## anonymous 4 years ago Find the Real Roots of: Equation in comment.

1. anonymous

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

2. anonymous

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

3. anonymous

Agreed.

4. anonymous

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

5. anonymous

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

6. 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.

7. anonymous

Okay. Same thing as you said.

8. anonymous

I rose faster than your treacherous spoiler "help"

9. anonymous

DO NOT "help" as long as not requested

10. anonymous

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

11. anonymous

Sorry. No help from now on.

12. anonymous

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

13. anonymous

Okay. Back to the problem if you may.

14. 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...

15. anonymous

What you have is the head of Exonential series

16. 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.

17. anonymous

Now these series are a generating function

18. anonymous

I beat you purple King as your other messenger

19. anonymous

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

20. KingGeorge

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

21. anonymous

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

22. anonymous

@KingGeorge By definition? @Mikael Let everybody try.

23. 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.

24. anonymous

Okay. Yes. Agreed.

25. anonymous

Soo, SOLVED !

26. anonymous

Here it is

27. anonymous

Your function diffrentiated is itself - last member

28. anonymous

f - f' = x^6/6!

29. anonymous

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

30. anonymous

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

31. anonymous

meant - x^6/6!

32. anonymous

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

33. 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. ?...

34. anonymous

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

35. anonymous

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

36. anonymous

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

37. anonymous

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

38. anonymous

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

39. anonymous

@siddhantsharan No real roots to that one

40. anonymous

Mister @KingGeorge - some encouragement here ?

41. 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

42. TuringTest

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

43. anonymous

So $f(x) - f'(x) = \frac{x^6}{6!} \text{which means that at the root } f'(x) = - \frac{x^6}{720} \text{which IS NEGATIVE}$

44. anonymous

But it is clear that the coefficient of x^6 which is the largest power is positive ==> therefore $\lim_{x \to -\infty} f(x) = + \infty$

45. 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.

46. 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.

47. anonymous

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

48. anonymous

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

49. 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.

50. TuringTest

it is a nice solution though :)

51. 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 :)

52. 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?

53. 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.

54. anonymous