Search results

Could you please explain the step of by assumption? Also biopia, how did you get this step:= (1+x)[(1+x)^k - 1 - kx] + kx^2

(1+x)^n>= 1+nx, x>0

If \frac{z-1+i}{z+1-i} = ki, where k is real Find with aid of a diagram the value of |z|

How many periods do you get a week? In terms of units how much is that? Ive got 4 units every 2 wks, same number of periods as a 1 unit subject :( Also, where is everyone up to? Were only started graphing, and has only done complex :(

I do not think the focal length changes as lens changes curvature in eye as image projected in focused. Unless anyone cares to expand.

Anyone here take supplements such as Berocca performance? What do you think about these pills. Do they really work? Also, anyone take anything for concentration and increased mental functioning?

Also Derive f(x) = sin2x -2x.cos2x, and us this to find S{pi/4,0} xsin2x dx

How did you integrate the cosx+xsinx? O crud, nvm

second one: 4(1+tan^2x) = 3tanx + 5\\ \text{taking Equation to one side will give }\\ 4tan^2x-3tanx-1 = 0\\ \text{Solving Quadratic Equation }\\ (4tanx+1)(tanx-1)=0\\ tanx = \frac{1}{4}\\ \text{OR}\\ tanx=1

4sin^2 x - 4sinx = 1\\ 4sin^2 x - 4sinx -1 = 0\\ \text{using quadratic formula to solve will give answer, with subsitution of sin x = x}

Find the derivative of f(x) = cos x + x sin x. Hence evaluate the definite integral \int_{0}^{\frac{\pi}{2}} x cos x dx

Is Latex Working. Somehow its not for me... I mean particularly the images :(

Please help with this trignometric Proof: \frac{1+cosec^2A tan^2C}{1+cosec^2B tan^2 C} = \frac{1+cot^2A sin^2 C}{1 + cot^2B sin ^2 C}

Can you just post your notes Cinnamonmonster? Attach it or something since they are so good.

Are you sure this is rite? I think youve done T_{n+1} wrong It should be [tex] -4(-2)^{k+1)[tex] Also, the whole thing is in brackets. Sorry about mistake. Re-did it in edit.

yep ur absolutely rite. Now how do you do it for n? And to check, it is on the LHS, it is -4 rite?

Prove using induction -4 + 8 - 16 + ... + -4(-2)^ {n-1} = \frac{4((-2)^{n} - 1)}{3} The original document said 4(-2)^(n-1). And that works perfectly for induction. However, a = -4, and thus Tn = -4(-2)^(n-1). Please help. EDIT: Fatal Error: 4((-2)^(n) - 1)/3 = RHS

Indeed alive. Hello ladies :sun:

Yeah... but some one told me complex is a tool for later chapters. But i do know your absoulutely rite and thxs for warning me.

Also just generally, when conducting complex trig, where root is in the form of \frac{k\pi}{4},\\ \text{k=0,1,2,.... } and the equation is cubed is it enough to say, k=0,1,2 since only first 3 root are distinct?