has two main drawbacks:

• The jack into which the analog cable fits, is poorly designed, and can cause bad connections,
• The single, AAA battery can only deliver a voltage of 1.5V, and if the actual voltage is any lower, either because a Ni-MH battery was used in place of an alkaline cell, or, because the battery is just plain low, the low-voltage equalizer chip will no longer work fully, resulting in sound that reveals the deficiencies in the voice-coil.

The second set of headphones overcomes both these limitations, and I fully expect that its equalizer chips will have uniform behaviour, that my ears will be able to adjust to in the long term, even when I use them for hours or days. Also, I’d tend to say that the way the equalizer arrangement worked in the first set of headphones, was not complete in fulfilling its job, even when the battery was fully charged. Therefore, If I only had the money to buy one of the headphones, I’d choose the second set, which I just received today.

But, having said that, I should also add that I have two 12,000BTU air conditioners running in the Summer months, which really require the noise-cancellation of the first set of headphones, that the second set does not provide.

Also, I have an observation of why the EQ chip in the second set of headphones may work better than the similarly purposed chip in the first set…

(Updated 9/28/2019, 19h05 … )

## Overheated Circuitry

One of the things which I do frequently, is ‘walk around’, or, ‘use public transit’, with my disposable earphones plugged in to my Samsung Galaxy S9 Smart-Phone, and listening to music. These earphones are clearly not the ones, which had the AKG seal of approval, and which shipped with the phone. But this week-end marks the second heat-wave this Summer, when outside daytime temperatures exceeded 31⁰C, with direct sunlight and not a cloud in the sky. And under those conditions, the battery of my phone starts to hit a temperature of 42⁰. One of the facts which I know is, that Lithium-Ion batteries like the one in my phone do not tolerate temperatures exceeding 41⁰C.

A peculiar behaviour which has set in for the second time, during this second heat-wave of the season, is that the music I was listening to would either back-space to the beginning of the song, or skip ahead one song, or just stop. So, a catastrophic sort of explanation I could think of would be, that the entire phone, with its battery, is finally just having a meltdown. But, a second possibility exists, that merely the chip in the earphone-cord could be malfunctioning. After all, the little pod in the earphone-cord has one button and a mike, and it’s actually cheaper to mass-produce the chip that makes it work, than it would be to mass-produce other sorts of discrete components. One cheap chip could just be malfunctioning in the extreme heat, and not the entire, complex circuitry of the phone. (:1)

The earphones cost me about \$15, while the phone is much more expensive than that.

But even if it was true, that only the little remote-control in the earphone-cord was malfunctioning, this can lead to impractical situations, because just random patterns, of unreal button-press-combinations, could also send the software of my phone into a confused state, and even so, if the circuitry in the smart-phone never malfunctioned. This behaviour could get misinterpreted by the security apps of the phone, let’s say, as though somebody had ripped the earphone-cord off my head, and thrown all my possessions around.

All that was really happening was that my music was no longer playing, as I was walking home normally, in the heat, with my overheated electronics. And when I got home, my actual phone never displayed any signs of having malfunctioned.

(Updated 8/17/2019, 17h50 … )


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# There are two spheres of same materials and radius . One is solid and the other is hollow . If they are heated to the same temperature ,the expansions of objects are

Grade:12th pass

## 1 Answers

Arun
25763 Points
2 years ago
Dear Lakshya

(i) As thermal expansion of isotropic solids is similar to true photographic enlargement, the expansion of a cavity is same as if it were a solid body of the same material i.e. ∆V = $\gamma$V ∆T. As here V, $\gamma$ and ∆T are same for both solid and hollow spheres, so the expansions of both will be equal.
(ii) If same amount of heat is given to the two spheres, then due to lesser mass, rise in temperature of hollow sphere will be more (as ∆T = Q/Mc) and hence the expansion will bemore as ∆V = $\gamma$V ∆T.

Regards
Arun(askIITians forum expert)

## ASK QUESTION

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# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Remark 9.10.5.4. In the situation of Example 9.10.5.3, the Kleisli category $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$ is equipped with a comparison functor $H: \operatorname{\mathcal{E}}\rightarrow \int ^{\operatorname{\mathcal{C}}} \mathscr {F}$, which carries each object of $\operatorname{\mathcal{E}}$ to itself and each morphism $u: X \rightarrow Y$ in $\operatorname{\mathcal{E}}$ to the composite morphism $X \xrightarrow {u} Y \xrightarrow { \epsilon (Y) } T(Y)$ (regarded as a morphism from $X$ to $Y$ in the category $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$). The following conditions are equivalent:

• The functor $H$ is an equivalence of categories.

• The functor $H$ is an isomorphism of categories.

• The unit map $\epsilon : \operatorname{id}_{\operatorname{\mathcal{E}}} \rightarrow T$ is an isomorphism of functors.


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# What is a term?

If expressions are combined by addition, then each expression is called a term. For example, the three terms of ${x}^{2} + 3 x - 5$ are ${x}^{2}$, $3 x$, and $- 5 x$.


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# Breaking the Equation ' Empirical Argument = Proof '

##### Age 7 to 18

Published 2011 Revised 2015

Trained as a primary teacher in Cyprus, Andreas Stylianides studied for a masters in maths education, as well as a masters in mathematics, in the United States. He followed these studies with a PhD in mathematics education, again in the United States. He has always wanted to combine his love of mathematics with his interest in the teaching and learning of mathematics, and feels that his research achieves this kind of integration. Andreas is currently a lecturer in mathematics education at the University of Cambridge.

Andreas' interest in proof developed in his third year of undergraduate studies when many of his peers struggled with the concept of proof whilst he was finding the challenges the course offered both fulfilling and exciting. He feels that, for engagement with proof to be meaningful, it has to be placed in the context of problem solving so that one experiences the emergence of ideas that can often lead to dead ends. Linked to this is his view that there is a gap between mathematics at school and university:

"In maths courses at the university the concept of proof is very central, but at school it is possible not even to encounter the concept. When students experience proof at the university, it seems alien and unfamiliar to them rather than being a natural extension of habits of mind they developed at school. There is a big gap in the teaching of mathematics between school and university, and students are not prepared well for the kind of mathematical work required at maths courses at the university."

The following article focuses on Andreas' interest in and recent research on the teaching of proof in schools. After reading the article you might like to read more. The notes section of this article contains extracts from a discussion between Jenny Piggott and Andreas about some of the issues that are raised here.

BREAKING THE EQUATION 'EMPIRICAL ARGUMENT = PROOF'

Empirical argument vs. proof

Consider the generalisation: "the sum of any two odd numbers is an even number." What argument would your students offer for it? Would that be a proof?

An overwhelming body of research shows that students of all levels of schooling including high-attaining secondary students "prove" mathematical generalisations such as the above by using empirical arguments (e.g., Coe and Ruthven, 1994). By empirical arguments I mean those that purport to show the truth of a generalisation by validating the generalisation in a proper subset of all possible cases. These arguments are clearly invalid, because they cannot exclude the possibility of the existence of a counterexample to the generalisation. Here are two examples of empirical arguments for the above generalisation:

Empirical argument 1: naive empiricism
I tried many different pairs of odd numbers and their sum was always an even number: $\bf 7 + 9 = 16$, $\bf 15 + 21 = 36$, $\bf 25 + 27 = 52$, etc. So the sum of any two odd numbers is an even number.

Empirical argument 2: crucial experiment
I checked different kinds of pairs of odd numbers: some with small odd numbers (e.g., $1 + 9 = 10$), some with big odd numbers (e.g., $213 + 399 = 612$), some with the same odd numbers (e.g., $25 + 25 = 50$), and some with prime odd numbers (e.g., $17 + 31 = 48$). No pair gave me a counterexample - the sum was always an even number. So the sum of any two odd numbers is an even number.

Even though both arguments are invalid, the second argument can be considered more advanced than the first, because, by seeking possible counterexamples, it communicates a concern that the generalisation may not be true. Balacheff (1998) used the terms naive empiricism and crucial experiment to describe the special categories of empirical arguments represented by the first and second examples, respectively. The search of possible counterexamples in crucial experiment requires a strategic selection of cases in contrast to the random (or convenience) sampling of cases in naive empiricism.

The fact that a generalisation is true in some cases does not guarantee and, thus, does not prove that the generalisation is true for all possible cases. This is the main limitation of any kind of kind of empirical argument that many students find difficult to understand. What would be a proof for the generalisation then? Figure 1 shows three possible proofs for the generalisation on the set of whole numbers.

Figure 1: Three possible proofs (on the set of whole numbers) for 'odd + odd = even.'

Notice the correspondences among the three arguments: they all seem to be saying the 'same thing' using different representations. Notice also how each argument can be used to help someone understand why the generalisation is true, but also convince someone that the generalisation is true for all cases without requiring that person to make a leap of faith. A proof's potential to promote understanding and conviction is one of the main reasons why proof is so important for students' learning of mathematics.

A question that arises at this point is: How can we help students overcome the misconception that 'empirical argument = proof '? Unless students realise the limitations of empirical arguments as methods for validating mathematical generalisations, they are unlikely to appreciate the importance of proof in mathematics.

Next I describe and discuss a mathematics lesson in a high-attaining Year 10 class that aimed to help the students begin to realise the limitations of empirical arguments. The lesson was an adapted version of one developed by a research project in the context of a university course (Stylianides and Stylianides, accepted). I worked with the teacher of the Year 10 class and another Year 10 teacher in the same school to adapt the lesson to the particular context of their two classes, and then I observed the lesson being taught in each class. The lesson plan, in the form of annotated PowerPoint slides, is available at www.atm.org.uk/mt213.

The lesson

The lesson was approximately 60 minutes long and was taught over two consecutive 45-minute periods. The lesson involved three activities: the Squares Problem, the Circle and Spots Problem, and the 'Monstrous Counterexample'. As you read the following sections, I invite you to pay attention to how each activity was used by the teacher to facilitate students' progression along the 'learning path' in Figure 2: from using naive empiricism as a method for validating patterns, to using crucial experiment, to feeling a need to learn about more secure methods for validating patterns (i.e., to learn about proofs). Note that a pattern is a kind of generalisation. The teacher and student names are pseudonyms.

Figure 2: The three activities and corresponding 'learning path.'

Activity 1: The squares problem

Kathy, the teacher, introduced the Squares Problem (Figure 3). The hardest part of the problem was the third: it asked students to find the number of different 3-by-3 squares in a case that was difficult for them to check practically and also to explain whether and why they were sure their answer was correct.

Figure 3: The Squares Problem (adapted from Zack, 1997).

Kathy made sure the students understood what the problem was saying and then she asked them to work on the problem in their small groups. The small group closest to myself had six students: Bob, Calvin, Dan, Lazarus, Robert, and Sharon. These students counted squares to answer parts 1 and 2 of the problem, and then Bob asked his peers: "Have you actually got a formula?" Dan responded: "It's the number of ... it's $n$ minus $2$, and then squared." Sharon showed excitement and confirmed with Dan that the answer for part 1 would be $4$. Robert asked how many 3-by-3 squares there were in a 60-by-60 square (part 3) and Dan used his calculator and the formula he had described earlier to find the answer: $(60 - 2)^2 = 3364$.

At some point Kathy visited the small group and the students explained their work. Kathy then asked the students whether they were sure their answer was correct. Lazarus replied "yes" with confidence and Kathy posed a new question: "And have you thought about why you are sure?" There was no response from the students. Kathy asked the students to think about this and write their ideas on paper.

Dan drew figures for the 4-by-4 and 5-by-5 squares showing the 3-by-3 squares in each of them. He wrote down $58^2 = 3364$ as the answer to part 3 and also the formula $(n - 2)^2$. He concluded: "We realised that if you took 2 away from the number of cubes along the top and then square the answer you will get the number of 3$\times$3 boxes in the grid.?" The other students in the small group wrote similar conclusions in their papers.

So, what has happened thus far in the small group? The students identified the pattern that the number of different 3-by-3 squares in an n-by-n square was given by the formula $(n - 2)^2$. They verified the pattern for $n=4$ and $n=5$ and, based on these results, they concluded that the pattern would hold true for all values of $n$ including $n=60$. Thus the students validated the pattern on the basis of naive empiricism (cf. Figure 2).

The whole group discussion that followed illustrated further the use of naive empiricism in the class, as all groups answered the three parts of the problem using the formula $(n - 2)^2$. After some discussion on the meaning of the formula, Kathy asked the class whether and why they could be sure that their answers based on this formula were correct. Emily said: "We tried it [the formula] for a 6-by-6 square and it worked for that too. " Kathy invited further comments but the students did not have anything to add to what Emily had said.

Kathy then asked the students to write down individually their thoughts: "I want to know what your feelings are about whether this [the answer to part 3] is correct or not. You may think it is correct, you may not. If you are sure, I want to learn why you are sure." Someone asked "what if you're not sure?" and Kathy responded "then put not sure, but say why you are not sure - what makes you doubt it?"

In the focal small group the students wrote:

• Bob: "Because we have found a formula and tried it against smaller squares so we can make sure that the formula is right."
• Calvin: "I am sure that this solution works because it worked for every one we did."
• Dan: "I am sure that the answer is correct because it has been proved for a number of smaller grids."
• Lazarus: "I am sure that the answer is correct because it has been tested and proved correct. The pattern will continue to $60\times60$."
• Robert: "I am sure it's correct because we did a test on the $6 \times 6$ grid and it worked."
• Sharon: "We are sure that it is right because we have tried it for a $6 \times 6$ square as well. So we assume that it would work."
Notice that the six students were convinced of the truth of the pattern on the basis of naive empiricism: the pattern worked for the first few cases and so, according to the students, it would work also for $n=60$. This reasoning was reflected in the writings of the rest of the class, something that we had anticipated in our planning and Kathy confirmed as she was circulating around and looking at students' papers.

Following the students' individual reflections, Kathy proceeded with the next item in the lesson plan, which was to summarise students' validation method thus far:

"I get a feeling that most of you have said 'Well, I think we have sort of answered this question that $58^2$ is the right answer: we have found a pattern by checking smaller grid sizes and then we have used that pattern, assuming that it would continue all the way up to 60-by- 60.' That's the stage where we are right now: we've seen a pattern working, somebody said they tried the 6-by-6 and it worked for that too, and so we continued our pattern up to the $58^2$."

Bob asked Kathy whether the pattern was correct and Kathy said that the class would come back to this issue later, but first they would work on a couple of other activities. Indeed, according to our lesson plan the issue about the correctness of the pattern in the Squares Problem would remain tentatively unresolved. The class would revisit and resolve the issue after the students had been assisted to realise the limitations of empirical arguments (both naive empiricism and crucial experiment). Had the issue been resolved at this point of the lesson, this would probably require a lot of 'telling' by the teacher, which was inconsistent with our goals in the lesson. We wanted the students to realise the limitations of empirical arguments on their own, by experiencing and reflecting on situations where the empirical validation method was inadequate. For the readers' information, I note that the $(n - 2)^2$ pattern was actually correct.

Activity 2: The circle and spots problem

Kathy introduced the Circle and Spots Problem (Figure 4) and helped the students understand what the problem was saying. Specifically, she discussed with them the meaning of the terms 'maximum' and 'non-overlapping regions' Also, she clarified that the phrase 'around the circle' referred to the circle's circumference and that the spots on the circumference did not have to be equidistant. Then Kathy asked the students to work on the problem in their small groups.

Figure 4: The Circle and Spots Problem (adapted from Mason et al, 1982).

Notice that, similar to part 3 of the Squares Problem, the question in the Circle and Spots Problem (pale grey box in Figure 4) was asking the students to make a statement about a case that was difficult for them to check practically. In our planning we had anticipated that the students, like they did in the Squares Problem, would check simpler cases, identify a pattern, trust the pattern based on naive empiricism, and apply it to offer a definite answer for $n=15$ (where $n$ stands for the number of spots). The main difference between the two problems is that the emerging pattern in the Circle and Spots Problem fails for $n=6$. Our plan was for Kathy to use the anticipated surprise that the students would experience with the failing pattern to help them move from naive empiricism towards crucial experiment (cf. Figure 2).

After about 10 minutes of small group work, Kathy brought the whole class together and said: "Circulating around I think there are some people who think they know what the answer will be for 15 [spots]. Is there anyone who is willing to tell us their number of regions, what it will be for 15 spots?"

Mac said that his group thought the formula for the problem was $(n - 1)^2$ but soon thereafter he corrected himself to say the formula included powers of $2$. Kathy asked the class to say the maximum number of non-overlapping regions they found for different spots, and she constructed a table on the board with the following numbers: $4$, $8$, and $16$, for $n = 3$, $4$, and $5$, respectively. Then she pointed out that, as Mac had mentioned earlier, the values were all powers of $2$ and that, in each case, the power was one less than the number of spots: $2^2$ (for $n=3$), $2^3$ (for $n=4$), and $2^4$ (for $n=5$). Kathy asked: "So what will it be for 15 spots then?"

Several students offered to answer Kathy's question. Based on what I had observed during these students' prior work in their small groups, I presumed they would propose the application of the $2^{n-1}$ formula for $n=15$. However, Ken said loudly: "Can I just say that is wrong because on $6$ [spots] there are only $30$ [regions]." Kathy said: "We were about to say that the answer would be $2$ to the power of $14$. However, you are telling me that for $6$ spots it doesn't work out to be... With this pattern for $6$ six spots it would be $2$ to the power of $5$, that would be $32$, but did anyone manage to find this number of spots?" Some students said they found $31$ spots.

Kathy continued:

"When we were back to the Squares Problem, we said that because the pattern worked for some of the different grids, the 5-by-5, 6-by-6 squares, and so on, we were willing to trust it. But this time we have shown that it works for $3$, it works for $4$, it works for $5$, but actually, Ken, you are right: if we had $6$ spots on a circle and we joined them all up, the number of nonoverlapping regions that we get is not what we expect to get, it's not $32$. It's actually $31$."

As she talked, Kathy used a PowerPoint slide to illustrate the counterexample for $n=6$. She noted also that, if one drew the spots in a regular hexagon, the maximum number of regions would be $30$, which is again smaller than $32$. Then, following the lesson plan, Kathy asked the students to write down their thoughts about what the Circle and Spots problem had taught them.

The students in the focal small group wrote:

• Bob: "You can't always trust a formula until you have tested it many times over for lots of different examples."
• Calvin: "This test has taught us that if you see a pattern doesn't make it correct."
• Dan: "The circle and spots tells us that we can't always trust a formula that works on the first few."
• Lazarus: "This teaches us that just because something works for one thing, that doesn't mean it will work for everything."
• Robert: "You can't always trust a formula until you have tested many times over for lots of different numbers of spots."
• Sharon: "You can't always trust a formula. You shouldn't presume it is correct because it worked for the first few."
Notice that the students began to move away from naive empiricism. For example, Dan, Lazarus, and Sharon started feeling uneasy to trust a pattern based on checks of the first few cases. Also, Bob and Robert's comments approximated the crucial experiment method of validation, as they appeared to raise a concern about the number ('many') and quality ('different') of cases that had to be checked before a pattern could be trusted.

Thus an important issue for many students at this stage of the lesson was how many cases would be enough for them to check before trusting a pattern. We had anticipated this issue in our planning and we prepared a PowerPoint slide with a fictional student comment on it that Kathy used in the lesson to organise a discussion around the issue. The fictional student comment said:

"The Circle and Spots Problem teaches me that checking $5$ cases is not enough to trust a pattern in a problem. Next time I work with a pattern problem, I'll check more cases to be sure."

Kathy invited reactions from her students on this comment. Dan suggested trying spread cases such as for $n = 1$, $75$, and $100$. Robert observed that "you can't always trust the formula, you have to test it." Kathy asked Robert how many times one had to test a formula and Robert said "more than like 5 times." Kathy invited more comments and Larry said: "you should test it as many times as you have time to do." Kathy asked Larry: "So when you have tested it as many times as you have time to do, can you then trust it?" Larry revised: "No ... not a 100%!" Then Pauline said: "try it out with smaller numbers and bigger numbers." Kathy observed that Pauline's comment was similar to Dan's earlier comment.

Indeed, the two comments were similar to one another and illustrative of the crucial experiment method for validating patterns (cf. Figure 2). As I noted earlier, crucial experiment can be considered to be a more advanced method than naive empiricism, but is still an invalid, for a counterexample may exist in a case that was not checked. Some students in the class were thinking along similar lines, as illustrated by their responses to Kathy's question: "And then do we trust it if it worked for all of those [cases, big and small ones]?" Silvia said in a low voice: "No, because you might have missed one." Another student was heard to say: "You could spend your whole life and still miss one!" These students' fear that a pattern can fail in a case that was not checked was manifested in the next activity we planned for the students.

Activity 3: The 'Monstrous Counterexample' illustration

Kathy introduced the PowerPoint slide in Figure 5 that shows what I call the 'Monstrous Counterexample' Illustration. Kathy did not use this name during the lesson. The slide was presented in segments to give students a chance to process the information in it. For example, there was a discussion about how one would check whether a given number was a square number using a calculator. Also, the students confirmed the statement for particular values of $n$ using their calculators.

Figure 5: The 'Monstrous Counterexample' Illustration (adapted from Davis,1981).

Once the students checked many different cases and were comfortable with the meaning of the statement, Kathy presented the counterexample. The students were amazed: they had not anticipated that a pattern that held for so many cases (of the order of septillions) could ultimately fail!

Kathy then directed the students' attention to their previous discussion: "We said in the Circle and Spots Problem that, okay, it's not enough to just check a few cases, you need to try different ones. Well, this expression, what does this tell us?" Emily said: "If you kept trying, you might have to go that high until you find one [a counterexample]." Kathy said: "But I can imagine that it took the computer quite a long time to check all of those cases. And when do you stop checking?" Larry said: "when you've found one!" Several students laughed with what Larry had said. Kathy continued: "And when do you trust a pattern then?" Adam said: "When you cannot find one, until you are dead!"

Notice that the students began to develop distrust in empirical arguments of any kind, including crucial experiment. Yet, although the students began to realise the limitations of empirical arguments, they lacked knowledge of more secure methods for validating patterns. This caused a feeling of frustration among some of them as illustrated in Adam's comment: one would die checking cases before being in a position to trust a pattern! Thus we may say that the students reached the point when they felt a need to learn about more secure validation methods (cf. Figure 2).

The misconception that 'empirical arguments = proofs' is deeply rooted in many students' thinking. Nevertheless, the story I presented in this article sends the optimistic message that it is possible to help students realise the limitations of empirical arguments and create a need in them to learn about more secure methods for validating patterns. Needless to say, it is not enough for teachers to create this need in students and then leave them in a state of frustration. Teachers have the responsibility to also help their students appreciate the role of proof as a secure method for validating patterns in mathematics, to teach them what is involved in developing a proof, and give them opportunities to develop and criticise proofs against a list of criteria that students can understand. This is precisely what happened in subsequent lessons in Kathy's class: she introduced her students to the notion of proof in mathematics and she took them back to the Squares Problem and helped them develop a proof for the pattern they had identified earlier. The next part of the story will appear in a future article!

Andreas J. Stylianides
Article taken from Mathematics Teaching 213 / March 2009

References

Balacheff, N. (1988) Aspects of proof in pupils' practice of school mathematics, in D. Pimm (Ed.), Mathematics, Teachers and Children (pp. 216-235), London, Hodder and Stoughton.

Coe, R. and Ruthven, K. (1994) Proof practices and constructs of advanced mathematics students, British Educational Research Journal, 20, 41-53.

Davis, P. J. (1981) Are there coincidences in mathematics? American Mathematical Monthly, 88, 311-320.

Mason, J., Burton, L. and Stacey, K. (1982) Thinking Mathematically, London, Addison-Wesley.

Stylianides, G. J. and Stylianides, A. J. (accepted) Facilitating the transition from empirical arguments to proof, Journal for Research in Mathematics Education.

Zack, V. (1997) 'You have to prove us wrong': proof at the elementary school level. In E. Pehkonen (Ed.), Proceedings of the 21st Conference of the International Group for the Psychology of Mathematics Education (Vol. 4, pp. 291-298), Lahti, University of Helsinki.

Jenny Piggott followed up reading the article by talking to Andreas about some questions it raised about her own thinking and teaching. To see this discussion go to the teachers' notes section of this resource (see tab at top of article)


text:
## [POJ2411]Mondriaan\'s Dream

 成绩 开启时间 2014年09月19日 星期五 10:08 折扣 0.8 折扣时间 2014年09月26日 星期五 10:08 允许迟交 是 关闭时间 2014年09月26日 星期五 10:08 输入文件 examfive.in 输出文件 examfive.out

### 【题目描述】

Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series' (where he had to use his toilet paper to draw on, for all of his paper was filled with squares and rectangles), he dreamt of filling a large rectangle with small rectangles of width 2 and height 1 in varying ways.

Expert as he was in this material, he saw at a glance that he'll need a computer to calculate the number of ways to fill the large rectangle whose dimensions were integer values, as well. Help him, so that his dream won't turn into a nightmare!

### 【输入格式】

The input contains several test cases. Each test case is made up of two integer numbers: the height h and the width w of the large rectangle. Input is terminated by h=w=0. Otherwise, 1<=h,w<=11.

### 【输出格式】

For each test case, output the number of different ways the given rectangle can be filled with small rectangles of size 2 times 1. Assume the given large rectangle is oriented, i.e. count symmetrical tilings multiple times.

### 【样例输入】

1 2
1 3
1 4
2 2
2 3
2 4
2 11
4 11
0 0

### 【样例输出】

1
0
1
2
3
5
144
51205


text:
# Macro that allows for Linear Problem notation to create Genetic Algorithm fitness functions

As a side toy project for my Genetic Algorithm, I decided to try to create a macro that lets you use maximize: ... subject to: ... linear problem notation to generate fitness functions. Those fitness functions can then be used in a genetic algorithm to solve linear problems through brute force.

Yes, I'm aware genetic algorithms are the (very) wrong tool for the job here, but it's for kicks, not actual use.

As an example of its use, say you have the following problem:

A calculator company produces a scientific calculator and a graphing calculator. Long-term projections indicate an expected demand of at least 100 scientific and 80 graphing calculators each day. Because of limitations on production capacity, no more than 200 scientific and 170 graphing calculators can be made daily. To satisfy a shipping contract, a total of at least 200 calculators much be shipped each day. If each scientific calculator sold results in a $2 loss, but each graphing calculator produces a$5 profit, how many of each type should be made daily to maximize net profits?

To create a fitness function that can solve this problem, you can write:

(fit-func-for [:s :g]
:maximize (+ (* :s -2) (* :g 5)) ; Profit function
:subject-to (<= 100 :s 200) ; Problem constraints
(<= 80 :g 170)
(<= 200 (+ :s :g) 1e6))))

Where :s is representing scientific calculators, and :g graphing calculators.

This works by, basically, translating the above code into:

(->Linear-Problem (fn [vars] ; Profit
(+ (* (:s vars) -2) (* (:g vars) 5)))

(fn [vars] ; Constraints
(total-range-error
[100 (:s vars) 200]
[80 (:g vars) 170]
[200 (+ (:s vars) (:g vars))])))

Then giving the resulting Linear-Problem to evaluate-problem, and sticking the call in a quoted function that accepts the genes to test.

Usage:

(def f
(fit-func-for [:s :g]
:maximize (+ (* :s -2) (* :g 5))
:subject-to (<= 100 :s 200)
(<= 80 :g 170)
(<= 200 (+ :s :g) 1e6)))
=> #'genetic-algorithm.test-fit-funcs.linear-calculator/f

(f [-1 1000]) ; Outright illegal
=> -18273

(f [150 120]) ; Legal, but not optimal
=> 300

(f [100 170]) ; The optimal answer
=> 650

(f [99 171])
=> 607

(f [100 180])
=> 450

Features:

• Can accept either <= or >= in constraints.
• Can handle an arbitrary number of variables to find for.
• Not a whole lot else. It's pretty limited currently.

It works actually far better than I expected. It generates fitness functions that can solve the problem perfectly in a couple generations, or less.

This is the most complicated macro I've ever written though, and the code gets a little odd feeling in some places. I'd like advice mainly on how I can write this better/most idiomatically, but also on a few other points:

• I'm doing some traversing of nested structures to find/modify parts of code (replace-keys-with-call, find-number) using explicit recursion. This works fine, but I'm wondering if there's a better way.

• If the profit for an illegal result is high enough, it can overcome the constraint penalty, and cause bad results to score highly. To get around this, I'm finding the largest number in the profit function, then multiplying the constraint penalty by it to make sure the error will always "win". This is overly simple though, and causes a couple significant problems:

• This forces use of only + and *, since - and / aren't commutative, so large number literals being present doesn't necessarily mean large profits.
• This prevents + and * from being mixed, which limits what it can be used for.
• Anything else. I don't mind minor suggestions.

If the above limitations are too great, you can just bypass the macro by constructing a Linear-Problem manually and specifying the error multiplier you want to use, but where's the fun in that?

(ns genetic-algorithm.text-ui.linear-generator)

(defrecord Linear-Problem [profit-f constraint-f])

(defn range-error
"Returns how out of bounds n is from [c-min c-max]. Bounds are inclusive.
c-min defaults to 0 if not supplied."
([c-min n c-max]
(cond
(<= c-min n c-max) 0
(> c-min n) (- c-min n)
:else (- n c-max)))

([n c-max]
(range-error 0 n c-max)))

(defn total-range-error
"Sums the total range error defined by each range-vec.
Each vec must be a seq of 2 or 3 items, representing either [range-min n range-max],
or [n range-max] which assumes a min of 0.
Delegates to range-error."
[& range-vecs]
(->> range-vecs
(map #(apply range-error %))
(apply +)))

(defn evaluate-problem
"Returns a fitness value based on the profit and constraint functions, and the var values via var-map.
Constraint-mult multiplies the constarint error to make up for the profit function overcoming constraint error.
Should be greater than the largest profit multiplier in the profit function."
[l-prob var-map constraint-mult]
(let [{:keys [profit-f constraint-f]} l-prob]
(+ (profit-f var-map)
(- (* constraint-mult (constraint-f var-map))))))

; TODO: Generalize the number? part
(defn find-number
"Searches through a nested coll looking for a number as decided by finder-f.
finder-f is a function taking a list of numbers, and returning a single number.
Returns default if no numbers are found."
[coll default finder-f]
(let [candidates (->> coll
(map (fn [e]
(cond
(number? e) e
(coll? e) (find-number e default finder-f))))

(filter number?))]

(if (empty? candidates)
default
(finder-f candidates))))

(defn max-number [coll]
(find-number coll 1 (partial apply max)))

(defn min-number [coll]
(find-number coll 1 (partial apply min)))

(defn replace-keys-with-call
"Replaces all keywords in the nested coll with (:keyword map-sym)."
[coll map-sym]
(->> coll
(map (fn [e]
(cond
(keyword? e) (list e map-sym)
(coll? e) (replace-keys-with-call e map-sym)
:else e)))

(apply list)))

(defn process-profit-expr
"Returns a processed profit function from the given profit expression."
[profit-expr map-sym]
(fn [~map-sym]
~(replace-keys-with-call profit-expr map-sym)))

; TODO: Throw parse error for too many args?
; TODO resolve both op syms so backticks can be used in testing?
; TODO: Generalize and allow for more ops? Accept a map of symbol->transformations?
(defn standardize-op [constraint-expr]
(let [correct-op '<=
[op & args] constraint-expr]
(if (= op correct-op)
constraint-expr
(apply list correct-op (reverse args)))))

(defn process-constraint-exprs [constraint-exprs map-sym]
(let [procd (->> constraint-exprs
(map #(replace-keys-with-call % map-sym))
(map standardize-op)
(mapv #(vec (drop 1 %))))]

(fn [~map-sym]
(apply total-range-error ~procd))))

(defn- linear-problem-expr
"Returns a expression that evaluates to a Linear-Problem object based on the given profit and constraint expressions."
[profit-expr constraint-exprs]
(let [map-sym (gensym 'lp-obj-for-map)
prof-f (process-profit-expr profit-expr map-sym)
cons-f (process-constraint-exprs constraint-exprs map-sym)]

(->Linear-Problem ~prof-f ~cons-f)))

(defn genes+vars->var-map
"Accepts a gene sequence and a list of keyword vars, and returns a map mapping each gene to a var."
[genes vars]
(into {}
(map vector vars genes)))

(defmacro fit-func-for
"Accepts a linear porgramming-like syntax, and evaluates to a profit-seeking fitness function that accepts a vector
of genes, and returns the calculated fitness."
[vars _ profit _ & constraints]
(let [; How much more significant errors are than profit. Arbitrary.
profit-error-mult 5

; Find the largest number in the profit expression so we can make sure the
; contraint error overcomes any profit from illegal results.
; Assumes only * and + will be used, and use of + and * won't be mixed.
max-prof-mult (max-number profit) ; Find the
error-mult (* profit-error-mult max-prof-mult)

lp (linear-problem-expr profit constraints)]

(fn [genes#]
(let [var-map# (genes+vars->var-map genes# ~vars)]
(evaluate-problem ~lp var-map# ~error-mult)))))

• Since posting this, I've gotten rid of Linear-Problem altogether, and allowed for a :minimize variant that prefixes the profit function with (- ...). This change basically turns a profit function into a cost function. – Carcigenicate Apr 12 '18 at 3:03


text:
# Driving blue LED with 3.3V using an op-amp

I'm trying to design a circuit that drives six LEDs - two red, two green, and two blue - off a 3.3V GPIO pin. Since the voltage drop on a blue LED is in that range, I'm attempting to step the voltage up with an LM318N and then drop it with a resistor. However, all the red and green ones are working when set up like this, but the blue ones are not.

Is this a reasonable setup to accomplish what I'm going for, or is there something wrong with the way I'm approaching this?

• If you just want to PWM/on-off it, you can simplify it greatly using just one NFET and a resistor. But sure, an opamp would do the trick too if it have enough current sourcing capability on the output. Also, +1 for schematic, propper punctation and topic/subject. – winny Jan 20 '17 at 6:39
• This comment is aimed at your edit from 2 hours ago. Change R3 from 15k to more like 1k. In this case, you are running Q1 as a saturated switch, so you want the base current to be around 1/10 or 1/20 of the collector current. If the collector current is 10mA, you want the base current to be around 1mA. Etc. With 15k, I would expect the circuit to run SOME current through the LED's, but maybe not enough to make them very bright. – mkeith Jan 21 '17 at 1:21
• @mkeith That ought to work, but it didn't, so I tried to isolate the problem. Even running voltage source to resistor to LED, pure and simple, nothing is happening. It's possible the LED isn't to spec - I just grabbed some from the lab. In any case, I seem to be stuck here, but thank you for your help. – Dinesh Jan 21 '17 at 3:49
• Maybe you had the LED backwards. It is a diode, after all. It just happens to emit light when forward current flows. – mkeith Jan 21 '17 at 5:59
• @Dinesh please, don't ask a second question in your first one. It's both unlikely to help future visitors and confusing to current answerers. The best course of action would now be for you to remove the follow-up question about the new circuit (completely unrelated to "driving LED with 3.3V using op-amp") from here, and post it as a new question. – user20088 Jan 21 '17 at 14:06

LM318 is only specified to drive to within 3v of either rail, it's not a rail to rail (R2R) output opamp. As you've got it supplied from the +5v pin, its output current into a 3v load will be low at best, and might be zero.

For a drop in replacement, find an amplifier that's got a R2R output.

It might be simpler to use a switch instead, a simple transistor or FET, or a 5v rail gate with 3.3v capable input like an HCT74xx part. LM339 makes a nice low current FET driver as well, output pulls close to ground, and the inputs are very flexible.

• Thank you! I've attempted to switch to a BJT switch, but I'm not too familiar with those and the LEDs still don't work... I think my resistor setup may be wrong. I've added the schematic to OP. – Dinesh Jan 20 '17 at 23:15

I don't think you need to boost the voltage to drive the LED. Just look for a blue LED with a low Vf. Also, there is no reason to use an op-amp.

Use this circuit:

simulate this circuit – Schematic created using CircuitLab

You may need to tune the resistor values a bit. This is a current source circuit. When the GPIO is high, you will get a fairly constant current of roughly 10mA through the LED over a wide range of voltages. This circuit performs much better than a simple current limiting resistor when you barely have enough voltage to drive the LED.

The basic idea is that R1 and R2 are a resistor divider from the GPIO voltage of 3.3V. They produce around 1.1V at the base of Q1. Q1 is not saturated. It is acting as an emitter follower, producing 1.1 - 0.6V at the emitter. This means that the emitter current is around 0.5V / R3.

The things you might want to do are lower R2 to get a smaller voltage at the base, and change R3 to change the operating current. Right now it is around 10mA, very roughly.

Also, since this circuit maintains a regulated current irrespective of voltage, you can potentially use an unregulated voltage for the LED. For example, a single cell lithium ion battery voltage, or 5V from USB. The LED will only turn on when the GPIO is high, no matter what voltage you use for the diode.

• I'm not too familiar with circuits like this, so it may take some fiddling to figure out the right resistor values, but I will give it a try! Thank you! – Dinesh Jan 20 '17 at 23:17
• Fiddle in a simulator first. Also see the equations. – mkeith Jan 21 '17 at 1:17

A single transistor configured as a common emitter is enough to allow you to drive the LED from a Pi GPIO and using the Pi +5v line.

Q1 can be any jellybean NPN part such as a 2N3904, 2N4401 etc.

simulate this circuit – Schematic created using CircuitLab

• This seemed to be the simplest approach so it's what I went with first, but it still doesn't seem to be working. I've added my schematic to OP -- I think my resistor values may be off. Thank you for the suggestion! – Dinesh Jan 20 '17 at 23:19

There are several flaws in the circuit.

Firstly you are loading the GPIO pin quite heavilly. When unloaded the pin will be at damn near 3.3V but when loaded down heavilly it may be substantially lower than that.

Secondly as others have pointed out your op-amp is nowhere near a rail to rail device.

In general using an op-amp for this seems rather silly. I would suggest you use a transitor switch configuration and use the transitor to control all the LEDs, not just the blue ones.

• The op-amp is probably unnecessary, you're absolutely right - I've only worked with op-amps in theoretical problems, not in a physical circuit, so their limitations are a little fuzzy to me. I'm trying to switch to a transistor switch now - my schematic has been added to OP. Thank you! – Dinesh Jan 20 '17 at 23:18


text:
0
Research Papers: Jets, Wakes, and Impingment Cooling

# Supersonic Two-Phase Impinging Jet Heat Transfer

[+] Author and Article Information
Richard R. Parker

e-mail: parkerrr@ufl.edu

James F. Klausner

Professor
Fellow ASME
e-mail: klaus@ufl.edu

Renwei Mei

Professor
Mem. ASME
e-mail: rwmei@ufl.edu
Department of Mechanical
and Aerospace Engineering,
University of Florida,
Gainesville, FL 32611

1Corresponding author.

Contributed by the Heat Transfer Division of ASME for publication in the Journal of Heat Transfer. Manuscript received June 2, 2011; final manuscript received April 2, 2012; published online December 26, 2012. Assoc. Editor: Frank Cunha.

J. Heat Transfer 135(2), 022201 (Dec 26, 2012) (9 pages) Paper No: HT-11-1286; doi: 10.1115/1.4007408 History: Received June 02, 2011; Revised April 02, 2012

## Abstract

The experimental heat transfer rates from a supersonic two-phase impinging air jet with disperse droplets are presented. The experimental configuration consists of an expanding disperse mixture of air and water through a converging–diverging nozzle, designed for Mach 3.26 with a liquid to air mass flow ratio ranging from 1.28% to 3.83%, impinging upon a thin film heater constructed of nichrome. The spatially varying heat transfer coefficient is measured, and peak values are on the order of $200,000 W/m2K$. Two distinct regions of heat transfer are identified, one dominated by the jet impingement flow and another dominated by thin film heat transfer. The heat transfer coefficient of an impinging jet with dry air and no droplets is measured during the investigation as well. The heat transfer results are compared, and it is demonstrated that the addition of disperse water droplets to the jet significantly increases the heat removal capability of the jet as well as smoothing the spatial temperature distribution of the heater surface. As much as an order of magnitude increase in heat transfer coefficient is observed near the centerline of the jet and a factor of 3–5 increase is seen at a distance of approximately 4 nozzle diameters from the jet. The fundamental heat transfer coefficient measurements should benefit applications involving supersonic two-phase jets for high heat flux thermal management.

<>

## References

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Alvi, F. S., Bower, W. W., and Ladd, J. A., 2002, “Experimental and Computational Investigation of Supersonic Impinging Jets,” AIAA J., 40(4), pp. 599–609.
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Heymann, F. J., 1969, “High-Speed Impact Between a Liquid Drop and a Solid Surface,” J. Appl. Phys., 40(13), pp. 5113–5122.
Engel, O. G., 1955, “Waterdrop Collisions With Solid Surfaces,” J. Res. Natl. Bur. Stand., 54(5), pp. 281–298.
Engel, O. G., 1960, “Note on Particle Velocity in Collisions Between Liquid Drops and Solids,” J. Res. Natl. Bur. Stand., Sect. A, 64A(6), pp. 497–498.

## Figures

Fig. 1

Jet impingement facility

Fig. 2

Heater assembly and temperature measurement

Fig. 5

Heater thermocouple temperature profiles for single-phase jet and different heat fluxes

Fig. 4

Radial variation of single-phase jet NuD for different nozzle heights and ReD

Fig. 3

Correlating single-phase jet NuD with r/D and ReD. (a) NuD(r/D) and (b) NuD/ReD(r/D).

Fig. 6

Radial variation of two-phase jet NuD for different water mass fractions

Fig. 7

Variation of two-phase jet NuD for different nozzle heights

Fig. 8

Heater thermocouple temperature profiles for two-phase jet and different heat fluxes

Fig. 9

Variation of heat transfer enhancement factor with water mass fraction and ReD

Fig. 10

Comparison of single-phase (w = 0) and two-phase heat transfer coefficient

Fig. 11

Variation of heat transfer enhancement factor for different nozzle heights

## Errata

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