anonymous
  • anonymous
Find the Real Roots of: Equation in comment.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[1 + x + x^2 /2! + x^3 /3! ..........x^6/6! = 0\]
anonymous
  • anonymous
Obviously real roots - if exist at all - must be negative numbers.
anonymous
  • anonymous
Agreed.

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anonymous
  • anonymous
Noe we have a war on our screens - the great war between the Good Evens (pos) and the bad Odds (neg)
anonymous
  • anonymous
clearly \[ \Sigma |Evens| < \Sigma |Odds|\] if x > 6 and we can improve this estim.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Okay. Same thing as you said.
anonymous
  • anonymous
I rose faster than your treacherous spoiler "help"
anonymous
  • anonymous
DO NOT "help" as long as not requested
anonymous
  • anonymous
Hmm. Although this estimate will help us calculate an approximate value, how will we actually find the roots?
anonymous
  • anonymous
Sorry. No help from now on.
anonymous
  • anonymous
addictive poisonous byproduct? I beg to differ. Okay, for now, shutted up. Continue.
anonymous
  • anonymous
Okay. Back to the problem if you may.
anonymous
  • anonymous
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...
anonymous
  • anonymous
What you have is the head of Exonential series
KingGeorge
  • KingGeorge
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.
anonymous
  • anonymous
Now these series are a generating function
anonymous
  • anonymous
I beat you purple King as your other messenger
anonymous
  • anonymous
Haha. Good guess. My first thoughts too. However I think this is only till the degree is even. @KingGeorge
KingGeorge
  • KingGeorge
If it's odd, by definition it must have at least one real root.
anonymous
  • anonymous
Look @KingGeorge please let me do my try ? can you
anonymous
  • anonymous
@KingGeorge By definition? @Mikael Let everybody try.
KingGeorge
  • KingGeorge
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.
anonymous
  • anonymous
Okay. Yes. Agreed.
anonymous
  • anonymous
Soo, SOLVED !
anonymous
  • anonymous
Here it is
anonymous
  • anonymous
Your function diffrentiated is itself - last member
anonymous
  • anonymous
f - f' = x^6/6!
anonymous
  • anonymous
It is some sort of generating unction type argument with exp(x) and its derivatives
anonymous
  • anonymous
If f(x_1) = 0 ==> f'(x_1) = = x^6/6!
anonymous
  • anonymous
meant - x^6/6!
anonymous
  • anonymous
So at the root the derivative (slope) must be negative
anonymous
  • anonymous
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. ?...
anonymous
  • anonymous
After all - at minus-infty it tends to +infty
anonymous
  • anonymous
So the leftmost root - the crosiing MUST be from negative to positive.
anonymous
  • anonymous
I am still claiming it does go +infty at x--> -infty and Wolfram approves it
anonymous
  • anonymous
So it seems NO roots becomes quite closer to reality - dont you say ?
anonymous
  • anonymous
Yes and so NO REAL roots is approved By Mister Stephan son of Hugo and Sybil
anonymous
  • anonymous
@siddhantsharan No real roots to that one
anonymous
  • anonymous
Mister @KingGeorge - some encouragement here ?
anonymous
  • anonymous
@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
TuringTest
  • TuringTest
I would give you acknowledgement if I had the time to check, but I don't right now sorry
anonymous
  • anonymous
So \[ f(x) - f'(x) = \frac{x^6}{6!} \text{which means that at the root } f'(x) = - \frac{x^6}{720} \text{which IS NEGATIVE} \]
anonymous
  • anonymous
But it is clear that the coefficient of x^6 which is the largest power is positive ==> therefore \[\lim_{x \to -\infty} f(x) = + \infty\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
@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.
anonymous
  • anonymous
Thank You @demitris I hope @TuringTest will also see it was ingenious enough for his appreciation of the effort and the breakthrough
anonymous
  • anonymous
Tahnks @TuringTest . Where are the people who asked the question in the first place ?... vanished JUST IN TIME don't you say ....
TuringTest
  • TuringTest
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.
TuringTest
  • TuringTest
it is a nice solution though :)
anonymous
  • anonymous
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 :)
anonymous
  • anonymous
@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?
anonymous
  • anonymous
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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