why is this problem 52 (e) singled out here?

A

allegansveritatem

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Here is problem:
13285:

Here is 52 (e):
13286

Here is what I did:
13287

I keep wondering why the author singles out 52 (e) here? Why is this particularly relevant in the problem? I didn't need it at all. All I really needed to know is that the greatest integer function is going to be a player in this job. So...what am I missing? I begin to wonder if my solution is correct.Seems to me it has to be.
 
allegansveritatem said:
Here is problem:Here is 52 (e):
View attachment 13286

I keep wondering why the author singles out 52 (e) here? Why is this particularly relevant in the problem? I didn't need it at all. All I really needed to know is that the greatest integer function is going to be a player in this job. So...what am I missing? I begin to wonder if my solution is correct.Seems to me it has to be.
Click to expand...
Have you actually studied these graphs?
GRAPH I GRAPH II GRAPH III GRAPH IV
 
allegansveritatem said:
I keep wondering why the author singles out 52 (e) here? Why is this particularly relevant in the problem? I didn't need it at all. All I really needed to know is that the greatest integer function is going to be a player in this job. So...what am I missing? I begin to wonder if my solution is correct.Seems to me it has to be.
Click to expand...

Your answer is [MATH]40 + 20\lfloor x\rfloor[/MATH], in terms of the more standard notation for the greatest integer function that your text denotes as [[x]].

Did you check your answer? Take their example, which says that f(30) = 80 and f(31) = 100. Your formula gives:
  • [MATH]f(30) = 40 + 20\lfloor 30\rfloor = 40 + 20(30) = 640[/MATH]
  • [MATH]f(31) = 40 + 20\lfloor 31\rfloor = 40 + 20(31) = 660[/MATH]
I do see that you defined x as something other than the number of minutes; but that means you haven't answered their question!

I think what you really mean is [MATH]40 + 20\lfloor \frac{x}{15}\rfloor[/MATH], where x is the number of minutes. But that doesn't work, either:
  • [MATH]f(30) = 40 + 20\lfloor \frac{30}{15}\rfloor = 40 + 20(2) = 80[/MATH]
  • [MATH]f(31) = 40 + 20\lfloor \frac{31}{15}\rfloor = 40 + 20(2) = 80[/MATH]
What you need there is not [MATH]\lfloor x\rfloor[/MATH], but what they hinted at, [MATH]-\lfloor -x\rfloor[/MATH]. To see why, do exercise 52e, as pka suggested. You clearly haven't seen the difference! (The WolframAlpha graphs show the main difference, but not the actual behavior at integers, which is also important. This is a defect of most graphing programs.)

Details matter. It is not just that the greatest integer function is involved somehow; it's how it's used!
 
Dr.Peterson said:
Your answer is [MATH]40 + 20\lfloor x\rfloor[/MATH], in terms of the more standard notation for the greatest integer function that your text denotes as [[x]].

Did you check your answer? Take their example, which says that f(30) = 80 and f(31) = 100. Your formula gives:
  • [MATH]f(30) = 40 + 20\lfloor 30\rfloor = 40 + 20(30) = 640[/MATH]
  • [MATH]f(31) = 40 + 20\lfloor 31\rfloor = 40 + 20(31) = 660[/MATH]
I do see that you defined x as something other than the number of minutes; but that means you haven't answered their question!

I think what you really mean is [MATH]40 + 20\lfloor \frac{x}{15}\rfloor[/MATH], where x is the number of minutes. But that doesn't work, either:
  • [MATH]f(30) = 40 + 20\lfloor \frac{30}{15}\rfloor = 40 + 20(2) = 80[/MATH]
  • [MATH]f(31) = 40 + 20\lfloor \frac{31}{15}\rfloor = 40 + 20(2) = 80[/MATH]
What you need there is not [MATH]\lfloor x\rfloor[/MATH], but what they hinted at, [MATH]-\lfloor -x\rfloor[/MATH]. To see why, do exercise 52e, as pka suggested. You clearly haven't seen the difference! (The WolframAlpha graphs show the main difference, but not the actual behavior at integers, which is also important. This is a defect of most graphing programs.)

Details matter. It is not just that the greatest integer function is involved somehow; it's how it's used!
Click to expand...
I will do 52 (e) tomorrow. I have already done it, but that was weeks ago and so it will be as if I am doing it for the first time--that is how it is when I come back to these problems after a few weeks. I have to learn them all over again--well, not quite. Something is retained. Thanks for the pointers. As of now I can't imagine why the double negative should be significant. Maybe doing 52 (e) will give me some ideas.
 
allegansveritatem said:
I looked at the links you provided. Thanks. Actually I did set up a few of the graphs while doing the problem on my calculator. When I found that graph (e) was the same as int x, I was puzzled and still am. Why the double negative?
Click to expand...
NO! The graph of -int(-x) is NOT the same as int(x). It's similar, but that's not enough.

Take lev's example: int(1.4) = 1, right? But -int(-1.4) = -(-2) = 2.

Math can't be done lightly. If you're going to work a problem, you must pay attention to the details, or you are wasting your time.
 
Actually, perhaps your calculator's int function is not the greatest integer function (properly called floor), which is what is under discussion. See this page: https://www.mathsisfun.com/sets/function-floor-ceiling.html

If so, then the problem is that you are depending on the calculator rather than on thinking about definitions. Do the work by hand in order to fully understand!

But from what I understand, if you are using a TI-84, that is not the issue: https://mathbits.com/MathBits/TISection/PreCalculus/GraphGreatestIntFunction.html
 
Dr.Peterson said:
Actually, perhaps your calculator's int function is not the greatest integer function (properly called floor), which is what is under discussion. See this page: https://www.mathsisfun.com/sets/function-floor-ceiling.html

If so, then the problem is that you are depending on the calculator rather than on thinking about definitions. Do the work by hand in order to fully understand!

But from what I understand, if you are using a TI-84, that is not the issue: https://mathbits.com/MathBits/TISection/PreCalculus/GraphGreatestIntFunction.html
Click to expand...
I am using TI Nspire CAS. The int() is indeed the greatest integer function. I agree that letting the calculator do the thinking is bad practice and I try to avoid that. I like to use the calculator to experiment with different ideas. For instance, I put up a graph, then change one of the parameters to see what the effect will be. With the calculator you can do a great many of these transformations in a very short time and I think this is a learning experience. But to use the calculator as a kind of black box...that's no good at all, certainly not for students, and, for better or for worse, student is what I am.
 
NO! The graph of -int(-x) is NOT the same as int(x). It's similar, but that's not enough.

Take lev's example: int(1.4) = 1, right? But -int(-1.4) = -(-2) = 2.

Math can't be done lightly. If you're going to work a problem, you must pay attention to the details, or you are wasting your time.
[/QUOTE]
Right. I set up both graphs on my calculator again today and saw the difference.I also used the trace function on the calculator and found that on the graph int(x) the greatest integer was found to the left on the number line and the graph - int(-x) was found to the right. And that is what I want for this problem, as you said.
Part of my problem is that I thought the greatest integer functions default procedure was to go to RIGHT to find the integer whereas I found today upon review that the default is to go to the left. Makes a big difference! Here is my new billing function:
b of t.PNG
 
Looks good.

It pays to check definitions, doesn't it? Hopefully, this will be memorable. The floor (int) function rounds down; -int(-x) rounds up. Both leave integers unchanged.

By the way, have you compared f(x) = -int(-x) to g(x) = int(x) + 1 to see the similarity and difference in their graphs? It's subtle but important.
 
Quote: Lev:
I meant this function: [[x]]. What does it do? What's the result if x is 1.4?

Yes, I see what you mean. the answer would be 1 and if it were a int -1.4 it would be -2 and if that were negatived it would be 2--and that is what is being called for here.
 
I am not familiar with the notation, but it appears that the answer is

[MATH]f(m) = 40 + 20 * \left \lceil \dfrac{m}{15} \right \rceil, \text { where } m = \text {number of minutes spent on repair.}[/MATH]
Now

[MATH]\lceil x \rceil = -\ \lfloor -\ x \rfloor.[/MATH]
For example,

[MATH] \lceil 1.2 \rceil = 2 = -\ (-\ 2) = \lfloor -\ 1.2 \rfloor.[/MATH]
[MATH] \lceil -\ 1.2 \rceil = -\ 1 = -\ (1) = -\ \lfloor 1.2 \rfloor = -\ \lfloor -\ (-\ 1.2) \rfloor.[/MATH]
So you can restate the answer as

[MATH]f(m) = 40 - 20 * \left \lfloor -\ \dfrac{m}{15} \right \rfloor.[/MATH][/MATH]
 
Last edited:
Dr.Peterson said:
Looks good.

It pays to check definitions, doesn't it? Hopefully, this will be memorable. The floor (int) function rounds down; -int(-x) rounds up. Both leave integers unchanged.

By the way, have you compared f(x) = -int(-x) to g(x) = int(x) + 1 to see the similarity and difference in their graphs? It's subtle but important.
Click to expand...
Yes, I saw the difference: The double negative graph seemed to be shifted over a little more to the right, which, I suppose as to do with the fact that all the action in this graph is in a rightward direction.
 
JeffM said:
I am not familiar with the notation, but it appears that the answer is

[MATH]f(m) = 40 + 20 * \left \lceil \dfrac{m}{15} \right \rceil, \text { where } m = \text {number of minutes spent on repair.}[/MATH]
Now

[MATH]\lceil x \rceil = -\ \lfloor -\ x \rfloor.[/MATH]
For example,

[MATH] \lceil 1.2 \rceil = 2 = -\ (-\ 2) = \lfloor -\ 1.2 \rfloor.[/MATH]
[MATH] \lceil -\ 1.2 \rceil = -\ 1 = -\ (1) = -\ \lfloor 1.2 \rfloor = -\ \lfloor -\ (-\ 1.2) \rfloor.[/MATH]
So you can restate the answer as

[MATH]f(m) = 40 - 20 * \left \lfloor -\ \dfrac{m}{15} \right \rfloor.[/MATH][/MATH]
Click to expand...
Right. As long as the two negatives are there, and one of them is the int(x), all is well.
 
It is a nuisance that many computer packages support the floor function, but not the ceiling function. It is also nuisance that they usually call it the "int function" because "floor function" is actually more descriptive. When you have only the floor function, but need the ceiling function, here is another way to get there in a program.

[MATH]\text {IF INT}(x) = x, \text { THEN }y \rightarrow x; \text { ELSE } y \rightarrow \text {INT}(x) + 1.[/MATH]
What is shown above is simple to understand, but is a less efficient algorithm than

[MATH]y \rightarrow - \text { INT}(-\ x)[/MATH],

which only involves the INT routine once and requires no branching.
 
JeffM said:
It is a nuisance that many computer packages support the floor function, but not the ceiling function. It is also nuisance that they usually call it the "int function" because "floor function" is actually more descriptive. When you have only the floor function, but need the ceiling function, here is another way to get there in a program.

[MATH]\text {IF INT}(x) = x, \text { THEN }y \rightarrow x; \text { ELSE } y \rightarrow \text {INT}(x) + 1.[/MATH]
What is shown above is simple to understand, but is a less efficient algorithm than

[MATH]y \rightarrow - \text { INT}(-\ x)[/MATH],

which only involves the INT routine once and requires no branching.
Click to expand...

Yes, it is odd that they omit the ceiling
Another thing missing is a dedicated program for finding inverse functions--but there are workarounds for this stuff and so, no big deal.
 
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