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siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0\[1 + x + x^2 /2! + x^3 /3! ..........x^6/6! = 0\]

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3Obviously real roots  if exist at all  must be negative numbers.

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3Noe we have a war on our screens  the great war between the Good Evens (pos) and the bad Odds (neg)

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3clearly \[ \Sigma Evens < \Sigma Odds\] if x > 6 and we can improve this estim.

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0Haha. 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.

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0Okay. Same thing as you said.

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3I rose faster than your treacherous spoiler "help"

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3DO NOT "help" as long as not requested

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0Hmm. Although this estimate will help us calculate an approximate value, how will we actually find the roots?

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry. No help from now on.

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0addictive poisonous byproduct? I beg to differ. Okay, for now, shutted up. Continue.

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0Okay. Back to the problem if you may.

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3well 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...

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3What you have is the head of Exonential series

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.0Just 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.

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3Now these series are a generating function

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3I beat you purple King as your other messenger

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0Haha. Good guess. My first thoughts too. However I think this is only till the degree is even. @KingGeorge

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.0If it's odd, by definition it must have at least one real root.

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3Look @KingGeorge please let me do my try ? can you

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0@KingGeorge By definition? @Mikael Let everybody try.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.0Perhaps 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.

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0Okay. Yes. Agreed.

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3Your function diffrentiated is itself  last member

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3It is some sort of generating unction type argument with exp(x) and its derivatives

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3If f(x_1) = 0 ==> f'(x_1) = = x^6/6!

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3So at the root the derivative (slope) must be negative

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3But this is either my mistake or Impossibele since we know that after the largest in absolvalue negative root the function Must become Poistive fo all largerabsvalue negative xs. ?...

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3After all  at minusinfty it tends to +infty

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3So the leftmost root  the crosiing MUST be from negative to positive.

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3I am still claiming it does go +infty at x> infty and Wolfram approves it

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3So it seems NO roots becomes quite closer to reality  dont you say ?

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3Yes and so NO REAL roots is approved By Mister Stephan son of Hugo and Sybil

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3@siddhantsharan No real roots to that one

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3Mister @KingGeorge  some encouragement here ?

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3@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
 2 years ago
Best ResponseYou've already chosen the best response.1I would give you acknowledgement if I had the time to check, but I don't right now sorry

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3So \[ f(x)  f'(x) = \frac{x^6}{6!} \text{which means that at the root } f'(x) =  \frac{x^6}{720} \text{which IS NEGATIVE} \]

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3But it is clear that the coefficient of x^6 which is the largest power is positive ==> therefore \[\lim_{x \to \infty} f(x) = + \infty\]

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3So the last crossing of the xaxis MUST be from negative ys to positive ys ==> 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.

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3@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.

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3Thank You @demitris I hope @TuringTest will also see it was ingenious enough for his appreciation of the effort and the breakthrough

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3Tahnks @TuringTest . Where are the people who asked the question in the first place ?... vanished JUST IN TIME don't you say ....

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1Okay 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 highranking people here. It just doesn't seem becoming.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1it is a nice solution though :)

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry @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 :)

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0@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?

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.3Yes that's almost so. But once again you Must use the eqn ff'=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 unheardof 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.

RaphaelFilgueiras
 2 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/505ebe90e4b0583d5cd1dfc6
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