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Find the Real Roots of: Equation in comment.

Physics
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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.

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Other answers:

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.
Haha. Yeah. Note that x^6 rises much faster than x^5 and so on the rest of the Good Evens And bad odds. So for some x no roots.
Okay. Same thing as you said.
I rose faster than your treacherous spoiler "help"
DO NOT "help" as long as not requested
Hmm. Although this estimate will help us calculate an approximate value, how will we actually find the roots?
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.
well now I see two ways (thinking aloud ) 1 A fixed point transformation which is abit elusive 2 Some ad hoc trickery - I got it I thik...
What you have is the head of Exonential series
Just thinking about it, but this is the first part of the Taylor expansion of \(e^x\) centered at \(x=0\), and it has an even power, so I'm going to guess it has no real roots.
Now these series are a generating function
I beat you purple King as your other messenger
Haha. Good guess. My first thoughts too. However I think this is only till the degree is even. @KingGeorge
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.
Perhaps not "by definition," but by "fundamental theorem of algebra." Any complex roots have to occur in pairs, and if you have an odd power, then no matter how you pair them up, you're always left with at least one not paired up, so it must be real.
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
But this is either my mistake or Impossibele- since we know that after the largest in absol-value negative root the function Must become Poistive fo all larger-abs-value negative x-s. ?...
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 ?
@TuringTest @Algebraic! Please acknowlege I proved this not easy problem. Please - the guy who gave it turned out to be a fox - he logged out at the second of solution. And so did other moderator
I would give you acknowledgement if I had the time to check, but I don't right now sorry
So \[ f(x) - f'(x) = \frac{x^6}{6!} \text{which means that at the root } f'(x) = - \frac{x^6}{720} \text{which IS NEGATIVE} \]
But it is clear that the coefficient of x^6 which is the largest power is positive ==> therefore \[\lim_{x \to -\infty} f(x) = + \infty\]
So the last crossing of the x-axis MUST be from negative y-s to positive y-s ==> UPWARDS so that means that f'(x) at the last crossing should be Positive - while the above says it is negative. Contradiction to existence of such crossing. No real roots exist. Q.E.D.
@TuringTest this is something of injustice. Please read the 20 lines I have rewritten clearly now. Is their some kind of problem acknowledging a solution of hard riddle ? The guy CALLED me to solve it after all.
Thank You @demitris I hope @TuringTest will also see it was ingenious enough for his appreciation of the effort and the breakthrough
Tahnks @TuringTest . Where are the people who asked the question in the first place ?... vanished JUST IN TIME don't you say ....
Okay I understand your solution, but I really would not clamor for appreciation like that People just drop out for various reasons frequently. I never have nor have most of the high-ranking people here. It just doesn't seem becoming.
it is a nice solution though :)
Sorry @Mikael It was just getting late so I shut down the pc. Doesnt mean I wont come back. I do appreciate the solution, nice reasoning. I didnt "vanish" knowing that you would have a solution. Sorry for the misunderstanding. Well done and cheers :)
@Mikael We could say that directly too. If f(x) has real roots, It must have at least 2 real roots ( Complex exist in pairs.) Since it's positive initially For two real roots, f'(x) must be zero when f(x) is negative. Clearly from f(x) - f'(x) > 0 that is not possible. Is this reasoning correct?
Yes that's almost so. But once again you Must use the eqn f-f'=x^6 because technically there is a possibilty od Double real root. then f=f'=0 at this root and the impossibility stems from the x^6 contradiction. All correct - \[ \bf \color{brown}{BUT \,\,LONGER\,\,proof\, ,and\,\,HEAVY \,\, WEAPONS !} \] A) It uses the Fundamental Theorem of Algebra which is UNNECESSARY in my proof, AND usually unproovable, and often even unheard-of to high school / early college which is overwhelmingly the people on this site. B) Well it is longer. Of course all that is not mathematics but simply style and purpose.

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