given:
f(x) =6/x a=-4

f(x) = [sum](....)
so my first step was to try to find the n^(th) derivative of f(-4)
so f'(x) = 6/x f''(x)= -6/x^(2) f'''(x) = 12/x^(3) f^(4)(x) = -36/x^(4) and so on
....f'(-4) = -6/4^(1) f''(-4)= -6/4^(2) f'''(-4) = -12/4^(3) f^(4)(-4)= -36/4^(4) and so on

so i think f^(n)(-4) = (-1)/4^(n) * something
I can't figure out how to fix the numerator for f^(n)(-4)

I would then take [sum] (f^(n)(-4)/n!) *(x+4)^(n) to find the radius of convergence

Question

given:  
f(x) =6/x    a=-4 

f(x) = [sum](....)  
so my first step was to try to find the n^(th) derivative of f(-4)  
so f'(x) = 6/x  f''(x)= -6/x^(2) f'''(x) = 12/x^(3)  f^(4)(x) = -36/x^(4)  and so on  
....f'(-4) = -6/4^(1) f''(-4)= -6/4^(2) f'''(-4) = -12/4^(3) f^(4)(-4)= -36/4^(4) and so on  

so i think f^(n)(-4) = (-1)/4^(n) * something  
I can't figure out how to fix the numerator for f^(n)(-4) 

I would then take [sum] (f^(n)(-4)/n!) *(x+4)^(n) to find the radius of convergence

anonymous · Answer

f'(x) is not 6/x ?

anonymous · Answer

$$f(x) = \frac{ 6 }{ x }$$

anonymous · Answer

Yeah, okay.
$$\Large f'(x) = -\frac{6}{x^2}$$

anonymous · Answer

@Supreme_Kurt do you know what a power series is? this is what the question is asking about

anonymous · Answer

I have a vague idea.
You also asked what the numerator of $\Large f^{(n)}(x)$ should be.
$$\Large (-1)^n6\ n!$$ seems to work...?