Question

\[ \text{Analyze the real valued roots of the function} \\ \quad f(x) = 1 +x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...+ \frac{x^6}{6!} \]

What are the real roots - or they do not exist ?

Answer 1

I will answer questions there or here

Answer 2

Hi @sauravshakya

Answer 3

Hi @Mikael ..... what is the question

Answer 4

What are the real roots - or they do not exist ?

Answer 5

f(x)=?

Answer 6

\[ \text{Is the} \text{\{set of solutions}\quad f(x) = 0\} \,\,=\,\, \emptyset\quad \text{or not} \]

Answer 7

I have just read about Talyour series..... some what looks like it....

Answer 8

Yes....please...

Answer 9

I think
f'(0)=f''(0)=f'''(0)=f''''(0)=f'''''(0)=f''''''(0)=1
f'''''''(0)=0

Answer 10

OK those are all true statements.

Answer 11

Well actually these are in a sense relevant. Also,though , notice you have a polynomial.

Answer 12

Does it have positive roots ?

Answer 13

looks like \(e^x\) but only to sixth power

Answer 14

Does it have positive roots ?

Answer 15

What abt negative roots ?

Answer 16

can't have positive roots.

Answer 17

Well ok, 4 the sake of others explain

Answer 18

this is increasing function ...

Answer 19

Pardon me, I have to go NOW. BAck in 90 minutes

Answer 20

i don't have nice method to do this prob either.

Answer 21

i think it always positive,it's look like exp(x) taylor series,since exp(x)>0

Answer 22

not sure

Answer 23

I can tell there aren't roots in 0 to -1, -2 to -3, and -4 to -5 and beyond -6 because the even terms are stronger. this seems more like analytic method. perhaps there should be nice algebra.

Answer 24

something like these stuff http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem but when this types of things come up I usually give up thinking it's not my domain.

Answer 25

No @experimentx Usolvability in radicals are completely irrelevant here. Excuse my directness.