maths 1B last minute questions (1 Viewer)

[leehuan]

Reason is due to which part of the graph you miss out on by using the continuous approximation. If X* is the continuous approximtion r.v. to X, then clearly, to approximate P(X > a), we need to use the continuity correction P(X* > a – 0.5), whereas for P(X < a), it'd be P(X* < a + 0.5). Don't need to guess.

Oh no I didn't mean that, choosing when to +0.5 as opposed to -0.5 is all good

Some of the questions he showed me just did +0 and that's what agitate me

 

[InteGrand]

Oh no I didn't mean that, choosing when to +0.5 as opposed to -0.5 is all good

Some of the questions he showed me just did +0 and that's what agitate me

Ah right. Were they for extremely large values, as Drsoccerball mentioned? In these cases, the continuity correction would make little difference.

 

[leehuan]

Ah right. Were they for extremely large values, as Drsoccerball mentioned? In these cases, the continuity correction would make little difference.

I had a look at both photos.

In both cases n = 100

(i.e. they did -0.5 in one of them and -0 in the other)

 

[Drsoccerball]

Yes that was my question so what value is considered "very large" that I don't have to at epsilon anymore? What If I say n =2 is very large and thus don't have to add an epsilon?

 

[InteGrand]

Yes that was my question so what value is considered "very large" that I don't have to at epsilon anymore? What If I say n =2 is very large and thus don't have to add an epsilon?

n = 2 is generally considered very small.

 

[Drsoccerball]

BTW:
Would the matrix be

0 1
-2 0
0 -1 theres no answers so am really not sure.

 

[Drsoccerball]

Or

0 1 0
-2 0 2
0 -1 0 ?

 

[InteGrand]

BTW:
Would the matrix be

0 1
-2 0
0 -1 theres no answers so am really not sure.

Note that P_2 has dimension 3 and P_3 has dimension 4, so the matrix should be a 4x3 one.

 

[Drsoccerball]

Note that P_2 has dimension 3 and P_3 has dimension 4, so the matrix should be a 4x3 one.

But didn't we say the Im(T) = 2 and the nullity = 1 ?

 

[InteGrand]

But didn't we say the Im(T) = 2 and the nullity = 1 ?

The dimensions of the matrix are based on the dimensions of the domain and codomain, not the image and kernel.

 

[leehuan]

But didn't we say the Im(T) = 2 and the nullity = 1 ?

The dimension of the vector space you map from (P₂) is 3.

So if rank(T) = 2 and nullity(T) = 1
2+1 = 3. Satisfying the rank-nullity theorem.

 

[Drsoccerball]

But arn't we putting in a vector from P_2 and getting out a vector from P_3 ? If we have a 4x3 matrix then that says we are putting in a vector from P_3?

 

[leehuan]

 

[InteGrand]

But arn't we putting in a vector from P_2 and getting out a vector from P_3 ? If we have a 4x3 matrix then that says we are putting in a vector from P_3?

No, remember P_n has dimension n+1. So P_2 (domain) has dimension 3. So the matrix will have three columns.

In general, if the linear map T has a domain with dimension d and codomain with dimension c (c,d finite) then a matrix for T will have dimensions c-by-d.

 

[Drsoccerball]

So do I T(<1,0,0,0>) , T(<0,1,0,0>), T(<0,0,1,0>) as the columns ?

 

[leehuan]

So do I T(<1,0,0,0>) , T(<0,1,0,0>), T(<0,0,1,0>) as the columns ?

How do you expect to multiply a 4x3 matrix with a 4x1 vector

You need to multiply a 4x3 matrix with a 3x1 vector

 

[Drsoccerball]

How do you expect to multiply a 4x3 matrix with a 4x1 vector

You need to multiply a 4x3 matrix with a 3x1 vector

Im just thinking the last row is going to be all zeros...

So confused right now

 

[InteGrand]

Im just thinking the last row is going to be all zeros...

So confused right now

Yeah it will be, reason being we can never achieve an actual cubic (degree 3 poly) in the range of T.

 

[Drsoccerball]

Yeah it will be, reason being we can never achieve an actual cubic (degree 3 poly) in the range of T.

Okay that makes sense now thank you

 

[Drsoccerball]

Thanks Integrand for saving my maths 1B :') and thanks everyone else who helped the exams tomorrow I hope I can get good for all the effort you guys put in for me I may have a few more questions and then thats it. (May also make a discrete maths forum but I went to most of my lectures so there won't be as many questions).