Math talent help me
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Question

Math talent help me
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phi · Answer

do you know about the taylor series ?

phi · Answer

use the first 2 terms of the Taylor series, with a= -8 http://en.wikipedia.org/wiki/Taylor_series#Definition

anonymous · Answer

Wow i have no idea about taylor series!
i dont think I learn this part?!

anonymous · Answer

@stamp are you familiar with this?

phi · Answer

Here is a plot that shows the cube root vs the linearized function (centered at x=-8)

anonymous · Answer

Thank you, but how do you put x=-8 into the equation given ?

phi · Answer

$$f(x)\approx f(a) + \frac{ f'(a) }{ 1! }(x-a)$$

where f'(a) is the first derivative of the function, evaluated at a= -8 in this case
$$ f(-8)= \sqrt[3]{-8} = -2 $$
$$\frac{ df}{dx} = \frac{1}{3} x^{-\frac{2}{3} }$$
evaluated at -8  is 1/12
we get
$$f(x)\approx -2 + \frac{1}{ 12 }(x- -8) = \frac{x}{12}-\frac{4}{3}$$

anonymous · Answer

Thats cool! 
how do you caculate exact number?

phi · Answer

a calculator or use wolfram. example: http://www.wolframalpha.com/input/?i=real+cube+root+x++where+x%3D+-7

anonymous · Answer

I used wolfarm but it didnt give me an answer but linear graph