UoM Maths Thread

[Deleted User]

Thank you! I understand now. 
So if E[X|Y=y] = 1/y, then E[X|Y]=1/Y.

[Deleted User]

Also can someone explain to me why:
a) the sum of x*(x+1) from x=1 to x=n is equal to 1/3*n*(n+1)*(n+2)? 
and
b) the sum of 2/[(n+2)*(n+1)] from x=0 to x=n is equal to 2/(n+2)?

Are the formulae for these that I am not aware of?

[notveryasian]

From the looks of those 2 sums they look like they can be shown using telescoping series. However that is only a wild guess lol

[melbunistudent]

I'm having trouble doing lines of plane questions in linear algebra, for example the line passes through the point p(2,4,5) and perpendicular to the plane 5x -5y -10z= 2 I know that the answer will be r=p + at but I have a problem getting the a part as its perpendicular and not parallel any help would be great cheers

[lzxnl]

Also can someone explain to me why:
a) the sum of x*(x+1) from x=1 to x=n is equal to 1/3*n*(n+1)*(n+2)? 
and
b) the sum of 2/[(n+2)*(n+1)] from x=0 to x=n is equal to 2/(n+2)?

Are the formulae for these that I am not aware of?

I did this and tried to upload my work but the file was too big. Sigh.
For the first one, use sum of x from x=1 to x=n is 1/2 (n^2 + n) by arithmetic series.
Then, you just need to find the sum of x^2 from x=1 to x=n. Use x^3 - (x-1)^3 = 3x^2 - 3x + 1 and that the sum of (x^3 - (x-1)^3) from x=1 to x=n is just n^3 to find the sum of x^2 from x=1 to x=n.

Then for the next one, use partial fractions. Then, try subbing numbers in and see what happens. It's a telescoping series in that lots of the terms will cancel.

I'm having trouble doing lines of plane questions in linear algebra, for example the line passes through the point p(2,4,5) and perpendicular to the plane 5x -5y -10z= 2 I know that the answer will be r=p + at but I have a problem getting the a part as its perpendicular and not parallel any help would be great cheers

So, your line is a normal to the plane. The normal to a plane of form ax + by + cz = d is the vector (a,b,c)
Therefore for any given position vector in the plane r, r-r0 (where r0 = (x,y,z), a general point) must be perpendicular to the normal vector. AKA n.(r-r0)=0
You have n. You also know that r = (2,4,5), so put in r0 = (x,y,z) and you're done

[Deleted User]

If I know the moment generating function of a random variable X, is there a way I can find the moment generating function of X^2?
 ie. if the MGF of X is given by e^(0.5*σ^2*t^2), how can I find the MGF of X^2?

Thank you in advance.

[TrueTears]

If I know the moment generating function of a random variable X, is there a way I can find the moment generating function of X^2?
 ie. if the MGF of X is given by e^(0.5*σ^2*t^2), how can I find the MGF of X^2?

Thank you in advance.

In general (and without any further information), to find the mgf of , you must compute the distribution of and then compute the mgf as per definition.

However, your question is a very good one and requires a bit of trick of get the mgf of . Note that your given mgf of is that of a normal random variable with mean 0 and variance , let , so is chi-square with 1 degree of freedom. Now note , since , then where and . Thus the mgf for is for .

[bonappler]

I'm having trouble doing lines of plane questions in linear algebra, for example the line passes through the point p(2,4,5) and perpendicular to the plane 5x -5y -10z= 2 I know that the answer will be r=p + at but I have a problem getting the a part as its perpendicular and not parallel any help would be great cheers

I think you can choose any vector that is perpendicular to the plane, you know what the normal will be (5,-5,-10) so use trial and error to get a vector that will be perpendicular to that hence parallel to your plane. For example (0,-2,1) i.e. the dot product is 0 that means a will be (0,-2,1)

[kinslayer]

Where do I start with this?

dy/dx - 4y = e^(4x).cosh^2(3x) with y(0) = 1

don't have time to take a good look at it atm, but try using an integrating factor

http://en.wikipedia.org/wiki/Integrating_factor

[ldunn]

That's only applicable for equations in the form dy/dx + P(x)y = Q(x)
[snip]

Without going through it - I think it is in that form, your P(x) just happens to be P(x) = -4

[kinslayer]

That's only applicable for equations in the form dy/dx + P(x)y = Q(x)

All this is is dy/dx + P(y) = Q(x)

I reckon I'm just missing something really basic here lol

It is in the right form, just set P(x) = -4.

Here is how I would do this question. The way I do integrating factors is a bit different to what's in wiki/wolfram so bear with me.



Since y(0) = 1 then C_2 = 12. So the solution is

[lzxnl]

Yeah this IS an integrating factor question
You want d(my)/dx = m dy/dx + y dm/dx to be a function multiplied by dy/dx - 4y
Or f dy/dx - 4fy
Comparing both terms to m dy/dx + y dm/dx, we see that m = f
and -4f = dm/dx = df/dx
If df/dx = -4f, one possibility is f = e^-4x

So dy/dx - 4y = d(ye^-4x)/dx * e^4x
Your DE becomes d(ye^-4x)/dx * e^4x = e^4x cosh^2(3x)
d(ye^-4x)/dx = cosh^2(3x) = 1/2 (1 + cosh 6x)
You can do the rest now. Integrate both sides to find y e^-4x

Damn. Someone beat me to it.

[lzxnl]

I love it! You can use this sort of calculus to solve VCE spesh mixing questions when the volume changes
Remember how in spesh, the rate of flow of the solution out of the container was always equal to the rate of flow of pure water into the container, for instance?

[lzxnl]

They save those for the exam. At least, my physics exams had lots of application questions last year that were quite interesting to do.

[Inside Out]

hello.
just wondering how i would use the sandwich theorem to figure out the limit of n!/n^n
thanks.