Local Linearization of f(x) = cosx at a = $\pi$/2

Mirole · 2009-11-01T20:50:27+00:00

f(x) = cosx, a = \pi/2 since, L(x)=f'(x)(x-a) -f(a) f'(x) = -sinx = -sin(\pi/2)(x-\pi/2) - f(a) I'm stuck as to where to go next, is this even right?

Thread 'Finding the nth roots of a complex number'

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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Thread 'Solve this problem that involves induction'

Hello, This is the attachment, the steps to solution are pretty clear. I guess there is a mistake on the highlighted part that prompts this thread. Ought to be ##3^{n+1} (n+2)-6## and not ##3^n(n+2)-6##. Unless i missed something, on another note, i find the first method (induction) better than second one (method of differences).

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Thread 'Finding the nth roots of a complex number'

Thread 'Solve this problem that involves induction'

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