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siddhantsharan

  • 2 years ago

Find the Real Roots of: Equation in comment.

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  1. siddhantsharan
    • 2 years ago
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    \[1 + x + x^2 /2! + x^3 /3! ..........x^6/6! = 0\]

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

  3. siddhantsharan
    • 2 years ago
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    Agreed.

  4. Mikael
    • 2 years ago
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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)

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

  6. siddhantsharan
    • 2 years ago
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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.

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

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

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

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

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

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

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

  14. Mikael
    • 2 years ago
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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...

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

  16. KingGeorge
    • 2 years ago
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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.

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

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

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

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

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

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

  23. KingGeorge
    • 2 years ago
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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.

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

  25. Mikael
    • 2 years ago
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    Soo, SOLVED !

  26. Mikael
    • 2 years ago
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    Here it is

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

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

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

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

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

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

  33. Mikael
    • 2 years ago
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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. ?...

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

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

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

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

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

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

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

  41. Mikael
    • 2 years ago
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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

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

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

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

  45. Mikael
    • 2 years ago
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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.

  46. Mikael
    • 2 years ago
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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.

  47. Mikael
    • 2 years ago
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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

  48. Mikael
    • 2 years ago
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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 ....

  49. TuringTest
    • 2 years ago
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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.

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

  51. siddhantsharan
    • 2 years ago
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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 :)

  52. siddhantsharan
    • 2 years ago
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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?

  53. Mikael
    • 2 years ago
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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.

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

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