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sjg13e

  • one year ago

Partial Functions - Help!

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  1. sjg13e
    • one year ago
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    please give me a second to upload the question and my work :)

  2. anonymous
    • one year ago
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    it's alright

  3. anonymous
    • one year ago
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    i wait ....

  4. sjg13e
    • one year ago
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    \[\int\limits_{}^{}\frac{ x^2 - 3 }{ x^2 - 4 }\]

  5. sjg13e
    • one year ago
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    my work (it's wrong obviously lol): \[(x^2 - 4) = (x - 2)(x +2)\] so \[\int\limits_{}^{}\frac{ x^2 - 3 }{ x^2 - 4 }= \frac{ x^2 - 3 }{ (x^2 - 2)(x^2 +2) } = \frac{ A }{ x -2 }+\frac{ B }{ x + 2 }\]

  6. amistre64
    • one year ago
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    apart from a typo or two its going fine so far

  7. sjg13e
    • one year ago
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    then \[x^2 - 3 = A(x+2) = B(x-2) \] \[x^2 - 3 = Ax + 2A + Bx - 2B\] \[x^2 - 3 = (A + B)x - 2A - 2B\]

  8. sjg13e
    • one year ago
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    so yes, sorry there are some typos!

  9. amistre64
    • one year ago
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    dont bother distributing .. let x=2 and -2 to solve for A and B \[(2)^2 - 3 = A(2+2) + B(2-2)\] \[(-2)^2 - 3 = A(-2+2) + B(-2-2)\]

  10. sjg13e
    • one year ago
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    From there, I tried to find the values of A and B A + B = 0 A = -B \[-2A - 2B = -3 \] \[-2A = -3 + 2(-A)\] \[-2A = -3 - 2A\] 0 = 3 (??)

  11. freckles
    • one year ago
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    Partial most likely won't work until you do division. We need to divide first because you have an improper fraction.

  12. amistre64
    • one year ago
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    ... it works fine, just dont do the distribution part; that way we can zero out a term to solve

  13. sjg13e
    • one year ago
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    @amistre64 Wait, can you explain a little more why I shouldn't bother distributing and should instead just use substitution. @freckles, How did you know that I need to divide first? That step didn't immediately pop out to me

  14. amistre64
    • one year ago
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    well, you saw why distribution cuases issues already :)

  15. amistre64
    • one year ago
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    A(x-2) can be eliminated when x=2 B(x+2) can be eliminated when x=-2

  16. amistre64
    • one year ago
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    (2)^2-3=A(2+2)+B(2-2) 4-3=A(4)+B(0) 1 = 4A ......................... (-2)^2-3=A(-2+2)+B(-2-2) 4-3=A(0)+B(-4) 1=-4B

  17. freckles
    • one year ago
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    Your top poly and bottom poly had equal degree. Whenever the top poly has equal or greater degree to that of the bottom poly. I think division then partial

  18. freckles
    • one year ago
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    Anyways i will let Amistre help you because it is hard to read on this phone

  19. amistre64
    • one year ago
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    x^2-3 + (1-1) = (x^2-4) + 1 if you really want to do the division first, but its not required

  20. sjg13e
    • one year ago
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    @amistre64 okay, i understand the following work, but how did you know to let x = 2? in my calculus class we haven't really discussed this method within partial functions and thank you @freckles for your help! i'll try doing division first

  21. amistre64
    • one year ago
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    well, how else do you zero out a term? A(x+2) + B(x-2) anything times 0 is 0 ... so by observation letting x=2 or -2 gets rid of a term so that we can solve for the other.

  22. sjg13e
    • one year ago
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    okay, that makes sense! thank you! i'll try that method and solve the problem again

  23. amistre64
    • one year ago
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    \[\frac{x^2-3}{x^2-4}\implies \frac{(x^2-4)+1}{x^2-4}\implies1+\frac{1}{x^2-4}\] we still have to work the partial fraction even after the division, so the division is not necessary to me.

  24. Jhannybean
    • one year ago
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    What i realized for most functions with polynomials in the denominator, it's either u-sub or partial fraction decomp. Hmm!

  25. amistre64
    • one year ago
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    Jhanny, its a conspiracy :)

  26. Jhannybean
    • one year ago
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    Knew it!

  27. amistre64
    • one year ago
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    hmm, working it to a proper fraction might be needed ... never considered it before but the wolf suggests that it might be needed.

  28. sjg13e
    • one year ago
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    okay, so i would have to do long division before i decompose it?

  29. freckles
    • one year ago
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    yes @sjg13e as we notice \[\frac{A}{x-2}+\frac{B}{x+2}=\frac{(A+B)x+(2A-2B)}{x^2-4} \neq \frac{x^2-3}{x^2-4} \text{ for any real choice } \\ A \text{ and } B\] so division is definitely required before doing the partial fractions

  30. sjg13e
    • one year ago
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    Thank you @freckles! I solved the problem and read through some calculus notes and realized that it's required to get the leading x degree in the numerator to be less than the leading x degree in the denominator, and for this problem, it required long division

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