A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

P(x) = 1 - x + (x²/2!) - ... + (-1)^n (x^n/n!) RTP that P(x) has no double zeroes for n≥2

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm using proof by contradiction. P'(x) = -1 + x - (x²/2!) + ... + (-1)^2 [nx^(n-1) / n!] Assume that there is a multiple root A, ie P(A) = 0 and P'(A) = 0 ie 1 - A + (A²/2!) - ... + (-1)^n (A^n/n!) = 0 ...<1> and -1 + A - (A²/2!) + ... + (-1)^2 [nA^(n-1) / n!] ...<2> Now, <1> + <2>: 1 - A + (A²/2!) - ... + (-1)^n (A^n/n!) + -1 + A - (A²/2!) + ... + (-1)^2 [nA^(n-1) / n!] = 0 All except for the last terms cancel out. (-1)^n (A^n/n!) + + (-1)^2 [nA^(n-1) / n!] = ? I'm also not sure if the last term in <2> should be negative or not. Help please?

  2. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    to me, P(x) looks like the Taylor polynomial of e^(-x)... just to throw out a possibility

  3. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    but I will try this as well

  4. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    actually, the last term in P'(x) should cancel with one of the terms in P(x), the reason is this \[P(x) = 1 + ... + (-1)^{n-1}{x^{n-1} \over (n-1)!}+ (-1)^n{x^n \over n!}\] and P'(x) is \[P'(x) = -1+...+(-1)^n{x^{n-1} \over (n-1)!} \]

  5. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so the last term of P'(x) and the second to last term of P(x) have opposite signs no matter what n is since one has (-1)^n-1 and the other has (-1)^n so they will cancel each other letting us know that P(x)+P'(x) = \[(-1)^n{x^n \over n!}\]

  6. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    does that help ?

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ah, awesome!! Thanks so much :D

  8. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.