Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

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

MikaelBest ResponseYou've already chosen the best response.3
Obviously real roots  if exist at all  must be negative numbers.
 one year ago

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

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

siddhantsharanBest ResponseYou've already chosen the best response.0
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.
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
Okay. Same thing as you said.
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
I rose faster than your treacherous spoiler "help"
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
DO NOT "help" as long as not requested
 one year ago

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

siddhantsharanBest ResponseYou've already chosen the best response.0
Sorry. No help from now on.
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
addictive poisonous byproduct? I beg to differ. Okay, for now, shutted up. Continue.
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
Okay. Back to the problem if you may.
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
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...
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
What you have is the head of Exonential series
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
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.
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
Now these series are a generating function
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
I beat you purple King as your other messenger
 one year ago

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

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

MikaelBest ResponseYou've already chosen the best response.3
Look @KingGeorge please let me do my try ? can you
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
@KingGeorge By definition? @Mikael Let everybody try.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
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.
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
Okay. Yes. Agreed.
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
Your function diffrentiated is itself  last member
 one year ago

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

MikaelBest ResponseYou've already chosen the best response.3
If f(x_1) = 0 ==> f'(x_1) = = x^6/6!
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
So at the root the derivative (slope) must be negative
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
But 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. ?...
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
After all  at minusinfty it tends to +infty
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
So the leftmost root  the crosiing MUST be from negative to positive.
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
I am still claiming it does go +infty at x> infty and Wolfram approves it
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
So it seems NO roots becomes quite closer to reality  dont you say ?
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
Yes and so NO REAL roots is approved By Mister Stephan son of Hugo and Sybil
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
@siddhantsharan No real roots to that one
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
Mister @KingGeorge  some encouragement here ?
 one year ago

MikaelBest 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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I would give you acknowledgement if I had the time to check, but I don't right now sorry
 one year ago

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

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

MikaelBest ResponseYou've already chosen the best response.3
So 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.
 one year ago

MikaelBest 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.
 one year ago

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

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

TuringTestBest ResponseYou've already chosen the best response.1
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 highranking people here. It just doesn't seem becoming.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
it is a nice solution though :)
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
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 :)
 one year ago

siddhantsharanBest 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?
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
Yes 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.
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
http://openstudy.com/study#/updates/505ebe90e4b0583d5cd1dfc6
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.