I have to perform complex partial fraction decomposition of the following term: (k^4)/((a k^3-1)^2) where a is a real positive number. and I would like to know if it is possible to reduce it to a sum of fractions of the type (A)/(k+-z), (B k)/(k^2+-z), (Ck)/(k^2-y) or similar. Where z is a complex number and y is a real number. If it is not possible to reduce it to the kind of fraction I listed above other type of decomposition might work too. Any hint on the process, or any reference would be nice. Thanks in advance

aanpalendmw 2022-07-22 Answered
partial fraction decomposition of k 4 ( a k 3 1 ) 2
I have to perform complex partial fraction decomposition of the following term:
k 4 ( a k 3 1 ) 2
where a is a real positive number.
and I would like to know if it is possible to reduce it to a sum of fractions of the type A k ± z , B k k 2 ± z , C k k 2 y or similar. Where z is a complex number and y is a real number.
If it is not possible to reduce it to the kind of fraction I listed above other type of decomposition might work too.
Any hint on the process, or any reference would be nice.
Thanks in advance
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

yelashwag8
Answered 2022-07-23 Author has 17 answers
I would start by making the change of variable a k 3 = x 3 , which transforms the expression into
a 3 / 4 x 4 ( x 3 1 ) 2 .
Now we factorise the denominator. This is easy since we have a difference of two cubes.
x 3 1 = ( x 1 ) ( x 2 + x + 1 ) .
Notice that the quadratic is irreducible over the reals. Then (ignoring the constant a 3 / 4 ) we have
x 4 ( x 1 ) 2 ( x 2 + x + 1 ) = A x 1 + B ( x 1 ) 2 + C x + D x 2 + x + 1 + E x + F ( x 2 + x + 1 ) 2 .
Did you like this example?
Subscribe for all access
Glenn Hopkins
Answered 2022-07-24 Author has 4 answers
Set a = b 3 and denominator will be factored as
( b 3 k 3 1 ) 2 = ( b k 1 ) 2 ( b 2 k 2 + b k + 1 ) 2
Then you decompose in partial fractions
e + f k b 2 k 2 + b k + 1 + g + h k ( b 2 k 2 + b k + 1 ) 2 + c b k 1 + d ( b k 1 ) 2
where c , d , e , f , g , h are unknowns and must be obtained solving a six equations linear system. First you add all the fractions and consider the numerator of the result
b 5 c k 5 + b 4 c k 4 + b 4 d k 4 + b 4 e k 4 + b 4 f k 5 + b 3 c k 3 + 2 b 3 d k 3 b 3 e k 3 b 3 f k 4 b 2 c k 2 + 3 b 2 d k 2 + b 2 g k 2 + b 2 h k 3 b c k + 2 b d k b e k b f k 2 2 b g k 2 b h k 2 c + d + e + f k + g + h k
Numerator must be identical to k 4 so coefficients (from degree 0 up to 5 must be { 0 , 0 , 0 , 0 , 1 , 0 }
The system has quite nice solutions
c = 2 9 b 4 , d = 1 9 b 4 , e = 4 9 b 4 , f = 2 9 b 3 , g = 1 3 b 4 , h = 0
Therefore the starting fraction can be written as
1 9 b 4 ( 2 ( b k 2 ) b 2 k 2 + b k + 1 3 ( b 2 k 2 + b k + 1 ) 2 + 2 b k 1 + 1 ( b k 1 ) 2 )
with b = a 1 3
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-06-20
Simplifying this fraction in a different base
Note: I would appreciate a solution that DOES NOT convert back to base 10.
How would one simplify 43 70 8 ? I assume, like in decimal, I must recognize a common factor and divide by that factor. Keyword is recognize because we are taught for example that 5 / 10 has a common factor of 5. Is it the same here? If so, must I ultimately either learn the base (impractical) or convert to base ten?
By the way, this question similar to another question I posted in this forum except that this one simplified the base (possibly directly from the decimal without going through simplification). I would like to mention that I have approximately 12 seconds to do this.\
asked 2022-05-25
Distribution of the sum reciprocal of primes 1
1 2 + 1 3 + 1 7 + 1 43 + 1
This is an interesting infinite summation. This is very closely resembling my other problem with has to do with the distribution of composites. That other problem is here. The same rules of that other summation apply, but it's good to repeat them :).
Rules of this summation:
1) All terms of the summation must be reciprocal of some prime. For example, 1 6 could never be a term because 6 is not a prime number.
2) The next term must be the largest reciprocal term that does not allow the entire sum to be > 1. The next term also must be < the term before it.
3) The sum must also be permanently less than 1 if you were to stop at a number < . If you were to stop on term , then the sum would amount to exactly 1 by definition.
Now that you know the rules, we must rewrite the summation in a different way in order to propose my question in a reasonable manner.
1 a + 1 b + 1 c + 1 d + 1
As a mote of clarification, a , b , c , d etc. represent 2 , 3 , 7 , 43 etc. respectively.
The question:
-Envision f ( 1 ) = a , f ( 2 ) = b , f ( 3 ) = c, and so on. Keep in mind there are an infinite number of terms, thus an infinite number of variables to represent primes. My question is what function, f ( n ), could hold this property and maintain true no matter what term you are at.
-One who is loving this must remember that a , b , c , d, etc. are integers rather than fractions in this case and therefor f ( 1 ) 1 / a , f ( 1 ) 1 / b, etc. The reciprocal however is the form in which the numbers have that interesting property.
-What is f ( n )? (The primary question)
-When answering this question, do the following, please provide some form of evidence or explain your case. If I missed something obvious, please let me know what that is with a website in the comments.
Have a great day! Thanks!
asked 2022-07-07
Rearranging exponential functions
Using two different strategies, I've derived an equation for a particular function f ( ϕ ). That equation is
f ( ϕ ) = 1 1 e T γ ( 1 e T γ ϕ ) .
However, the paper whose result I'm trying to replicate is telling me that the function is
f ( ϕ ) = ( 1 e γ ) ( 1 e γ ϕ ) .
Is there a way to rearrange the result I got so that it matches the result in the paper?
EDIT:
What's confusing me about this is that the function needs to be such that f ( 0 ) = 0 and f ( 1 ) = 1. This is true of the equation I derived. But for the equation in the paper, f ( 1 ) = 1 only if γ = ln ( 1 / 2 ), which is not the case in the specific example given later in the paper. So something weird is afoot (either on my part or on the paper's).
asked 2022-06-24
Solve quadratic fraction
I would like to simplify the fraction
x 2 2 x x 2 + x 6
I know from Mathematica that it should equal 3 3 x but how do I get there?
asked 2022-02-05
Paul hit 33 tennis balls during practice. He hit 13 of them out of the tennis court. How many balls did he hit out of the court?
asked 2021-11-20
For Exercise, solve the equation.
45x23=710x2
asked 2022-05-28
Ratio of sums vs sum of ratio
Is anyone aware of any general (or perhaps not so general) relationship (inequality for instance) relating
A ( x , y ) = z f ( x , y , z ) z g ( y , z )
and
B ( x , y ) = z ( f ( x , y , z ) g ( y , z ) )
?
Specific context (for what I'm dealing with, but not necessarily the question) is that x , y , x f ( x , y , z ) = 1 and y , x g ( y , z ) = 1 and f ( x , y , z ) 0 and g ( y , z ) 0 x , y , z. I.e. probabilities (or more generally, I guess, measures).
It seems like it could 'vaguely' be related to log sum inequalities (when transformed) or Jensen's inequality perhaps?

New questions