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siddhantsharan Group Title

Find the Real Roots of: Equation in comment.

  • 2 years ago
  • 2 years ago

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

    • 2 years ago
  2. Mikael Group Title
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    Obviously real roots - if exist at all - must be negative numbers.

    • 2 years ago
  3. siddhantsharan Group Title
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    Agreed.

    • 2 years ago
  4. Mikael Group Title
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    Noe we have a war on our screens - the great war between the Good Evens (pos) and the bad Odds (neg)

    • 2 years ago
  5. Mikael Group Title
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    clearly \[ \Sigma |Evens| < \Sigma |Odds|\] if x > 6 and we can improve this estim.

    • 2 years ago
  6. siddhantsharan Group Title
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    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.

    • 2 years ago
  7. siddhantsharan Group Title
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    Okay. Same thing as you said.

    • 2 years ago
  8. Mikael Group Title
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    I rose faster than your treacherous spoiler "help"

    • 2 years ago
  9. Mikael Group Title
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    DO NOT "help" as long as not requested

    • 2 years ago
  10. siddhantsharan Group Title
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    Hmm. Although this estimate will help us calculate an approximate value, how will we actually find the roots?

    • 2 years ago
  11. siddhantsharan Group Title
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    Sorry. No help from now on.

    • 2 years ago
  12. siddhantsharan Group Title
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    addictive poisonous byproduct? I beg to differ. Okay, for now, shutted up. Continue.

    • 2 years ago
  13. siddhantsharan Group Title
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    Okay. Back to the problem if you may.

    • 2 years ago
  14. Mikael Group Title
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    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...

    • 2 years ago
  15. Mikael Group Title
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    What you have is the head of Exonential series

    • 2 years ago
  16. KingGeorge Group Title
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    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.

    • 2 years ago
  17. Mikael Group Title
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    Now these series are a generating function

    • 2 years ago
  18. Mikael Group Title
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    I beat you purple King as your other messenger

    • 2 years ago
  19. siddhantsharan Group Title
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    Haha. Good guess. My first thoughts too. However I think this is only till the degree is even. @KingGeorge

    • 2 years ago
  20. KingGeorge Group Title
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    If it's odd, by definition it must have at least one real root.

    • 2 years ago
  21. Mikael Group Title
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    Look @KingGeorge please let me do my try ? can you

    • 2 years ago
  22. siddhantsharan Group Title
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    @KingGeorge By definition? @Mikael Let everybody try.

    • 2 years ago
  23. KingGeorge Group Title
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    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.

    • 2 years ago
  24. siddhantsharan Group Title
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    Okay. Yes. Agreed.

    • 2 years ago
  25. Mikael Group Title
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    Soo, SOLVED !

    • 2 years ago
  26. Mikael Group Title
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    Here it is

    • 2 years ago
  27. Mikael Group Title
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    Your function diffrentiated is itself - last member

    • 2 years ago
  28. Mikael Group Title
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    f - f' = x^6/6!

    • 2 years ago
  29. Mikael Group Title
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    It is some sort of generating unction type argument with exp(x) and its derivatives

    • 2 years ago
  30. Mikael Group Title
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    If f(x_1) = 0 ==> f'(x_1) = = x^6/6!

    • 2 years ago
  31. Mikael Group Title
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    meant - x^6/6!

    • 2 years ago
  32. Mikael Group Title
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    So at the root the derivative (slope) must be negative

    • 2 years ago
  33. Mikael Group Title
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    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. ?...

    • 2 years ago
  34. Mikael Group Title
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    After all - at minus-infty it tends to +infty

    • 2 years ago
  35. Mikael Group Title
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    So the leftmost root - the crosiing MUST be from negative to positive.

    • 2 years ago
  36. Mikael Group Title
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    I am still claiming it does go +infty at x--> -infty and Wolfram approves it

    • 2 years ago
  37. Mikael Group Title
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    So it seems NO roots becomes quite closer to reality - dont you say ?

    • 2 years ago
  38. Mikael Group Title
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    Yes and so NO REAL roots is approved By Mister Stephan son of Hugo and Sybil

    • 2 years ago
  39. Mikael Group Title
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    @siddhantsharan No real roots to that one

    • 2 years ago
  40. Mikael Group Title
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    Mister @KingGeorge - some encouragement here ?

    • 2 years ago
  41. Mikael Group Title
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    @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

    • 2 years ago
  42. TuringTest Group Title
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    I would give you acknowledgement if I had the time to check, but I don't right now sorry

    • 2 years ago
  43. Mikael Group Title
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    So \[ f(x) - f'(x) = \frac{x^6}{6!} \text{which means that at the root } f'(x) = - \frac{x^6}{720} \text{which IS NEGATIVE} \]

    • 2 years ago
  44. Mikael Group Title
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    But it is clear that the coefficient of x^6 which is the largest power is positive ==> therefore \[\lim_{x \to -\infty} f(x) = + \infty\]

    • 2 years ago
  45. Mikael Group Title
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    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.

    • 2 years ago
  46. Mikael Group Title
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    @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.

    • 2 years ago
  47. Mikael Group Title
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    Thank You @demitris I hope @TuringTest will also see it was ingenious enough for his appreciation of the effort and the breakthrough

    • 2 years ago
  48. Mikael Group Title
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    Tahnks @TuringTest . Where are the people who asked the question in the first place ?... vanished JUST IN TIME don't you say ....

    • 2 years ago
  49. TuringTest Group Title
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    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.

    • 2 years ago
  50. TuringTest Group Title
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    it is a nice solution though :)

    • 2 years ago
  51. siddhantsharan Group Title
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    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 :)

    • 2 years ago
  52. siddhantsharan Group Title
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    @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?

    • 2 years ago
  53. Mikael Group Title
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    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.

    • 2 years ago
  54. RaphaelFilgueiras Group Title
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    http://openstudy.com/study#/updates/505ebe90e4b0583d5cd1dfc6

    • 2 years ago
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