First Year Mathematics A (Differentiation & Linear Algebra) (4 Viewers)

[InteGrand]

Re: MATH1131 help thread

Yeah I understand that rule, thanks.

So I'm just not understanding why the sqrt on the bottom is making y have to be positive. I get that it involves the x but how does it involve the y?

A way of doing this, do we rearrange the function so that x is the subject, and from that we can easily see that y cannot be negative (and is this a good method for solving other functions like this?)?

Because of the square root, the denominator is by definition positive (it can't be 0 since we can't have 0 on a denominator; and remember, the symbol √(t) refers to the positive positive square root of t, where t is a positive real number).

Since y is equal to 1/(something positive), y itself is positive. In other words, 1/t > 0 whenever t > 0.

 

[leehuan]

Re: MATH1131 help thread

sqrt(1) = 1
sqrt(2) = sqrt(2)
sqrt(3) = sqrt(3)
sqrt(4) = 2
...

I just picked some positive integers.

Effectively, you cannot use the square root to force out a negative y value.

 

[Flop21]

Re: MATH1131 help thread

Because of the square root, the denominator is by definition positive (it can't be 0 since we can't have 0 on a denominator; and remember, the symbol √(t) refers to the positive positive square root of t, where t is a positive real number).

Since y is equal to 1/(something positive), y itself is positive. In other words, 1/t > 0 whenever t > 0.

Ohhh haha right, I understand.

Thanks all

 

[Flop21]

Re: MATH1131 help thread

Additional input.

Sorry can someone further explain this.

The part I don't understand is actually getting the points to draw, I understand enough to figure out what quadrant etc though.

So you say sketching y<=2x, but how do you do that?


E.g. I have another similar question: sketch the set of points in the (x,y) plane satisfying 0<x<3y and 0<y<2

 

[InteGrand]

Re: MATH1131 help thread

Sorry can someone further explain this.

The part I don't understand is actually getting the points to draw, I understand enough to figure out what quadrant etc though.

So you say sketching y<=2x, but how do you do that?


E.g. I have another similar question: sketch the set of points in the (x,y) plane satisfying 0<x<3y and 0<y<2

To sketch y ≤ 2x, first sketch y = 2x. This is the boundary of the region. Then the places where y is less than (or equal to) 2x will be the places below (or on) this line. If the restriction is 0 ≤ x ≤ 2, then we only take the part where x is between 0 and 2.

To sketch the set of points in the (x,y) plane satisfying 0<x<3y and 0<y<2, first we note 0 < y < 2. Therefore, the region must be in the horizontal band strictly between the lines y = 0 and y = 2, so you can draw these lines (in dashed form), as our region will be between these.

Then sketch 0 < x < 3y, taking only the part in the horizontal band. To sketch 0 < x < 3y, first sketch x = 3y (this is the line y = x/3). Its intersection with the dashed line y = 2 is at the point (6, 2). Then, since we want only the places where x > 0 and x < 3y, take the part to the left of this line, but to the right of the y-axis (as x > 0).

So the final region is the interior of the triangle formed by the points (0,0), (0,2), (6,2), and excluding the boundaries.

 

[Flop21]

Re: MATH1131 help thread

How do I show that f is continuous at 0... f(x) = |x|

 

[InteGrand]

Re: MATH1131 help thread

How do I show that f is continuous at 0... f(x) = |x|

Note f(0) = 0, so we need to show that f(x) -> f(0) = 0 as x -> 0. So we need to show that given eps > 0, there exists a delta > 0 such that if 0 < |x-0| < delta (i.e. 0 < |x| < delta), then |f(x) - 0| < eps.

To do this, let delta = eps, for any given eps > 0. Then suppose 0 < |x-0| < delta.

Then we have |x| < delta. Since |x| = ||x|| = ||x| - 0|, we have ||x| - 0| < delta, i.e. |f(x) - 0| < delta. But delta = eps, so |f(x) - 0| < eps, as we wanted.

Hence f is continuous at 0.

 

[Flop21]

Re: MATH1131 help thread

Note f(0) = 0, so we need to show that f(x) -> f(0) = 0 as x -> 0. So we need to show that given eps > 0, there exists a delta > 0 such that if 0 < |x-0| < delta (i.e. 0 < |x| < delta), then |f(x) - 0| < eps.

To do this, let delta = eps, for any given eps > 0. Then suppose 0 < |x-0| < delta.

Then we have |x| < delta. Since |x| = ||x|| = ||x| - 0|, we have ||x| - 0| < delta, i.e. |f(x) - 0| < delta. But delta = eps, so |f(x) - 0| < eps, as we wanted.

Hence f is continuous at 0.

Is that using the formula proof of a limit? [fml I was hoping that was just one small part of the topic and have skipped over it].

 

[InteGrand]

Re: MATH1131 help thread

Is that using the formula proof of a limit? [fml I was hoping that was just one small part of the topic and have skipped over it].

I did it (proved the required limit) via the epsilon-delta definition of a limit.

 

[Flop21]

Re: MATH1131 help thread

Note f(0) = 0, so we need to show that f(x) -> f(0) = 0 as x -> 0. So we need to show that given eps > 0, there exists a delta > 0 such that if 0 < |x-0| < delta (i.e. 0 < |x| < delta), then |f(x) - 0| < eps.

To do this, let delta = eps, for any given eps > 0. Then suppose 0 < |x-0| < delta.

Then we have |x| < delta. Since |x| = ||x|| = ||x| - 0|, we have ||x| - 0| < delta, i.e. |f(x) - 0| < delta. But delta = eps, so |f(x) - 0| < eps, as we wanted.

Hence f is continuous at 0.

So knowing the definition of a continuous function, can you just say since the its lim = 0 from both the left and right hand side, it is continuous since as x->a f(x) = f(a).

Would writing just that be acceptable as an answer instead of using the eps-delta method?

 

[InteGrand]

Re: MATH1131 help thread

So knowing the definition of a continuous function, can you just say since the its lim = 0 from both the left and right hand side, it is continuous since as x->a f(x) = f(a).

Would writing just that be acceptable as an answer instead of using the eps-delta method?

Idk, by saying that the limits are 0 from the left and right, it seems (in my opinion) too close to assuming what they're asking you to prove.

 

[leehuan]

Re: MATH1131 help thread

They defined continuity for us somewhere along the lines of this

 

[turntaker]

Re: MATH1131 help thread

(a) Show that if A is a 2 × 2 matrix, and its two rows are identical, then det(A) = 0.

(b) Prove by induction that if A is an n × n matrix which has two identical rows, then det(A) = 0.
[Hint: expand along one of the other rows].


How would one approach this question? Part a is quite easy, though I am not sure about part b.

 

[iforgotmyname]

Re: MATH1131 help thread

(a) Show that if A is a 2 × 2 matrix, and its two rows are identical, then det(A) = 0.

(b) Prove by induction that if A is an n × n matrix which has two identical rows, then det(A) = 0.
[Hint: expand along one of the other rows].


How would one approach this question? Part a is quite easy, though I am not sure about part b.

|a al
|b bl = ab-ba

 

[turntaker]

Re: MATH1131 help thread

|a al
|b bl = ab-ba

that's part a

 

[InteGrand]

Re: MATH1131 help thread

(a) Show that if A is a 2 × 2 matrix, and its two rows are identical, then det(A) = 0.

(b) Prove by induction that if A is an n × n matrix which has two identical rows, then det(A) = 0.
[Hint: expand along one of the other rows].


How would one approach this question? Part a is quite easy, though I am not sure about part b.

(Parity of an integer refers to whether it is odd or even.)

 

[Flop21]

Re: MATH1131 help thread

Okay so with parametric vector forms of lines...

I was under the impression it was x = a + lambda(v). And I though V was simply AB (given two points A and B), so to get the vector you go B-A.

But in the solutions for my practice test it has A-B. Why have they got this, and if that is correct, why?

 

[leehuan]

Re: MATH1131 help thread

Okay so with parametric vector forms of lines...

I was under the impression it was x = a + lambda(v). And I though V was simply AB (given two points A and B), so to get the vector you go B-A.

But in the solutions for my practice test it has A-B. Why have they got this, and if that is correct, why?

Doesn't matter. Let mu = -lambda and donezo

Convention is to use B-A because A-B is technically a negative case

 

[InteGrand]

Re: MATH1131 help thread

Okay so with parametric vector forms of lines...

I was under the impression it was x = a + lambda(v). And I though V was simply AB (given two points A and B), so to get the vector you go B-A.

But in the solutions for my practice test it has A-B. Why have they got this, and if that is correct, why?

 

[Flop21]

Re: MATH1131 help thread

Oh right, because you can have multiple parametric vector forms for the same line? Because I was getting the signs of my vectors opposite to the answers (since they used A-B).