First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (1 Viewer)

[seanieg89]

Re: MATH1231/1241/1251 SOS Thread

 

[leehuan]

Re: MATH1231/1241/1251 SOS Thread

Yeah, true. I think I just forgot how to matrix multiply because I got scared by the fact I had vectors

 

[leehuan]

Re: MATH1231/1241/1251 SOS Thread

Hang on, now I know where the confusion was.


I figured that to use the fact they are orthonormal we'd have to introduce the dot product. Which makes sense as then the entries do correspond to the identity.

But when performing the matrix multiplication, is there a reason why we tend to the dot product and not something else?

 

[InteGrand]

Re: MATH1231/1241/1251 SOS Thread

Hang on, now I know where the confusion was.


I figured that to use the fact they are orthonormal we'd have to introduce the dot product. Which makes sense as then the entries do correspond to the identity.

But when performing the matrix multiplication, is there a reason why we tend to the dot product and not something else?

(Note that row i of U-transpose is column i of U, due to the transposing just basically turning rows into columns.)

 

[seanieg89]

Re: MATH1231/1241/1251 SOS Thread

Here is a fun question to think about for those who like algebra. (Not an ideal thread to post this in, but first years doing some linear algebra have learned enough to approach this question...it came up in a discussion in a tute I was teaching at ANU. The question is open to all though.)

I will assume you know what fields are.

A ring is like a field, but we drop the assumptions of multiplication being commutative, and arbitrary elements need not have multiplicative inverses. A commutative ring is a ring with commutative multiplication.

See https://en.wikipedia.org/wiki/Ring_(mathematics)#Definition to remind yourself of axioms if you like.

Some examples of rings include the set of all integers, the set of all polynomials with real coefficients and the set of all (nxn)-matrices with real coefficients. The integers/polynomials are commutative rings, but the matrices are not.

Now by definition arbitrary elements of a ring R need not be invertible, but some can be. We call such invertible elements units of the ring. Eg/ the multiplicative identity 1 of a ring is trivially a unit as it is its own inverse.

1. Prove that if a,b are units of the ring R, then ab is also a unit of R.

2. Suppose a,b are elements of a commutative ring R such that ab is a unit. Is it true that a and b must be units?

3. Repeat 2. for the ring of nxn matrices with real coefficients.

4. Repeat 2. for arbitrary rings.

 

[Flop21]

Re: MATH1231/1241/1251 SOS Thread

What are they doing here:

 

[InteGrand]

Re: MATH1231/1241/1251 SOS Thread

What are they doing here:

They got the kernel of that matrix. The kernel can be found by inspection (as they did), but if you aren't comfortable with that, note that further row-reduction makes the matrix into:

[2 5]
[0 0].

So set x_2 = 2*alpha say (since second column is non-leading; you could just use alpha but then you end up with fractions for x_1, which is messier, and 2*alpha is just as arbitrary as alpha anyway), then the first row gives us 2x_1 = -5*2 alpha, so x_1 = -5*alpha.

So the kernel is vectors of the form (x_1, x_2) = alpha*(-5, 2). So the kernel is span{(-5,2)}.

 

[Flop21]

Re: MATH1231/1241/1251 SOS Thread

They got the kernel of that matrix. The kernel can be found by inspection (as they did), but if you aren't comfortable with that, note that further row-reduction makes the matrix into:

[2 5]
[0 0].

So set x_2 = 2*alpha say (since second column is non-leading; you could just use alpha but then you end up with fractions for x_1, which is messier, and 2*alpha is just as arbitrary as alpha anyway), then the first row gives us 2x_1 = -5*2 alpha, so x_1 = -5*alpha.

So the kernel is vectors of the form (x_1, x_2) = alpha*(-5, 2). So the kernel is span{(-5,2)}.

Ohhh right, okay thanks! The part after row reducing was what was confusing me / I didn't know.

 

[leehuan]

MATH1231/1241/1251 SOS Thread

I've gone dumb again.


Q19

Edit: supposedly the answer is 8/15

 

[InteGrand]

Re: MATH1231/1241/1251 SOS Thread

I've gone dumb again.


Q19View attachment 33525

Edit: supposedly the answer is 8/15

The question's asking for a distribution, the answer isn't a number.

Hint: hypergeometric distribution (recall this models sampling without replacement. Check Wiki for more details.).

 

[Flop21]

Re: MATH1231/1241/1251 SOS Thread

When doing questions involving Taylor's polynomials, they answers do not expand and simplify fully. Is this done simply because, or is there an actual reason and should I NOT be expanding and simplifying my answers?

e.g. 3 + 3(x-1)

 

[InteGrand]

Re: MATH1231/1241/1251 SOS Thread

When doing questions involving Taylor's polynomials, they answers do not expand and simplify fully. Is this done simply because, or is there an actual reason and should I NOT be expanding and simplifying my answers?

e.g. 3 + 3(x-1)

 

[Flop21]

Re: MATH1231/1241/1251 SOS Thread

Why is the answer to this 'describe the limiting behaviour of the sequence' question: (n*sin([n*Pi]/4))/sqr(n^2 + 1)

equal to "boundedly divergent"


All I can see is that the top becomes very big negative number when going to infinity, and the bottom becomes big positive number.

 

[seanieg89]

Re: MATH1231/1241/1251 SOS Thread

Why should the top be negative in the large n limit? The trig factor oscillates and changes sign.

This fraction is boundedly divergent because the size of the numerator and denominator are both ~n where "~" formally means "bounded between constant positive multiples of". This means that the overall expression is bounded between constant multiples of the oscillatory trig term and is hence boundedly divergent.

 

[Flop21]

Re: MATH1231/1241/1251 SOS Thread

"By using an appropriate test, determine whether each of the following series converges or diverges:"

"SUM sign, with k = 1 .. infinity, k/(k+1)"

What on earth does this answer mean... "No, since as k->infinity, ak = k/k+1 -> 1 != 0".

What's it saying "no" to? Doesn't that mean it converges?

 

[InteGrand]

Re: MATH1231/1241/1251 SOS Thread

"By using an appropriate test, determine whether each of the following series converges or diverges:"

"SUM sign, with k = 1 .. infinity, k/(k+1)"

What on earth does this answer mean... "No, since as k->infinity, ak = k/k+1 -> 1 != 0".

What's it saying "no" to? Doesn't that mean it converges?

No as in, no it doesn't converge.

Remember, for a series to converge, it is necessary that the k-th term tends to 0 as k -> oo. In this case, we have a_k -> 1 (not 0) as k -> oo, so the series can't converge (i.e. it's divergent).

 

[Flop21]

Re: MATH1231/1241/1251 SOS Thread

No as in, no it doesn't converge.

Remember, for a series to converge, it is necessary that the k-th term tends to 0 as k -> oo. In this case, we have a_k -> 1 (not 0) as k -> oo, so the series can't converge (i.e. it's divergent).

Oh I thought converge meant it tends to any real number.

So converge = 0, diverge = oo


thx

 

[InteGrand]

Re: MATH1231/1241/1251 SOS Thread

Oh I thought converge meant it tends to any real number.

So converge = 0, diverge = oo


thx

No, converge means the partial sums tend to a real number (can be any real number, yes). What I was talking about was the terms of the series going to 0. If a series is convergent, then its k-th term must tend to 0 as k -> oo (so if the k-th term of the series doesn't tend to 0, then the series is necessarily divergent).

Also, divergent doesn't mean the series is infinite, it just means not convergent (i.e. the partial sums don't tend to a real number). In general, a series being divergent need not mean the partial sums go to infinity: we could have boundedly divergent partial sums too (e.g. consider the series 1 – 1 + 1 – 1 + ... . Here, the partial sums are 1,0,1,0,1,0,... , so the partial sums don't converge to any real number, so the series is divergent (but doesn't blow up to infinity since the partial sums just oscillate between 0 and 1.)).

If a series has only non-negative terms though, then divergence is equivalent to the partial sums going to infinity.

Also keep in mind that the k-th term of a series going to 0 is not sufficient for convergence (it's just necessary). E.g. consider the sum of 1/k from k = 1 to oo. Here the k-th term (which is 1/k) tends to 0 as k -> oo, but the series still diverges.

(In your original question, the a_k = k/(k+1) referred to the k-th term of the series, not the k-th partial sum.)

 

[Paradoxica]

Re: MATH1231/1241/1251 SOS Thread

No as in, no it doesn't converge.

Remember, for a series to converge, it is necessary that the k-th term tends to 0 as k -> oo. In this case, we have a_k -> 1 (not 0) as k -> oo, so the series can't converge (i.e. it's divergent).

don't mind me, just highlighting significant terms....

 

[iforgotmyname]

Re: MATH1231/1241/1251 SOS Thread

Hi everyone, someone please help me out with this question. How did they turn the maclaurin series for 1/(x+1) into series 1/(x^2+1) just like that...