## siddhantsharan Group Title Find the Real Roots of: Equation in comment. one year ago one year ago

1. siddhantsharan Group Title

$1 + x + x^2 /2! + x^3 /3! ..........x^6/6! = 0$

2. Mikael Group Title

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

3. siddhantsharan Group Title

Agreed.

4. Mikael Group Title

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

5. Mikael Group Title

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

6. siddhantsharan Group Title

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 Group Title

Okay. Same thing as you said.

8. Mikael Group Title

I rose faster than your treacherous spoiler "help"

9. Mikael Group Title

DO NOT "help" as long as not requested

10. siddhantsharan Group Title

Hmm. Although this estimate will help us calculate an approximate value, how will we actually find the roots?

11. siddhantsharan Group Title

Sorry. No help from now on.

12. siddhantsharan Group Title

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

13. siddhantsharan Group Title

Okay. Back to the problem if you may.

14. Mikael Group Title

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 Group Title

What you have is the head of Exonential series

16. KingGeorge Group Title

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 Group Title

Now these series are a generating function

18. Mikael Group Title

I beat you purple King as your other messenger

19. siddhantsharan Group Title

Haha. Good guess. My first thoughts too. However I think this is only till the degree is even. @KingGeorge

20. KingGeorge Group Title

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

21. Mikael Group Title

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

22. siddhantsharan Group Title

@KingGeorge By definition? @Mikael Let everybody try.

23. KingGeorge Group Title

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 Group Title

Okay. Yes. Agreed.

25. Mikael Group Title

Soo, SOLVED !

26. Mikael Group Title

Here it is

27. Mikael Group Title

Your function diffrentiated is itself - last member

28. Mikael Group Title

f - f' = x^6/6!

29. Mikael Group Title

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

30. Mikael Group Title

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

31. Mikael Group Title

meant - x^6/6!

32. Mikael Group Title

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

33. Mikael Group Title

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 Group Title

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

35. Mikael Group Title

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

36. Mikael Group Title

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

37. Mikael Group Title

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

38. Mikael Group Title

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

39. Mikael Group Title

@siddhantsharan No real roots to that one

40. Mikael Group Title

Mister @KingGeorge - some encouragement here ?

41. Mikael Group Title

@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 Group Title

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

43. Mikael Group Title

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 Group Title

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 Group Title

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 Group Title

@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 Group Title

Thank You @demitris I hope @TuringTest will also see it was ingenious enough for his appreciation of the effort and the breakthrough

48. Mikael Group Title

Tahnks @TuringTest . Where are the people who asked the question in the first place ?... vanished JUST IN TIME don't you say ....

49. TuringTest Group Title

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 Group Title

it is a nice solution though :)

51. siddhantsharan Group Title

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 Group Title

@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 Group Title

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 Group Title