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Math Help - Without L.R.,Long division,definition of derivative and eplison-delta prove, evaluate

  1. #1
    Super Member General's Avatar
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    Without L.R.,Long division,definition of derivative and eplison-delta proof, evaluate

    Without what I said, Evaluate :
    \lim_{x \rightarrow 1} \frac{x^5-1}{x-1}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by General View Post
    Without what I said, Evaluate :
    \lim_{x \rightarrow 1} \frac{x^5-1}{x-1}
    factor: x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1), the bad guys cancel, evaluate at 1 and you're done.
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  3. #3
    Super Member General's Avatar
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    Quote Originally Posted by Jhevon View Post
    factor: x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1), the bad guys cancel, evaluate at 1 and you're done.
    How did you factor it ?
    1 is zero of x^5-1
    then (x-1) is a factor .. and you find x^4+x^3+x^2+x+1 by long division
    which is not allowed to use.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by General View Post
    How do you factor it ?
    It's a known fact that a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2}b + a^{n - 3}b^2 + \cdots + ab^{n - 2} + b^{n - 1}) when n is an odd natural number
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  5. #5
    Super Member General's Avatar
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    Quote Originally Posted by Jhevon View Post
    It's a known fact that a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2}b + a^{n - 3}b^2 + \cdots + ab^{n - 2} + b^{n - 1}) when n is an odd natural number
    OK.
    Waiting for the other solutions.
    I want to see another ideas

    Thanks.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by General View Post
    OK.
    Waiting for the other solutions.
    I want to see another ideas

    Thanks.
    Ok, do a change of variable. u = x - 1, so that as x \to 1, u \to 0. the limit becomes

    \lim_{u \to 0} \frac {(u + 1)^5 - 1}u

    now expand the top, cancel the u's, evaluate. do you like that one?
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  7. #7
    Super Member General's Avatar
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    Quote Originally Posted by Jhevon View Post
    Ok, do a change of variable. u = x - 1, so that as x \to 1, u \to 0. the limit becomes

    \lim_{u \to 0} \frac {(u + 1)^5 - 1}u

    now expand the top, cancel the u's, evaluate
    Its the same idea
    Thank you.
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    Ok, do a change of variable. u = x - 1, so that as x \to 1, u \to 0. the limit becomes

    \lim_{u \to 0} \frac {(u + 1)^5 - 1}u

    now expand the top, cancel the u's, evaluate. do you like that one?
    I wouldn't recommend that if having \frac{x^{n}-1}{x-1}, you wouldn't expand.
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    I wouldn't recommend that if having \frac{x^{n}-1}{x-1}, you wouldn't expand.
    Haha, point taken. for arbitrary n i would use one of the methods we have been restricted from using. here it's ok though

    do you have another method for arbitrary (large) n values?
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by General View Post
    How did you factor it ?
    1 is zero of x^5-1
    then (x-1) is a factor .. and you find x^4+x^3+x^2+x+1 by long division
    which is not allowed to use.
    You don't find this by long division you observe that (x-1)(x^{n-1}+x^{n-2}+... +1) gives a telescoping sum. Or equivalently:

    x^n-1=[x^n+x^{n-1}+ ... +x] - [x^{n-1}+x^{n-2}+ ... +1]

    .........  = (x^n-x^{n-1}) + (x^{n-1}-x^{n-2})+ ... + (x-1)

    ......... = x^{n-1}(x-1) + x^{n-2}(x-1) + ... + (x-1)

    ......... = (x-1)[x^{n-1} + x^{n-2} + ... +1]



    CB
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  11. #11
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    I think I can do it without what you said but using the ideas from calculus. Your limit represent the slope of the tangent of y = x^5 at x = 1. Let

    \lim_{x \to 1} \frac{x^5-1}{x-1} = m.

    The tangent line is y-1 = m(x-1) or y = mx -m +1

    The intersection of the tangent line and the quintic

    x^5 = mx - m + 1 or x^5 -mx + m -1 = 0 .\;\;(1.1)

    There should be two intersection points, at x = 1 and x = a (some a) where the first will be a double root so eqn. (1.1) has the form

     <br />
(x-1)^2(x-a)(x^2+px+q) = 0 \;\;(1.2)<br />
for some p and q.

    Expanding and equating the coefficients of (1.1) and (1.2) gives

    (2.1)\;\;m + aq - 1 = 0,
    (2.2)\;\;-m + ap - 2aq - q = 0,
     <br />
(2.3)\;\;aq - 2ap + 2q + a - p = 0,<br />
     <br />
(2.4)\;\;ap + 2p - q - 2a - 1 =0,<br />
    (2.5)\;\;a-p+2 = 0.

    If we add/substract in the following order

     <br />
(2.1) - (2.2)- 3 \times (2.3) - 5 \times (2.4) - 7 \times (2.5) = 0<br />

    gives

    2m - 10 = 0 or m = 5, our desired limit.
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  12. #12
    Newbie I4talent's Avatar
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    Quote Originally Posted by Jhevon View Post
    It's a known fact that a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2}b + a^{n - 3}b^2 + \cdots + ab^{n - 2} + b^{n - 1}) when n is an odd natural number
    a^n - b^n = (a-b)\left(\sum_{k=0}^{n-1}a^{n-k-1}b^{k}\right), \forall \ n\in \mathbb{N}.

    It's a^n + b^n = (a+b)\left(\sum_{k=0}^{n-1}(a^{n-k-1})(-b)^k\right) that only holds when n is an odd positive integer.
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