## anonymous 5 years ago partial fractions fail

1. anonymous

ok i have 1/ (x (x^2-4)) = A/ x + (bx+c) / (x^2-4)

2. anonymous

im breaking the rule of irreducible factoring, but it still works

3. anonymous

you can factor x^2-4 = (x-2)(x+2)

4. anonymous

oh i know

5. myininaya

i don't think that will always work

6. anonymous

$\frac{1}{x(x-2)(x+2)}=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-2}$\]

7. anonymous

myin, can you think of a counterexample

8. myininaya

i will think one sec

9. anonymous

here is another example int ( 1/ ( x ( x^2 - 15))

10. anonymous

here we have reducible over reals, irreducible over rationals (integers)

11. myininaya

so this form only right? where you have 1/(a*(x^2-b))

12. myininaya

i mean that a is an x

13. myininaya

i mean you want me to find a counterexample of that form?

14. anonymous

right,

15. anonymous

( 1/ ( x ( x^2 - 15)) = A / x + (bx + c) / (x^2 - 15)

16. anonymous

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

i guess you may be right

18. myininaya

still thinking

19. anonymous

see my book keeps saying "irreducible"

20. anonymous

yes it works A = -1/b B = 1/b

21. anonymous

one must use quadratics that are irreducible, but im finding that is not the case

22. anonymous

now i remember what a pain this is. i got A = -1/4

23. anonymous

right we have 1/ (x (x^2-4)) = (-1/4) / x + (1/4 * x ) / (x^2 - 4)

24. anonymous

1/ (x (x^2-4)) = (-1/4) / x + (1/4 * x + 0 ) / (x^2 - 4)

25. anonymous

B=C and A+B+C=0 -1/4+2B=0 making B = 1/8

26. anonymous

but it is late and i could have made a mistake

27. anonymous

no you do not always have to reduce quadratics, as long as there is a product in denominator

28. anonymous

satellite, you did it out the normal way, we have same solution though

29. anonymous

whew. glad to know i can still solve a simple system

30. anonymous

i was just making a point, you can relax the rule about quadratics being irreducible

31. anonymous

sure if you use complex you can even do it with $x^2+1$ and even use it to integrate

32. anonymous

right but i mean , sometimes you have, say this case int 1 / ( x ( x^2 - 15)

33. anonymous

$\int \frac{1}{x^2+1} dx$ can use partial fractions.

34. anonymous

technically, yes over the complex numbers

35. anonymous

i think the issue boils down, what field you are restricting yourself to

36. anonymous

restricting to rationals , to reals, or to complex

37. anonymous

myinanay, are you in agrement?

38. myininaya

yes i just made a proof lol i will scan it

39. myininaya

for when we were talking about 1/[x(x^2-a)] form sorry i doubted you cantorset :)

40. myininaya

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

ok im back

42. myininaya

43. anonymous

so you got the same answer doing it both ways?

44. myininaya

right! so since we proved we can make it a theorem lol

45. myininaya

it is no longer a conjunction to us

46. anonymous

how do you know they are equal though?

47. myininaya

do you in the square i combined the fractions?

48. myininaya

from the reduced way

49. myininaya

it came out to be the same as the non reduced way

50. anonymous

oh i see

51. anonymous

you can combine the last two fractions ,

52. myininaya

right

53. myininaya

i can write it more neatly if you like

54. anonymous

thats fine, just that last step i did on paper

55. anonymous

so why does my book insist on reducing only the irreducible

56. anonymous

irreducible actually depends on the field i think, since 1 / (x^2 +1) is reducible in C

57. myininaya

well i only know this true so far for the form 1/[x*(x^2-a)] which is reduceable

58. anonymous

hmmmm

59. anonymous

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

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

thats ugly to me cantorset lol

62. anonymous

no it has a nice pattern

63. anonymous

The property of irreducibility depends on the field F; a polynomial may be irreducible over some fields but reducible over others

64. anonymous

right, the field F, but we can just change the field to suit our problem

65. anonymous

yes...we are dealing with real analysis here

66. anonymous

so F is the field of R

67. anonymous

like integral 1 / ( x^2 + 1) = int 1 / (x+ i ) + 1/ (x-i) i believe

68. myininaya

tan inverse of x

69. anonymous

1/ x^2 + 1 = A / x+i + B / x- i

70. myininaya

+C

71. anonymous

Well thats complex analysis cantorset!

72. myininaya

what is e^(ix) in terms of sine and cosine again i cant remember something like sin(x)+icos(x)

73. anonymous

yes, ok i see your point

74. anonymous

so you want to stay in field R, ok

75. anonymous

by the way x+i and x-i are not polynomials in real analysis

76. myininaya

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

i actually didnt do the A , and B thingy

78. anonymous

that was a guess

79. anonymous

1/ x^2 + 1 = A / x+i + B / x- i

80. myininaya

so A=-1/2i and B=1/2i lets see what happens

81. anonymous

brb

82. myininaya

i/2 *ln{(x-i)/(x+i)} i don't know if we can do anything with this

83. anonymous

right

84. anonymous

youre right about the arctangent though

85. anonymous

so my point was, can we relax our condition about being reducible , or irreducible?

86. myininaya

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

87. anonymous

is e^(arctanx)==((x-i)/(x+i))^(1/2i) true?

88. anonymous

irreducible over the reals, you mean

89. myininaya

i don't know saubihik i fail at complex analysis

90. myininaya

or whatever that is

91. anonymous

i can check my calculator

92. anonymous

but it might be off by a constant

93. myininaya

how do you plug imaginarys into your calculator?

94. myininaya

can we just pretend since they are imaginary lol

95. anonymous

i have TI 84

96. anonymous

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

only the answer will come in complex terms but dont woory thats same!!

98. anonymous

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

really? thats neat

100. myininaya

man math is deep

101. anonymous

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

but whats the field?

103. myininaya

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

I think there is no field restrictions for partial fracs. its C

105. anonymous

oh,

106. anonymous

where did you get that quote from?

107. anonymous

which one?

108. anonymous

the big one

109. anonymous

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

wikipedia rocks!!:)

111. anonymous

But integration is limited to R. Recall the fundamental theorems of calculus!!!

112. anonymous

hmmmm

113. anonymous

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

for eg. A/(x-i) is not real valued so integral of this is not defined.!!

115. anonymous

but it should come out the same

116. anonymous

which type of integral are you using? U cannot use the general Newton-Leibniz integral!

117. anonymous

I think if you want to avoid the complex analysis stuff (residue theorem,contour integration) avoid splitting to complex rational functions!

118. anonymous

ok, but i dont have to reduce things like x^2 - 15

119. anonymous

you may/maynot

120. anonymous

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

actually its not

122. anonymous

it makes it much more difficult

123. anonymous

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

that may be "partial" partial frac decomposition :D

125. anonymous

well try to do partial fract, it will be harder

126. anonymous

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

127. anonymous

yah, but that's not partial frac decompositon

128. anonymous

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

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

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

yes

132. anonymous

it is the full fractional decomposition over the integers (or rationals

133. watchmath

@Cantorset: can you always integrate (bx + c ) / (x^2-15) without splitting x^2-15