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The integration generates another pi.. so 1 is definitely missing...

There's a fundamental flaw to this question. Remember you don't have a definition for e^z where z is complex. Which means that you will either need an explicit defintion for e^z to define the derivative or you can define exp(z) to be a function such that d/dz exp(z) = exp(z) The first...

same problem.. and you don't even have a definition for e^z where z is complex

Computer science course at a reputable univeristy picking hardcore subjects + have your own projects. Doubling up with mathematics is probably useful too given that modern games usually requires some maths to work out the graphics and physics

Banks pay higher interest on their online accounts.. not much reason for a private individual to buy government bonds prices(yields) are published.

Your programming will get better once you stop worrying about programming languages

it is 225pi^2/4

1. substitute 1 into the polynomial and it's derivative, you get a + b + 1 = 0, (n+1)a + nb = 0, etc. 2.i. 0 obviously doesn't work.. and if you test 1 or -1, the given condition will rule them out. ii.substitute 1/a into the polynomial then multiply through by a^4 iii. Will leave you this...

It should be 255(pi^2)/4

You don't need to solve for Q in this question.

Another problem with many degree programs is that they spend alot of time justifying their existence and professing their importance.

1.) almost no maths compared to engineering or maths or physics. 2.) Remuneration.. well its not bad.. not very high either.. normal job.

there are 6 possibilities q = ip q = -ip q = (1+i)p q = (1-i)p q = p/(1+i) q = p/(1-i)

You don't learn that much in finance on top of accounting.

you can reduce the amount of calculations by a substitution u = x-3 V = 2\pi\int_1^5 \left( (4-(x-3)^2) + \sqrt{4 - (x-3)^2} \right) x \, dx\\ =2\pi\int_{-2}^2 \left( (4-u^2) + \sqrt{4 - u^2} \right) (u+3) \, du \\ = 2\pi\left( \int_{-2}^2 u\left( (4-u^2) + \sqrt{4 - u^2} \right) \, du +...

If I remember correctly, the matrices which satisfy A^2 = A^-1 are precisely the ones which are diagonalizable with eigen values which are cube roots of unity

you can transform it to \sqrt{a}\left(\frac{a}{b}\right)^{n+1} \int_0^1 x^n (1-x)^{1/2}\, dx Then you look up the beta function on wikipedia, then the gamma function comes out as something ugly like \sqrt{a}\left(\frac{a}{b}\right)^{n+1} \frac{3n!(n+1)!2^{2n+1}}{(2n+3)!}

This does not work unless you know more about Q. You are essentially assuming that Q = f(a,b,c) = f(a,b,1-a-b) = g(a+b) for some function g. In many cases this is false so you are forced to test Q for (a,b) ranging over a grid of points in a triangle in [0,1]x[0,1] (a+b<1) So again you are...

You need to know something about Q to answer that.. not the formula, but atleast some properties of Q.. is it smooth for example... If you know nothing about Q then you can't deduce anything... even in the 2 variable case, because you could try 0.1,0.9;0.2;0.8 ... but Q can have a spike at...

coop vs BCom/BEng.. hard pick.... Coop > cadet BCom/BEng > cadet