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Haha - Yeh I remember that. In first year maths they talk about how you can rearrange such sums to make them equal to anything you want to. The problem here is the infinite sum does not uniformly converge! You may want to look up the Bolzano-Wiestrass Theorem (spelling may be incorrect)

Hey there. The question is easy - all you need to do is work out the radius of the circle and since the tangent to the circle is perpendicular to the circle, you can use the perpendicular distance formula from the point (0,0) which gives you 9/sqrt(10) so the equation of the circle must be...

Let V(a,a^3). The gradient at V is 3a^2 so using the point gradient form of a line, the equation of the tangent is y-a^3=3a^2(x-a) which simplifies to y=3a^2x-2a^3 Now if the the tangent intersects the curve y=x^3, the equation becomes , upon simplification, x^3-(3a^2)x+2a^3=0 . Now...

Try doin this instead I agree with the previous posts about treating the quartic (p(x) of degree 4) as quadratic. But instead of using of solving the roots using the quadratic formulae you are better off splitting the middle term or the cross method depending on what you feel comfortable with...

Hey...try doin this Hello, I think this may help. Firstly, I think this question is a bit unfair considering polynomials, e.g. p(x) = ax^n+a2x^n-1+...c , are in the 3/4 U maths courses. Nevertheless I will offer a variety of solutions that may assist you and that you can understand. Now...

correction for response If you're gonna prove it using algebra try this. Firstly, 2*Arg(z+1) = [Arg{(z+1)^2} = Arg (z^2 +2z + 1) Taking out z = Arg {z (z +2 +1/z) i) Now with 1/z multiply both the numerator and the denominator by the conjugate of z...