Monash Maths thread

[kinslayer]

Hello,

I have a question I don't know how to do...so I'm hoping someone could help.

I have to prove this statement true or either counter-example to demonstrate it is false:

"For any function ."

Thank you!

What does mean? Can you say it in words?

[Phy124]

Edited to clarify and give a proper answer as kinslayer was correct in saying:

Strictly speaking, in Phy's example, equality does hold if x=y=0, so you would need to allow for that, at least on the last line.

[kinslayer]

Assuming the above interpretation is correct, a constant function would also work as a counter-example. If f(x) = c for all x and some  c which is not equal to 1 or 0, then f(x)f(y) = c^2 for all x,y, but f(xy) = c. 

c^2 = c only when c = 1 or 0; a contradiction. Therefore, with this f, f(x)f(y) does not equal f(xy).

Example: if f(x) = 2, then f(x)f(y) = 4, but f(xy) = 2.

Strictly speaking, in Phy's example, equality does hold if x=y=0, so you would need to allow for that, at least on the last line.

[keltingmeith]

Hey guys, this is for the MTH2010 assignment, so I'd rather not have the answer given to me. However, I'm a little unsure as to my reasoning.

I need to show that by the definition of a limit.

So, since delta in this case represents the radius of a circle closing in at (0, 1), I substituted that into the bottom, giving and from this, I let . However, then if I let epsilon get infinitely small, delta must be infinitely large. Similarly, if delta is infinitely small, epsilon become infinitely large, and that would sort of destroy the whole proof.

[kinslayer]

I don't follow your logic here. In particular, I'm not sure what you are doing in the step after you substituted delta. Also, delta should be a function of epsilon, not the other way around.

I recommend starting with and then write down a bunch of inequalities. What you want ultimately is a function of delta on the RHS (which you can then set equal to epsilon and solve for delta). Hint: the triangle inequality may help.

Doing it this way, I came up with .

PS: No more 2015?

[keltingmeith]

I don't follow your logic here. In particular, I'm not sure what you are doing in the step after you substituted delta. Also, delta should be a function of epsilon, not the other way around.

I recommend starting with and then write down a bunch of inequalities. What you want ultimately is a function of delta on the RHS (which you can then set equal to epsilon and solve for delta). Hint: the triangle inequality may help.

Doing it this way, I came up with .

PS: No more 2015?

Yeah, every example I've found online seems to include the triangle inequality, but I haven't been able to include it. Guess I'll just have to play around with adding random things and see what comes out, hahah.

And yeah, I had to drop it for health reasons. But I may appear in random workshops to learn fun things.

[BigAl]

Hey guys, this is for the MTH2010 assignment, so I'd rather not have the answer given to me. However, I'm a little unsure as to my reasoning.

I need to show that by the definition of a limit.

So, since delta in this case represents the radius of a circle closing in at (0, 1), I substituted that into the bottom, giving and from this, I let . However, then if I let epsilon get infinitely small, delta must be infinitely large. Similarly, if delta is infinitely small, epsilon become infinitely large, and that would sort of destroy the whole proof.

This question is same as last year's assignment...I had so much difficulty with Cartesian coordinates..but polar coordinates simplify this question..give it a go!

[keltingmeith]

That's what a friend of mine said, and I was unsure because I've never worked in polar co-ordinates before bar complex numbers in VCE. I still have a week until it's due, though, so I think I'll play around with it in polar and see what I can do.

[BigAl]

While I'm here..I have a question about differential equations..feeling a bit rusty
so I'm solving a separable differential equation and the left hand side has this form
so I have to exclude and But the wording in the assignment confuses me..it's insisting as if exists and wants me to draw the solutions given two initial conditions...Not really sure what to do

[keltingmeith]

Sorry for the double post, but I forgot to ask - what does this actually mean? Namely the "min{...}" part, I've never seen it before.

EDIT: Then you save me from the double post, BA.

While I'm here..I have a question about differential equations..feeling a bit rusty
so I'm solving a separable differential equation and the left hand side has this form
so I have to exclude and But the wording in the assignment confuses me..it's insisting as if exists and wants me to draw the solutions given two initial conditions...Not really sure what to do

Depends on how you got to that form, pretty sure (granted, haven't done MTH2032, so I don't know all the technicalities there).

Basically, y= -1 and y = 1 COULD be solutions if y(c) = 1 or -1 are an initial condition. Just that the integral won't include those solutions. There are three possible solution curves in all, the solution to that integral, y=1 and y=-1. Which of the three you have depends on the initial condition.

Hope all that made sense. n.n;

[BigAl]

Depends on how you got to that form, pretty sure (granted, haven't done MTH2032, so I don't know all the technicalities there).

Basically, y= -1 and y = 1 COULD be solutions if y(c) = 1 or -1 are an initial condition. Just that the integral won't include those solutions. There are three possible solution curves in all, the solution to that integral, y=1 and y=-1. Which of the three you have depends on the initial condition.

Hope all that made sense. n.n;

that's the problem..there is no c that satisfies that condition

[keltingmeith]

that's the problem..there is no c that satisfies that condition

If may be easier to explain if I do this:



In this case, if y = -7, we have a gradient of zero. Nothing wrong with that, but then I go to solve by separation of variables:



Now, when I divided by y+7, we lost the solution y=-7. More specifically, we lost the solution y+7=0. So, we say that y+7=0 could still be a solution, since there was nothing wrong with that solution at the very start. So, if we have an initial condition of y(c)=-7, it will have the solution curve y=-7, as the possible solutions of the original differential equation are and .

If you're still unsure of what I'm trying to say, wait for someone else to come in and clarify, sorry. n.n;

[kinslayer]

Sorry for the double post, but I forgot to ask - what does this actually mean? Namely the "min{...}" part, I've never seen it before.

It just means the minimum. When I was doing the algebra I restricted delta to values between 0 and 1 to make it easier to deal with. If a given value of delta works for some epsilon, then it will also work for any larger epsilon.

Here, for any , works. For , we take .

[keltingmeith]

It just means the minimum. When I was doing the algebra I restricted delta to values between 0 and 1 to make it easier to deal with. If a given value of delta works for some epsilon, then it will also work for any larger epsilon.

Here, for any , works. For , we take .

Don't we want epsilon and delta to be infinitely small, though? Would we still claim that the limit exists if we can only have it for delta = 1 or epsilon > 2?

[kinslayer]

Don't we want epsilon and delta to be infinitely small, though? Would we still claim that the limit exists if we can only have it for delta = 1 or epsilon > 2?

Epsilon can be whatever positive number, it's just that is only small enough when . But that's okay because we know that for anything bigger than that, will work anyway.