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anonymous

  • 5 years ago

partial fractions fail

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  1. anonymous
    • 5 years ago
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    ok i have 1/ (x (x^2-4)) = A/ x + (bx+c) / (x^2-4)

  2. anonymous
    • 5 years ago
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    im breaking the rule of irreducible factoring, but it still works

  3. dumbcow
    • 5 years ago
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    you can factor x^2-4 = (x-2)(x+2)

  4. anonymous
    • 5 years ago
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    oh i know

  5. myininaya
    • 5 years ago
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    i don't think that will always work

  6. anonymous
    • 5 years ago
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    \[\frac{1}{x(x-2)(x+2)}=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-2}\]\]

  7. anonymous
    • 5 years ago
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    myin, can you think of a counterexample

  8. myininaya
    • 5 years ago
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    i will think one sec

  9. anonymous
    • 5 years ago
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    here is another example int ( 1/ ( x ( x^2 - 15))

  10. anonymous
    • 5 years ago
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    here we have reducible over reals, irreducible over rationals (integers)

  11. myininaya
    • 5 years ago
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    so this form only right? where you have 1/(a*(x^2-b))

  12. myininaya
    • 5 years ago
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    i mean that a is an x

  13. myininaya
    • 5 years ago
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    i mean you want me to find a counterexample of that form?

  14. anonymous
    • 5 years ago
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    right,

  15. anonymous
    • 5 years ago
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    ( 1/ ( x ( x^2 - 15)) = A / x + (bx + c) / (x^2 - 15)

  16. anonymous
    • 5 years ago
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    in general 1 / [(x-a)^n (x^2+bx+c)^n] = A1/(x-a) + A2/ ( x-a)^2 + ... An/(x-a)^n + (B1x+C1)/ (x^2+bx+c) + (B2x + C2) / (x^2+bx+c)^2 + ... + (Bn x + Cn ) / (x^2+bx+c)^n

  17. myininaya
    • 5 years ago
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    i guess you may be right

  18. myininaya
    • 5 years ago
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    still thinking

  19. anonymous
    • 5 years ago
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    see my book keeps saying "irreducible"

  20. dumbcow
    • 5 years ago
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    yes it works A = -1/b B = 1/b

  21. anonymous
    • 5 years ago
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    one must use quadratics that are irreducible, but im finding that is not the case

  22. anonymous
    • 5 years ago
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    now i remember what a pain this is. i got A = -1/4

  23. anonymous
    • 5 years ago
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    right we have 1/ (x (x^2-4)) = (-1/4) / x + (1/4 * x ) / (x^2 - 4)

  24. anonymous
    • 5 years ago
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    1/ (x (x^2-4)) = (-1/4) / x + (1/4 * x + 0 ) / (x^2 - 4)

  25. anonymous
    • 5 years ago
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    B=C and A+B+C=0 -1/4+2B=0 making B = 1/8

  26. anonymous
    • 5 years ago
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    but it is late and i could have made a mistake

  27. dumbcow
    • 5 years ago
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    no you do not always have to reduce quadratics, as long as there is a product in denominator

  28. anonymous
    • 5 years ago
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    satellite, you did it out the normal way, we have same solution though

  29. anonymous
    • 5 years ago
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    whew. glad to know i can still solve a simple system

  30. anonymous
    • 5 years ago
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    i was just making a point, you can relax the rule about quadratics being irreducible

  31. anonymous
    • 5 years ago
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    sure if you use complex you can even do it with \[x^2+1\] and even use it to integrate

  32. anonymous
    • 5 years ago
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    right but i mean , sometimes you have, say this case int 1 / ( x ( x^2 - 15)

  33. anonymous
    • 5 years ago
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    \[\int \frac{1}{x^2+1} dx\] can use partial fractions.

  34. anonymous
    • 5 years ago
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    technically, yes over the complex numbers

  35. anonymous
    • 5 years ago
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    i think the issue boils down, what field you are restricting yourself to

  36. anonymous
    • 5 years ago
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    restricting to rationals , to reals, or to complex

  37. anonymous
    • 5 years ago
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    myinanay, are you in agrement?

  38. myininaya
    • 5 years ago
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    yes i just made a proof lol i will scan it

  39. myininaya
    • 5 years ago
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    for when we were talking about 1/[x(x^2-a)] form sorry i doubted you cantorset :)

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  40. myininaya
    • 5 years ago
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    its alittle sloppy and i wanted it on one page so after i got to the end of the page i went back up and drew a square to finish showing the combination of fractions from i reduced the polynomial x^2-a would still give you the same combination of frations if you did not reduce

  41. anonymous
    • 5 years ago
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    ok im back

  42. myininaya
    • 5 years ago
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    can you read it?

  43. anonymous
    • 5 years ago
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    so you got the same answer doing it both ways?

  44. myininaya
    • 5 years ago
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    right! so since we proved we can make it a theorem lol

  45. myininaya
    • 5 years ago
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    it is no longer a conjunction to us

  46. anonymous
    • 5 years ago
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    how do you know they are equal though?

  47. myininaya
    • 5 years ago
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    do you in the square i combined the fractions?

  48. myininaya
    • 5 years ago
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    from the reduced way

  49. myininaya
    • 5 years ago
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    it came out to be the same as the non reduced way

  50. anonymous
    • 5 years ago
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    oh i see

  51. anonymous
    • 5 years ago
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    you can combine the last two fractions ,

  52. myininaya
    • 5 years ago
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    right

  53. myininaya
    • 5 years ago
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    i can write it more neatly if you like

  54. anonymous
    • 5 years ago
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    thats fine, just that last step i did on paper

  55. anonymous
    • 5 years ago
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    so why does my book insist on reducing only the irreducible

  56. anonymous
    • 5 years ago
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    irreducible actually depends on the field i think, since 1 / (x^2 +1) is reducible in C

  57. myininaya
    • 5 years ago
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    well i only know this true so far for the form 1/[x*(x^2-a)] which is reduceable

  58. anonymous
    • 5 years ago
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    hmmmm

  59. anonymous
    • 5 years ago
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    you dont think 1 / [(x-a)^n (x^2+bx+c)^n] = A1/(x-a) + A2/ ( x-a)^2 + ... An/(x-a)^n + (B1x+C1)/ (x^2+bx+c) + (B2x + C2) / (x^2+bx+c)^2 + ... + (Bn x + Cn ) / (x^2+bx+c)^n

  60. myininaya
    • 5 years ago
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    maybe it is easier just to memorize just reduce just in case instead of remembering for certain forms it is akay not to reduce

  61. myininaya
    • 5 years ago
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    thats ugly to me cantorset lol

  62. anonymous
    • 5 years ago
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    no it has a nice pattern

  63. anonymous
    • 5 years ago
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    The property of irreducibility depends on the field F; a polynomial may be irreducible over some fields but reducible over others

  64. anonymous
    • 5 years ago
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    right, the field F, but we can just change the field to suit our problem

  65. anonymous
    • 5 years ago
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    yes...we are dealing with real analysis here

  66. anonymous
    • 5 years ago
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    so F is the field of R

  67. anonymous
    • 5 years ago
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    like integral 1 / ( x^2 + 1) = int 1 / (x+ i ) + 1/ (x-i) i believe

  68. myininaya
    • 5 years ago
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    tan inverse of x

  69. anonymous
    • 5 years ago
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    1/ x^2 + 1 = A / x+i + B / x- i

  70. myininaya
    • 5 years ago
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    +C

  71. anonymous
    • 5 years ago
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    Well thats complex analysis cantorset!

  72. myininaya
    • 5 years ago
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    what is e^(ix) in terms of sine and cosine again i cant remember something like sin(x)+icos(x)

  73. anonymous
    • 5 years ago
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    yes, ok i see your point

  74. anonymous
    • 5 years ago
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    so you want to stay in field R, ok

  75. anonymous
    • 5 years ago
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    by the way x+i and x-i are not polynomials in real analysis

  76. myininaya
    • 5 years ago
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    also if we assume int(1/(x^2+1))=int(1/(x+i))+int(1/(x-i))=then we get ln(x^2+1) but this gives us 2x/(x^2+1) when we take derivative which is not 1/(x^2+1)

  77. anonymous
    • 5 years ago
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    i actually didnt do the A , and B thingy

  78. anonymous
    • 5 years ago
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    that was a guess

  79. anonymous
    • 5 years ago
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    1/ x^2 + 1 = A / x+i + B / x- i

  80. myininaya
    • 5 years ago
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    so A=-1/2i and B=1/2i lets see what happens

  81. anonymous
    • 5 years ago
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    brb

  82. myininaya
    • 5 years ago
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    i/2 *ln{(x-i)/(x+i)} i don't know if we can do anything with this

  83. anonymous
    • 5 years ago
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    right

  84. anonymous
    • 5 years ago
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    youre right about the arctangent though

  85. anonymous
    • 5 years ago
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    so my point was, can we relax our condition about being reducible , or irreducible?

  86. myininaya
    • 5 years ago
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    so far we can relax it for some reducibles like 1/[x*(x^2-a)] but not for forms that are irreducible

  87. anonymous
    • 5 years ago
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    is e^(arctanx)==((x-i)/(x+i))^(1/2i) true?

  88. anonymous
    • 5 years ago
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    irreducible over the reals, you mean

  89. myininaya
    • 5 years ago
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    i don't know saubihik i fail at complex analysis

  90. myininaya
    • 5 years ago
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    or whatever that is

  91. anonymous
    • 5 years ago
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    i can check my calculator

  92. anonymous
    • 5 years ago
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    but it might be off by a constant

  93. myininaya
    • 5 years ago
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    how do you plug imaginarys into your calculator?

  94. myininaya
    • 5 years ago
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    can we just pretend since they are imaginary lol

  95. anonymous
    • 5 years ago
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    i have TI 84

  96. anonymous
    • 5 years ago
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    i think its true and may be by some constant and so yah u can do partial fracs with rational functions having COMPLEX VARIABLES

  97. anonymous
    • 5 years ago
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    only the answer will come in complex terms but dont woory thats same!!

  98. anonymous
    • 5 years ago
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    So our rule is: partial fraction expansions provide an approach to integrating a general rational function. Any rational function of a real variable can be written as the sum of a polynomial function and a finite number of fractions. Each fraction in the expansion has as its denominator a polynomial function of degree 1 or 2, or some positive integer power of such a polynomial. (In the case of rational function of a complex variable, all denominators will have a polynomial of degree 1, or some positive integer power of such a polynomial.) If the denominator is a 1st-degree polynomial or a power of such a polynomial, then the numerator is a constant. If the denominator is a 2nd-degree polynomial or a power of such a polynomial, then the numerator is a 1st-degree polynomial.

  99. myininaya
    • 5 years ago
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    really? thats neat

  100. myininaya
    • 5 years ago
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    man math is deep

  101. anonymous
    • 5 years ago
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    So we have a rule of partial fractions now with the prerequisite: ANY RATIONAL FUNCTION WITH REAL OR COMPLEX VARIABLES. that is very good generalization!!

  102. anonymous
    • 5 years ago
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    but whats the field?

  103. myininaya
    • 5 years ago
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    saubhik i didn't realize i made a mistake until u posted the 1/(2i) dont know i put the i on top

  104. anonymous
    • 5 years ago
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    I think there is no field restrictions for partial fracs. its C

  105. anonymous
    • 5 years ago
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    oh,

  106. anonymous
    • 5 years ago
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    where did you get that quote from?

  107. anonymous
    • 5 years ago
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    which one?

  108. anonymous
    • 5 years ago
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    the big one

  109. anonymous
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    partial fraction expansions provide an approach to integrating a general rational function. Any rational function of a real variable can be written as the sum of a polynomial function and a finite number of fractions. Each fraction in the expansion has as its denominator a polynomial function of degree 1 or 2, or some positive integer power of such a polynomial. (In the case of rational function of a complex variable, all denominators will have a polynomial of degree 1, or some positive integer power of such a polynomial.) If the denominator is a 1st-degree polynomial or a power of such a polynomial, then the numerator is a constant. If the denominator is a 2nd-degree polynomial or a power of such a polynomial, then the numerator is a 1st-degree polynomial.

  110. anonymous
    • 5 years ago
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    wikipedia rocks!!:)

  111. anonymous
    • 5 years ago
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    But integration is limited to R. Recall the fundamental theorems of calculus!!!

  112. anonymous
    • 5 years ago
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    hmmmm

  113. anonymous
    • 5 years ago
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    every function to be integrated/differentiated (if such operation is allowed by the function) has to be real-valued. So we have a doubt here: Applying partial fracs if we get complex fracs. then how we are integrating??

  114. anonymous
    • 5 years ago
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    for eg. A/(x-i) is not real valued so integral of this is not defined.!!

  115. anonymous
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    but it should come out the same

  116. anonymous
    • 5 years ago
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    which type of integral are you using? U cannot use the general Newton-Leibniz integral!

  117. anonymous
    • 5 years ago
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    I think if you want to avoid the complex analysis stuff (residue theorem,contour integration) avoid splitting to complex rational functions!

  118. anonymous
    • 5 years ago
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    ok, but i dont have to reduce things like x^2 - 15

  119. anonymous
    • 5 years ago
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    you may/maynot

  120. anonymous
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    reducing x^2-15=(x-rt(15))(x+rt(15)) is recommended and the PROPER procedure to partial fraction decomposition. i.e. 1/(x^2-15)=A/(x-rt(15))+B/(x+rt(15))

  121. anonymous
    • 5 years ago
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    actually its not

  122. anonymous
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    it makes it much more difficult

  123. anonymous
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    that is the PROPER way. See that x^2-15 is reducible. So why not use this advantage. But if u use cx+d thing then don't call this partial frac decomposition.

  124. anonymous
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    that may be "partial" partial frac decomposition :D

  125. anonymous
    • 5 years ago
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    well try to do partial fract, it will be harder

  126. anonymous
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    its much faster 1/ (x(x^2-15) = A/x + (bx + c ) / (x^2-15)

  127. anonymous
    • 5 years ago
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    yah, but that's not partial frac decompositon

  128. anonymous
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    the degree of the rational function (i.e. the degree of both the numerator and denominator polynomials) must be reduced to the utmost extent

  129. anonymous
    • 5 years ago
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    the outcome of a full partial fraction expansion expresses that function as a sum of fractions, where: the denominator of each term is a power of an irreducible (not factorable) polynomial and the numerator is a polynomial of smaller degree than that irreducible polynomial.

  130. anonymous
    • 5 years ago
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    but in the case of "your" expansion: (bx+c)/(x^2-15) term has a reducible denominator and so your expansion is not the FULL partial frac decomposition. Understood?

  131. anonymous
    • 5 years ago
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    yes

  132. anonymous
    • 5 years ago
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    it is the full fractional decomposition over the integers (or rationals

  133. watchmath
    • 5 years ago
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    @Cantorset: can you always integrate (bx + c ) / (x^2-15) without splitting x^2-15

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