Show that the series $\sum\limits_{k=0}^\infty \frac{(-1)^k(x^{2k+1})}{(2k + 1)!}$ is greater than zero for $0 For a function to show something was greater than zero over an interval I would differentiate, show there was a maximum (or it is an increasing function) and then show that the two end points were also greater than zero. Since this function is the same as the sine function, I imagine there is an analogous way to do this for a series but can't think how. Can someone explain to me the method I should use, is differentiating again the right option? (Also I wish to do this particularly using series, and do not wish to show by the method I described above using the sine function). Thanks
Saturday, 9 April 2016
taylor expansion - How to prove a series is greater than zero over an interval?
Subscribe to:
Post Comments (Atom)
real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$
How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
-
How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
-
$$ 3x+6y+5z=7 $$ The general solution to this linear Diophantine equation is as described here (Page 7-8) is: $$ x = 5k+2l+14 $$ $$ y = -l $...
-
I need help to compute the following integral: $$\int_{-\infty}^{\infty}\frac{z^4}{1+z^8}dz$$ I need to use Cauchy's residue theorem. I ...
No comments:
Post a Comment