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\[1 + x + x^2 /2! + x^3 /3! ..........x^6/6! = 0\]

Obviously real roots - if exist at all - must be negative numbers.

Agreed.

Noe we have a war on our screens - the great war between the Good Evens (pos) and the bad Odds (neg)

clearly \[ \Sigma |Evens| < \Sigma |Odds|\] if x > 6 and we can improve this estim.

Okay. Same thing as you said.

I rose faster than your treacherous spoiler "help"

DO NOT "help" as long as not requested

Sorry. No help from now on.

addictive poisonous byproduct?
I beg to differ.
Okay, for now, shutted up. Continue.

Okay.
Back to the problem if you may.

What you have is the head of Exonential series

Now these series are a generating function

I beat you purple King as your other messenger

If it's odd, by definition it must have at least one real root.

Look @KingGeorge please let me do my try ? can you

@KingGeorge By definition?
@Mikael Let everybody try.

Okay. Yes. Agreed.

Soo, SOLVED !

Here it is

Your function diffrentiated is itself - last member

f - f' = x^6/6!

It is some sort of generating unction type argument with exp(x) and its derivatives

If f(x_1) = 0 ==> f'(x_1) = = x^6/6!

meant - x^6/6!

So at the root the derivative (slope) must be negative

After all - at minus-infty it tends to +infty

So the leftmost root - the crosiing MUST be from negative to positive.

I am still claiming it does go +infty at x--> -infty and Wolfram approves it

So it seems NO roots becomes quite closer to reality - dont you say ?

Yes and so NO REAL roots is approved By Mister Stephan son of Hugo and Sybil

@siddhantsharan No real roots to that one

Mister @KingGeorge - some encouragement here ?

I would give you acknowledgement if I had the time to check, but I don't right now
sorry

it is a nice solution though :)