leehuan's All-Levels-Of-Maths SOS thread (1 Viewer)

[leehuan]

Another question, but mainly to do with the fact that I haven't been fully taught how to apply the extreme value theorem yet

 

[InteGrand]

Another question, but mainly to do with the fact that I haven't been fully taught how to apply the extreme value theorem yet

The extreme value theorem applies to closed and bounded intervals only. The interval in your question is not closed, so the extreme value theorem doesn't apply here. You can see what's going on by sketching the graphs of those functions on the given interval.

 

[leehuan]

The extreme value theorem applies to closed and bounded intervals only. The interval in your question is not closed, so the extreme value theorem doesn't apply here. You can see what's going on by sketching the graphs of those functions on the given interval.

Fair enough. I knew I should've believed that it has to be a closed interval.

 

[seanieg89]

A remark that is useful in some situations:

You can work your way around the "closed" part of the assumption in the extreme value theorem to some extent in some situations.

Say you had a continuous function f(x) on the interval (a,b) that tends to limiting values at a and b. Then you can define the function g on [a,b] to be f(x) on the interior of the interval, and the limits of f at the the endpoints a and b.

This function is continuous on [a,b] and hence has extrema.

This of course does not guarantee the existence of extrema in (a,b), but it makes it easier to prove sometimes. Eg, suppose you knew that cos vanished at +-pi/2 and that it was positive somewhere between -pi/2 and pi/2. Then since its endpoints aren't maxima for the continuous extension, there must exist an interior maximum.

Another trivial application of this trick is to show that a function f which is continuous on (a,b) and tends to limits at the endpoints is bounded.

 

[leehuan]

Disappointing. What would be the inverse to:

f: R->R, f(x) = x^3
then, if g(x) = x^(1/3) is undefined for x<0?

Sorry guys, have to bump this one because this is now screwing my head around a bit too much

 

[InteGrand]

Sorry guys, have to bump this one because this is now screwing my head around a bit too much

The inverse is simply the function g: R -> R given by g(x) = cube root of x, where by cube root we mean the real cube root.

 

[seanieg89]

Sorry guys, have to bump this one because this is now screwing my head around a bit too much

x^{1/3} is perfectly well defined for x<0, even if you have never heard of complex numbers.

x^3 is a bijection from R to itself.

 

[leehuan]

x^{1/3} is perfectly well defined for x<0, even if you have never heard of complex numbers.

x^3 is a bijection from R to itself.

That's what I thought until Wolfram and GeoGebra messed up my brain.
http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=x^(1/3)

Slight edit: My Mac's GeoGebra... cause it seems like the web version is still fine.

 

[seanieg89]

That WA input is being interpreted as the principal cube root (take cube root of modulus, and divide principal argument by 3). This doesn't map R to R though.


Because cubing is a bijection on R, if we are only talking about reals, the natural definition of the cube root is just the inverse.
Ie x^(1/3) is the unique real number y such that y^3 =x.

 

[leehuan]

The answers for these are not uploaded, so I am requesting a marking service. (Sorry in advance as this is probably wasting people's time now.) Also some help on Q3 because I didn't quite understand it.

But to save space (and time) I'll drop out a lot of working out and just give the final results

Q1





___________________________________________
Q2

Note: The answer isn't concerned with a diagram being to scale so my triangle was arbitrarily drawn, just with points labelled

Answer: The entire region bounded by triangle ABC, where A is (1,2,-2), B is (3,-2,-2) and C is (1,-8,0)

Method: The boundary cases are when lambda1=lambda2=0, lambda1=2, lambda2=0 and lambda1=lambda2=2
___________________________________________
Q3


Answer: I don't know what a direction vector is anymore so I just converted...

___________________________________________
Q4



Row-echelon form:


Answer:
a) m=8 otherwise contradiction in row 3
b) never, all variables are leading variables provided m can't equal 8
c) m \neq 8

 

[InteGrand]

The answers for these are not uploaded, so I am requesting a marking service. (Sorry in advance as this is probably wasting people's time now.) Also some help on Q3 because I didn't quite understand it.

But to save space (and time) I'll drop out a lot of working out and just give the final results

Q1





___________________________________________
Q2

Note: The answer isn't concerned with a diagram being to scale so my triangle was arbitrarily drawn, just with points labelled

Answer: The entire region bounded by triangle ABC, where A is (1,2,-2), B is (3,-2,-2) and C is (1,-8,0)

Method: The boundary cases are when lambda1=lambda2=0, lambda1=2, lambda2=0 and lambda1=lambda2=2
___________________________________________
Q3


Answer: I don't know what a direction vector is anymore so I just converted...

___________________________________________
Q4



Row-echelon form:


Answer:
a) m=8 otherwise contradiction in row 3
b) never, all variables are leading variables provided m can't equal 8
c) m \neq 8

For Q3, a direction vector for the line is the vector that the lambda is multiplying in your parametric form (that's the direction the line points in).

 

[leehuan]

For Q3, a direction vector for the line is the vector that the lambda is multiplying in your parametric form (that's the direction the line points in).

...oh

So it's just (0,-1,5)?

 

[InteGrand]

...oh

So it's just (0,-1,5)?

Yeah.

 

[Drongoski]

Q1) - if still not done - sorry; it's been done!


1 way (using a for lambda and b for mu):

0a + 2b = x-4 . . . (1)
4a - 6b = y-2 . . . (2)
a = z+1 . . . (3)

Now, by (3): a = z+1
.: by (2); 6b = 4(z+1)- (y-2) = 4z - y + 6
.: by (1): 6b = 3(x-4) = 4z - y + 6

Solving we get: 3x + y - 4z = 18

Another way: From the vector eqn, a normal to the plane is: [6, 2, -8]' or [3, 1, -4]'. You can get this by taking the cross-product of the 2 vectors in the given eqn of the plane.

Since a point on the plane is (4, 2, -1): equation of plane is:

3(x-4) + 1(y-2) - 4(z+1) = 0

i.e. 3x + y - 4z = 18

 

[Drongoski]

leehuan

Your answer to Q1 is: 3x - y + 4z = 6

Mine is: 3x + y - 4z = 18

Obviously at least one of these answers must be wrong.

A(4, 2, -1) (lambda = mu = 0) and B(4, 6, 0) (lambda = 1, mu = 0) are 2 points on the given plane.

Both A and B satisfy our 2 answers.

Now try: C(8, -6, 0) (lambda = 1, mu = 2) and D(6, 4, 1) (lambda = 2, mu = 1)


Is your given answer from the book??

 

[leehuan]

No like I said the answers were not given. That's why I requested a marking service.

But I think I figured what happened. There was a transcription error with the first equation.



Edit: There was certainly a transcription error. For some reason I wrote x = 4 - 2lambda. Thanks Drongoski

 

[leehuan]

I still don't fully know what I'm doing. Another demo please, but on a trivial statement this time



All I can say is



I have no clue on the method

 

[InteGrand]

I still don't fully know what I'm doing. Another demo please, but on a trivial statement this time



All I can say is



I have no clue on the method

Let eps > 0. Take delta = min(2, eps/4). Let 0 < |x - 1| < delta. Then |x-1| < eps/4. Also, |x-1|<2, so -1 < x < 3, whence 0 < x+1 < 4. So |x+1| < 4.

Factorise the x^2-1 into (x+1)(x-1). So

|x^2 -1| = |(x+1)(x-1)|

= |x+1||x-1|
< 4*(eps/4)
= eps.

Thus the limit in question is 1.

 

[leehuan]

Take delta = min(2, eps/4).

Sorry, forgot what that meant again

 

[InteGrand]

Sorry, forgot what that meant again

Delta is the smaller of the two. The importance of this is that if we make something less than delta, it'll be simultaneously less than 2 and eps/4.