Recent content by CSNabeel

ahhh I see. Thanks you been really helpful

yeah I think so, they still cancel out even though 1/0 tends to infinity

I think it's [(1/(n+m)-1/(n-m)] - [(1/(n+m)-1/(n-m)] which I think cancels out?

oh well in that case cos(k*2*pi) = 1

the integral of cos(k*2*pi) is -sin(k*2*pi)/(2*pi)

yeah sorry I made a mistake cancelling that because cos 0 = 1 and forget that the denominator where different so it becomes [ cos 2*pi*(n+m) / n + m ] - [ cos 2*pi*(n-m) / n - m ] - [1 / n + m] - [ 1 / n - m] right?

so in that case does it become [ cos 2*pi*(n+m) / n + m ] - [ cos 2*pi*(n-m) / n - m ] ?

Homework Statement Calculate the gradient of the scalar field f(x,y) = x^{2} - y^{2} . Sketch the gradient for a few point on two straight lines y = x and y = -x on the plane and comment on the properties of the sketch. Homework Equations The Attempt at a Solution So I worked...

Homework Statement Show that provided that m and n are arbitrary integers, the two functions f(t) = sin nt and g(t) = cos mt are orthogonal over the interval (0,2\pi). Explain the significance of this result in Fourier series analysis. Hint: you may find the following trigonometric identity...

a) 1p + 2q = 0 (1) 2p +3q +r = 0 (2) q - r = 0 (3) 2p + 4q = 0 (4) (3) q = r (1) p = -2q put (3)and(1) into (2) 2(-2q) + 3(q) +q = -4q +3q + q = 0 p=-2 q = 1 r = 1 vectors are dependentb) 1p + 2q = 3 (1) 2p +3q +r = 5 (2) q - r = 1 (3) 2p + 4q = 6...

so with that being said which of the two do I follow from below to work out the answer? a) 1p + 2q = 0 2p +3q +r = 0 q - r = 0 2p + 4q = 0 b) 1p + 2q = 3 2p +3q +r = 5 q - r = 1 2p + 4q = 6 and if I follow b I'm I right to think that p = 1 q =2 and r = 0

Homework Statement Considering the following vectors R^{4}: v1 = (1,2,0,2) v2 = (2,3,1,4) v3 = (0,1,-1,0) Determine if these vectors are linearly independent. Let S be the linear span of the three vectors. Define a basis and the dimensions of S. Express the vector v=(3,5,1,6) as a linear...