The Mathematics Thread

[juddles]

Re: The totally off-topic thread

..... Test review made me explaining the concept of estimation Pi~3, so the answer has to be about 30. Therefore if your answer is 276 it must be wrong.....

Excellent support of her development

I think that is exactly the sort of reality check that so many younger generation people are unable to do.

 

[medhead]

Re: The totally off-topic thread

Excellent support of her development

I think that is exactly the sort of reality check that so many younger generation people are unable to do.

High school maths with her is a real struggle for me. At that age, I wasn't that bothered by maths, didn't put in an effort, didn't really understand it. But later on a had a good teacher who explained things so I could understand and grasp it. She struggles a bit, probably doesn't understand, as I didn't. Swmbo reckons I'm a natural at maths, I don't understand why others can't get it. I can't expect daughter to be good at maths. School wants to pop her down to lower level maths. Part of me says just do that, another part says push on she'll get there.

Who knows, but I gotta put in some effort to help.

 

[anat0l]

Re: The totally off-topic thread

High school maths with her is a real struggle for me. At that age, I wasn't that bothered by maths, didn't put in an effort, didn't really understand it. But later on a had a good teacher who explained things so I could understand and grasp it. She struggles a bit, probably doesn't understand, as I didn't. Swmbo reckons I'm a natural at maths, I don't understand why others can't get it. I can't expect daughter to be good at maths. School wants to pop her down to lower level maths. Part of me says just do that, another part says push on she'll get there.

Who knows, but I gotta put in some effort to help.

Do you know what is troubling her the most?

Maths is really difficult to help others understand. It's like they need a light bulb moment to get it, but making that light bulb go is really difficult and random. Not having an effective teacher does not help in this regard, though sometimes it can be trying as you really need patience to help some. Most will try to get by through "parrot" learning (this limits the ability to solve application problems).

For example, one time I was tutoring a grade 8 student and one of the problems was something along the lines of: "Rewrite each of these numbers in increasing order: 0.70, 7, 0.07, 7/1000, 10/7, √7, 0.77, 0.707". Sounds trivial enough a problem, but attempting to explain it took a very long time. He got there in the end, and was able to do another similar problem.

Ultimately, it also depends what a student wants in the end, i.e. just to pass it (get a C) or do well for some other reason (get an A). Certainly when they transition to senior high school where maths is not compulsory any more (I believe this is the predominant case in Australia), they'll be very glad to be rid of it, and it will show (at least until they may have to take remedial classes if preparing for other purposes, e.g. the Queensland Core Skills Test).

 

[medhead]

Re: The totally off-topic thread

On the subject of maths, I remember what my late mum used to recall when she was at school, and struggling with maths.

"When the teacher used to say something like "Let X equal 5", I used to say, "But what if it isn't? or ... "Why X and not Z?" "

That explains a lot.

Hopefully the teacher replied "Well it does as I've just set it" or "because Z =7".

 

[RAM]

Re: The totally off-topic thread

On the subject of maths, I remember what my late mum used to recall when she was at school, and struggling with maths.

"When the teacher used to say something like "Let X equal 5", I used to say, "But what if it isn't? or ... "Why X and not Z?" "

Some times you can do well DESPITE the teacher.

Physics - Yr12 - "why can't a capacitor achieve more than a 50% conversion of energy through an improvement in the storage material or gate for example?"

"The best it can ever achieve yesterday, today or ever is 50% the rest will always be lost as heat."

"Why? That sounds like what people said about the world being flat, they did not know better and hey presto knowledge improved and they were found to be wrong."

"It can NEVER be better than 50%, accept it!"

Just a few years later a break through happened and the rest is history. All the money I possibly could have made if I had persevered....

Or my personal favourite,

Maths: Why since 1/1 =1, -1/-1 = 1, 1^(1/infinity) / 1 ^(1/infinity) = 1, -1^(1/infinity) / -1 ^(1/infinity) = 1 then 0 /0 must =1.

"If you graph the function x = y for all values and calculate the slope anywhere along that function then it is also 1, or taking the limits approaching 1/infinity at any point of the function other than 0 it is 1 then why is it not 1 at 0?"

"Because it is a discontinuous function."

"Why? Where is the proof? If the definition of zero does not change than zero / zero must equal the identity, 1 by definition?"

"It doesn't. Just accept it, that's how mathematics is, there are rules you have to accept."

"However every other rule can be proven or not disproven. Here I have disproven in the case of graphically depicting the function of x=y, taking limits also disproves it at every other point."

"Get out!"

It was one way to leave school and go have a look at the museum though!

 

[TheRealTMA]

Re: The totally off-topic thread

For all you Maths PhDs out there, a questions from a 6 yo "Is infinity odd or even?"

 

[medhead]

Re: The totally off-topic thread

Some times you can do well DESPITE the teacher.

Physics - Yr12 - "why can't a capacitor achieve more than a 50% conversion of energy through an improvement in the storage material or gate for example?"

"The best it can ever achieve yesterday, today or ever is 50% the rest will always be lost as heat."

"Why? That sounds like what people said about the world being flat, they did not know better and hey presto knowledge improved and they were found to be wrong."

"It can NEVER be better than 50%, accept it!"

Just a few years later a break through happened and the rest is history. All the money I possibly could have made if I had persevered....

Or my personal favourite,

Maths: Why since 1/1 =1, -1/-1 = 1, 1^(1/infinity) / 1 ^(1/infinity) = 1, -1^(1/infinity) / -1 ^(1/infinity) = 1 then 0 /0 must =1.

"If you graph the function x = y for all values and calculate the slope anywhere along that function then it is also 1, or taking the limits approaching 1/infinity at any point of the function other than 0 it is 1 then why is it not 1 at 0?"

"Because it is a discontinuous function."

"Why? Where is the proof? If the definition of zero does not change than zero / zero must equal the identity, 1 by definition?"

"It doesn't. Just accept it, that's how mathematics is, there are rules you have to accept."

"However every other rule can be proven or not disproven. Here I have disproven in the case of graphically depicting the function of x=y, taking limits also disproves it at every other point."

"Get out!"

It was one way to leave school and go have a look at the museum though!

Seems to be a bit of a misunderstanding of slope or gradient. Determination of slope requires 2 points on the line. Change in Y/change in X if that is 0/0 then it's a point, even a singularity. Considering slope at any single point on the graphed function will result in 0/0. But the function y=x has a slope of 1

 

[RAM]

Re: The totally off-topic thread

Seems to be a bit of a misunderstanding of slope or gradient. Determination of slope requires 2 points on the line. Change in Y/change in X if that is 0/0 then it's a point, even a singularity. Considering slope at any single point on the graphed function will result in 0/0. But the function y=x has a slope of 1

Don't think there is a misunderstanding just different attempts to demonstrate why 0 / 0 = 1 vs not undefined.

About the slope or gradient. As the slope of this function never changes then by calculating the slope/gradient by taking an infinitesimally small distance above and below any point of the function will generate the same result for all readings of the function other than X = 0, but only through accepting that 0 / 0 does not equal 1.

That is why you take the limits approaching a point from above and below. Taking the limits at a almost non-existent distance above and below the point achieves the same result for any point along the function other than when X=0.

So where do you stand? Should 0 / 0 = 1?

 

[david1]

Re: The totally off-topic thread

Maths: Why since 1/1 =1, -1/-1 = 1, 1^(1/infinity) / 1 ^(1/infinity) = 1, -1^(1/infinity) / -1 ^(1/infinity) = 1 then 0 /0 must =1.

For all you Maths PhDs out there, a questions from a 6 yo "Is infinity odd or even?"

Same answer to both of these, infinity is not a number, it's a concept (infinity is quantity larger than any number, so it can't be a number itself). This means it doesn't behave like a number (for example this: infinity+1=infinity makes no sense). Odd/even-ness is only meaningful to positive integers (strictly), so you can say infinity is odd or even (you can't say whether 1.543 id odd or even either). This also means you can't write 1/infinity, the best you can do is take the limit as x approaches infinity of 1/x, which is zero. Also in that proof, you have assumed that -(1^(1/infinity))=(-1)^(1/infinity), which you can't really do, even forgetting about using infinity as a number.

About the slope or gradient. As the slope of this function never changes then by calculating the slope/gradient by taking an infinitesimally small distance above and below any point of the function will generate the same result for all readings of the function other than X = 0, but only through accepting that 0 / 0 does not equal 1.

That is why you take the limits approaching a point from above and below. Taking the limits at a almost non-existent distance above and below the point achieves the same result for any point along the function other than when X=0.

The slope is one everywhere, it does not depend on x. You shouldn't need to worry about 0/0 to get this, taking the limit at x=0 is no different from taking the limit anywhere else.

0/0 is undefined, and reasonably so I think. Division should be continuous with respect to the numerator and denominator, but the limit of x/y, as x and y both go to 0, is not defined as depending on your path to x=y=0, you can get the limit to go to any real number you want, or diverge to infinity or negative infinity, so you get an essential discontinuity. That's the problem with defining 0/0 (it's a similar thing for 0^0).

 

[trixydoza]

Re: The totally off-topic thread

depending on your path to x=y=0, you can get the limit to go to any real number you want, or diverge to infinity or negative infinity, so you get an essential discontinuity. That's the problem with defining 0/0 (it's a similar thing for 0^0).

Yep, the limits of x/x^2 and x^2/x as x goes to zero being good examples.

 

[RAM]

Re: The totally off-topic thread

I do not think you will find any mathematical definition that does not state 'infinity is a number'.

"a number greater than any assignable quantity or countable number (symbol ∞)."

or
"Infinity, most often denoted as

, is an unbounded quantity that is greater than every real number. The symbol

had been used as an alternative to M (

) in Roman numerals until 1655, when John Wallis suggested it be used instead for infinity.
Infinity is a very tricky concept to work with, as evidenced by some of the counterintuitive results that follow from Georg Cantor's treatment of infinite sets.
Informally,

, a statement that can be made rigorous using the limit concept,
[TABLE="align: center"]
[TR]
[TD="align: left"]

[/TD]
[/TR]
[/TABLE]

Similarly,
[TABLE="align: center"]
[TR]
[TD="align: left"]

[/TD]
[/TR]
[/TABLE]

where the notation

indicates that the limit is taken from the positive side of the real line.
In the Wolfram Language,

is represented using the symbol Infinity. "

Same answer to both of these, infinity is not a number, it's a concept (infinity is quantity larger than any number, so it can't be a number itself). This means it doesn't behave like a number (for example this: infinity+1=infinity makes no sense). Odd/even-ness is only meaningful to positive integers (strictly), so you can say infinity is odd or even (you can't say whether 1.543 id odd or even either). This also means you can't write 1/infinity, the best you can do is take the limit as x approaches infinity of 1/x, which is zero. Also in that proof, you have assumed that -(1^(1/infinity))=(-1)^(1/infinity), which you can't really do, even forgetting about using infinity as a number.

However, if infinity can only be defined with reference to numbers.....

Now in the example of taking the limit above and below, raising something to the same exponential power of 1/X will create the same resultant deviation from that number.

So that is why I was talking about taking the limit of X approaching ∞. That is effectively zero.

Also, in my proof I did not assume -(1^(1/infinity))=(-1)^(1/infinity).

Instead I stated that 1^(1/infinity) / 1 ^(1/infinity) = 1, -1^(1/infinity) / -1 ^(1/infinity) = 1 I did not equate them I used each (the +ve and -ve) in isolation.

The slope is one everywhere, it does not depend on x. You shouldn't need to worry about 0/0 to get this, taking the limit at x=0 is no different from taking the limit anywhere else.

And the slope at 0 is 1 so....

Instead of using 1/infinity I could have used +/- (1/10^1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000) you get my point that the gradient remains 1 despite the calculation (to even 20 decimal places being). The formula for calculating the gradient being{(X-X')/(Y-Y')}

so 0.00000000000000000000 / 0.00000000000000000000 = 1

0/0 is undefined, and reasonably so I think. Division should be continuous with respect to the numerator and denominator, but the limit of x/y, as x and y both go to 0, is not defined as depending on your path to x=y=0, you can get the limit to go to any real number you want, or diverge to infinity or negative infinity, so you get an essential discontinuity. That's the problem with defining 0/0 (it's a similar thing for 0^0).

Why is it reasonably so? What is the reason?

True you can generate numerous functions where when X=0 then Y=0 but in that case the gradient of the function is not always 1. In this instance it is. Surely you could then use the same argument to state that when X=Y but does not equal zero then the same reasoning should apply?

Unless you assume a discontinuity as your out. Which of course completes the circle once more, leading to why assume a discontinuity?

 

[RAM]

Re: The totally off-topic thread

Yep, the limits of x/x^2 and x^2/x as x goes to zero being good examples.

Yes, quite correct but the slope/gradient is not 1 everywhere. So if you use the formula to calculate the gradient at X=0, Y=0 with the limit X approaching ∞ you get 0 / 0.

 

[trixydoza]

Re: The totally off-topic thread

Yes, quite correct but the slope/gradient is not 1 everywhere. So if you use the formula to calculate the gradient at X=0, Y=0 with the limit X approaching ∞ you get 0 / 0.

The fact that the gradient of the function is 1 at all points has nothing whatsoever to do with any ratio of one number to another. Your insistence on trying to equate 1 to a formula of your choosing is ridiculous. Why not do the same with y=2x and argue that 0/0 equals 2?

I think your teacher exercised admirable restraint.

 

[david1]

Re: The totally off-topic thread

I do not think you will find any mathematical definition that does not state 'infinity is a number'.

Your second definition does not state that infinity is a number. Here's another definition:
"Infinity (symbol: ∞) is an abstract concept describing something without any limit and is relevant in a number of fields, predominantly mathematics and physics. In mathematics, "infinity" is often treated as if it were a number (i.e., it counts or measures things: "an infinite number of terms") but it is not the same sort of number as natural or real numbers."

And the slope at 0 is 1 so....

Instead of using 1/infinity I could have used +/- (1/10^1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000) you get my point that the gradient remains 1 despite the calculation (to even 20 decimal places being). The formula for calculating the gradient being{(X-X')/(Y-Y')}

so 0.00000000000000000000 / 0.00000000000000000000 = 1

Why is it reasonably so? What is the reason?

The reason is that in this case, as you make the change in x smaller and smaller, the change in y decreases at the same rate. It's a limit, and so you do this:
y'(x)=limit as h approaches 0 of ((x+h)-x)/h=limit as h approaches 0 of h/h=1, and it didn't matter what x was, and I never considered 0/0. (or limit of ((x+h)-(x-h))/(2h) as h->0 or whatever) Since y=x has constant slope, you can do y-step/x-step, but changing x by and an amount changes y by exactly the same amount, so you always get. I you make the amount smaller you are effectively doing limit as h->0 of h/h=1, but this limit does not imply that 0/0 should be 1, because by similar arguments you can "show" that 0/0 should be any number you can think of.

True you can generate numerous functions where when X=0 then Y=0 but in that case the gradient of the function is not always 1. In this instance it is. Surely you could then use the same argument to state that when X=Y but does not equal zero then the same reasoning should apply?

Unless you assume a discontinuity as your out. Which of course completes the circle once more, leading to why assume a discontinuity?

I'm saying that, for example, you take 0/1=0, then 0/0.5=0, then 0/0.25=0, and so on, you always get 0, so using that you would say that 0/0=0. But if you do 2/1=2, 1/0.5=2, 0.5/0.25=2 and so on, you would say that 0/0=2. You can't resolve that kind of inconsistency.

 

[RAM]

Re: The totally off-topic thread

The fact that the gradient of the function is 1 at all points has nothing whatsoever to do with any ratio of one number to another.

What is the formula to calculate a gradient at a specific point on a function?

Your insistence on trying to equate 1 to a formula of your choosing is ridiculous. Why not do the same with y=2x and argue that 0/0 equals 2?

When the function is Y = X, and the gradient is to be calculated for the specific point of X = 0, so Y = 0?

Its already been stated that at all points along that function the gradient = 1.

I think your teacher exercised admirable restraint.

So did he!

Just because something cannot be disproved should not cause aggravation, just like asking the question about capacitor efficiency - knowledge is always being extended. If one cannot explain a point without resorting to the equivalent of "it just is" then there is a knowledge or understanding problem.

What is convenient is not always what is correct.

 

[david1]

Re: The totally off-topic thread

What is the formula to calculate a gradient at a specific point on a function?

For a function defined as y=f(x) (like y=x, y=x^2+2*x+1, y=sin(x) or whatever), the gradient at a point c is denoted y'(c), and is defined as the limit, as h approaches 0, of (y(c+h)-y(c))/h.

 

[OATEK]

Re: The totally off-topic thread

For a function defined as y=f(x) (like y=x, y=x^2+2*x+1, y=sin(x) or whatever), the gradient at a point c is denoted y'(c), and is defined as the limit, as h approaches 0, of (y(c+h)-y(c))/h.

What is the probability that AFFers will return to speaking English soon?

 

[RAM]

How to get the blood pumping...

Same answer to both of these, infinity is not a number, it's a concept (infinity is quantity larger than any number, so it can't be a number itself).

Your second definition does not state that infinity is a number. Here's another definition:
"Infinity (symbol: ∞) is an abstract concept describing something without any limit and is relevant in a number of fields, predominantly mathematics and physics. In mathematics, "infinity" is often treated as if it were a number (i.e., it counts or measures things: "an infinite number of terms") but it is not the same sort of number as natural or real numbers."

In the 2nd definition it defines it with reference to numbers.

In the definition you have cited it again describes it as a number, but a 'different' sort of number. So I think it is fair enough to say infinity is a number, and it is a generally accepted principle that a number divided by itself generates 1. Similarly wrt limits.

The reason is that in this case, as you make the change in x smaller and smaller, the change in y decreases at the same rate. It's a limit, and so you do this:
y'(x)=limit as h approaches 0 of ((x+h)-x)/h=limit as h approaches 0 of h/h=1, and it didn't matter what x was, and I never considered 0/0. (or limit of ((x+h)-(x-h))/(2h) as h->0 or whatever) Since y=x has constant slope, you can do y-step/x-step, but changing x by and an amount changes y by exactly the same amount, so you always get. I you make the amount smaller you are effectively doing limit as h->0 of h/h=1, but this limit does not imply that 0/0 should be 1, because by similar arguments you can "show" that 0/0 should be any number you can think of.

Which is a given as the function is Y = X, and as you say will always get 1.

Yes totally agree, so in calculating the gradient at the point of X = 0 and Y = 0 as given by the function Y = X, taking the limits approaching infinity gets you to 0 / 0.

Taking this up at UNSW led to some great debates with the head of faculty and others with the result that several changed their tune as it does suggest an inconsistency. Black holes anyone?

I'm saying that, for example, you take 0/1=0, then 0/0.5=0, then 0/0.25=0, and so on, you always get 0, so using that you would say that 0/0=0. But if you do 2/1=2, 1/0.5=2, 0.5/0.25=2 and so on, you would say that 0/0=2. You can't resolve that kind of inconsistency.

Now your getting to the point. You have demonstrated a reason why it should not be defined. In the example I used, I demonstrated where it can be defined with reference to that function = the conundrum.

Why is it so? (Apologies to Sir Julius!)

 

[RAM]

Re: The totally off-topic thread

For a function defined as y=f(x) (like y=x, y=x^2+2*x+1, y=sin(x) or whatever), the gradient at a point c is denoted y'(c), and is defined as the limit, as h approaches 0, of (y(c+h)-y(c))/h.

Or it can be given by:

The gradient of the line is

Going from [FONT=Times New Roman, Times, serif]A[/FONT] to [FONT=Times New Roman, Times, serif]B[/FONT] the rise is y - y[SUB]1[/SUB], and the run is x - x[SUB]1[/SUB],

So,
gradient =

gradient = [FONT=Times New Roman, Times, serif]m[/FONT], so
[FONT=Times New Roman, Times, serif]m[/FONT] =

[FONT=Times New Roman, Times, serif]m[/FONT](x - x[SUB]1[/SUB]) = y - y[SUB]1[/SUB]
Let us look at this formula more closely.
It is telling us the equation of the line with gradient m which goes through the point (x[SUB]1[/SUB], y[SUB]1[/SUB]) is

 

[david1]

Re: How to get the blood pumping...

I fear I've taken the off-topic thread too far off-topic! This will be my last post on the matter.

Yes totally agree, so in calculating the gradient at the point of X = 0 and Y = 0 as given by the function Y = X, taking the limits approaching infinity gets you to 0 / 0.

No, taking a limit avoids calculating 0/0, that's the whole reason for using limits. Using your reasoning, if you considered y=2*x not y=x, you would be claiming that 0/0=2. (A technical terms for 0/0, 0*infinity and other such thins is "indeterminate forms")