can sum1 help explain/show why the sum from i= 0 to 4 of x^i/i! > 0 for all real numbers x?

(x^0 / 0!) + (x^1 / 1!) + (x^2 / 2!) + (x^3 / 3!) + (x^4 / 4!)
(1) + (x) + (x^2 / 2) + (x^3 / 6) + (x^4 / 24)

i think it's pretty obvious for x >= 0 since every term is non-negative.

just look at a graph to show it. you can see the highest power is x^4, so it has a somewhat parabola feel to it. you can take the derivative and set it equal to zero (and also the second derivative) to find the min/max and any inflection points. you'll find that it's always positive

https://mathway.com/graph

This post was edited by carteblanche on Oct 14 2015 09:31pm

i get the derivative to be 1 + x + x^2/2 + x^3/6

setting it equal to 0 how do i solve it without using any technology?

I'm confused on why this is a question. It is clear by definition that for I=1:4 that x^i is positive and clearly i! Must be positive based on conditions of I. Thus it's clearly positive.



Edit: You are simply wasting your time taking deriv and trying to solve for x when it is clear by definition the every part of the problem is positive.

This post was edited by Xx Shin3d0wn xX on Oct 19 2015 04:55pm

x^i isnt always positive. ex: i is odd

that's a cubic, which has a general solution.


There might be a more algebraic way, but i dont care enough to look lol. perhaps wiki will you give an idea. https://en.wikipedia.org/wiki/Cubic_function


it's negative for x < 0 and odd i

This post was edited by carteblanche on Oct 19 2015 06:32pm

i read the problem wrong I thought it said for a positive integer X.

Here's a little trick you can try using when tackling these types of problems (complete the square)

So you start off with
1 + x + x^2/2! + x^3/3! + x^4/4!

First off multiply the entire equation by 4! to get

24 + 24x + 12x^2 + 4x^3 + x^4

rewrite as
(x^4 + 4x^3 + 6x^2 + 4x + 1) + 6x^2 + 20x + 23
= (x+1)^4 + 6x^2 + 20x + 23
add and subtract 400/24
= (x+1)^4 + (6x^2 + 20x + 400/24) - 400/24 + 23
= (x+1)^4 + (√(6)x + 20/(2√6))^2 + 19/3

the first 2 terms are ≥0 since they are both raised to an even power. Clearly 19/3 is positive, when you add them up to get something >0

thx



thx. never liked completing hte squre, but i guess it can be useful

you can always use newton-raphson to converge on the solution as well in a few iterations, which would be quicker than completing the squares