Monash Maths thread

[keltingmeith]

Ahkay, I think I've got a handle on it now, thank you.

[emchun]

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

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.

Thank you both for replying, but in Phy's example I don't really understand where you got from?
And for kinslayer's example, I also don't understand how you got and
Sorry, these are probably really stupid questions but I'm not very good in maths. 

[m.Chemia]

Thank you both for replying, but in Phy's example I don't really understand where you got from?
And for kinslayer's example, I also don't understand how you got and
Sorry, these are probably really stupid questions but I'm not very good in maths. 

In Phy's example, the function is , that means the argument (variable) of the function is . What means is the value of the function when it is evaluated at the point . Substituting into the function as its argument, you get .

In kinslayer's example, the function is for all , that means no matter what the argument of the function becomes, the function always has value . In particular, , and . Therefore and as mentioned .

Hope this clarifies a bit

[emchun]

In Phy's example, the function is , that means the argument (variable) of the function is . What means is the value of the function when it is evaluated at the point . Substituting into the function as its argument, you get .

In kinslayer's example, the function is for all , that means no matter what the argument of the function becomes, the function always has value . In particular, , and . Therefore and as mentioned .

Hope this clarifies a bit

Thanks! I finally understand both of them now.

If it's not too much trouble, could anybody please explain how to write as a piecewise defined function? I'm assuming piecewise is also called hybrid functions? The in there is what is confusing me, none of the examples given in lecture notes have this kind of example. I tried doing it and I got when and when , which I don't think is right..

[keltingmeith]

Thanks! I finally understand both of them now.

If it's not too much trouble, could anybody please explain how to write as a piecewise defined function? I'm assuming piecewise is also called hybrid functions? The in there is what is confusing me, none of the examples given in lecture notes have this kind of example. I tried doing it and I got when and when , which I don't think is right..

You're right - a piecewise function is also called a hybrid function!

Instead of looking at the function as a whole, let's break it up like so:


Okay, so now we know that for g(x), when x>-1, we have the graph y=x+1. When x<-1, we have the graph y=-x-1. Now, let's put that into the first function and see what we get:



And that's your piecewise function (although you should write it with the super-cool fancy curly brackets. )

[simba]

I'm currently totally lost on how to prove the same question eulerfan101 mentioned.. I've gotten the first few steps but am confused at how to actually apply the triangle inequality! Any help would be fantastic!

[kinslayer]

Try this:

[emchun]

You're right - a piecewise function is also called a hybrid function!

Instead of looking at the function as a whole, let's break it up like so:


Okay, so now we know that for g(x), when x>-1, we have the graph y=x+1. When x<-1, we have the graph y=-x-1. Now, let's put that into the first function and see what we get:



And that's your piecewise function (although you should write it with the super-cool fancy curly brackets. )

I get it now! Thank you so much~

[Chazef]

forgive me for not using latex but can somebody explain how to get the limit as x -> 0 of 1/x - 1/(x*sqrt(x+1))

[e^1]

Realise that when you apply limit laws and continuity theorems, you get the indeterminate form 0/0. So you can use L'Hopital's rule to find the limit from there.

[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.

Back to this again.

I tried putting it into polar form, and it didn't work... Then I tried it again, and it did work after I had a better understanding of what I was doing. Now, I'm fine with all of my steps, except for my very first one.



Basically, my reasoning here is that since we're closing in on the point (0,1), I want to centre my coordinate system at (0,1). So, instead of doing your usual polar conversion of x=blah and y=similar blah, I've let x=blah and y-1=similar blah.

At first this seemed perfectly fine to me, but then a 4th year saw it, didn't realise what I was doing, and said it was all wrong... Then after explaining it, he got extremely confused and wasn't sure what was happening. So, I thought I'd ask here if what I'm doing is all right.

[kinslayer]

I don't see why it wouldn't work. If you're not sure about converting directly to polar, just make the shift explicit in the cartesian plane first, then convert.

I didn't switch to polar coordinates so I haven't tried, but I'm sure you could also do it just as easily without translating first.

[keltingmeith]

The only issue I had with not translating first is that there is no effective limit for r if I don't, but I've got a general consensus that doing my conversion is fine.

Should see some of the methods for it, though - I've seen some really weird stuff to solve this. Best part is, that all the methods are right, thus the fifty different relations for delta and epsilon should be right. This assignment has taught me that delta-epsilon proofs are thoroughly stupid, hahah.

[simba]

I ended up separating the limit into two separate limits and it was much easier from there!! although my answer was delta=sqrt(epsilon), which is different to kinslayer's but oh well haha!

[keltingmeith]

I ended up separating the limit into two separate limits and it was much easier from there!! although my answer was delta=sqrt(epsilon), which is different to kinslayer's but oh well haha!

I have legit seen about 5 different answers, all with praise from 3rd year pure maths guys. I honestly don't think there's one right answer to this anymore, unless someone wants to say otherwise, hahah.