Partial Fractions - Deduce the Equation for given fractions

In summary: Daniel: Uh, no, the question specifically said to solve for A and B.Marlon: sorry, I misunderstood. Thanks for your help.
  • #1
whkoh
29
0
Given

[tex]\frac{2+5x+15x^2}{\left (2-x\right )\left (1+2x^2\right )}=\frac{8}{2-x} + \frac{x-3}{1+2x^2}
[/tex]

I am asked to deduce the partial fractions of:

[tex]\frac{1+5x+30x^2}{\left (1-x\right )\left (1+8x^2\right )}[/tex]

I can solve it using my usual method, but that's not what the question requires. Any help? I can't see the link between the two.
 
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  • #2
what is your USUAL method, and can people help you without knowing how UNUSUAL is not your USUAL?

[tex]\frac{1+5x+30x^2}{\left (1-x\right )\left (1+8x^2\right )}=\frac{A}{1-x} + \frac{B}{1+8x^2}[/tex]

and solve for A and B... this is the standard approach, is my method UNUSUAL?
 
  • #3
i have a not so usual method... substitude x=2u in your first equation... and you will get the second one. is this what you looking for?
 
  • #4
Yes, that's my usual way of solving such questions. But I still don't get it...
 
  • #5
vincentchan said:
what is your USUAL method, and can people help you without knowing how UNUSUAL is not your USUAL?

[tex]\frac{1+5x+30x^2}{\left (1-x\right )\left (1+8x^2\right )}=\frac{A}{1-x} + \frac{B}{1+8x^2}[/tex]

and solve for A and B... this is the standard approach, is my method UNUSUAL?


This is wrong
it should be :
[tex]\frac{1+5x+30x^2}{\left (1-x\right )\left (1+8x^2\right )}=\frac{A}{1-x} + \frac{Bx +C}{1+8x^2}[/tex]

Now determin A,B and C by adding up these fractions and then compare the numerator with the given fraction...

marlon
 
Last edited:
  • #6
Marlon: thanks for your help, but the question specified that I deduce and not solve.

Vincent: substituting x=2u works. Thanks.
 
  • #7
whkoh said:
Marlon: thanks for your help, but the question specified that I deduce and not solve.

Vincent: substituting x=2u works. Thanks.

i know, but that answer is already given. I just wanted to point out that the given partial fractions were wrong,...that's all

marlon
 
  • #8
me wrong?
did I say B is constant?
 
  • #9
Compute A and B with your formula...

Daniel.
 

Related to Partial Fractions - Deduce the Equation for given fractions

1. What are partial fractions?

Partial fractions are a method used in algebra to simplify complex fractions into smaller, more manageable fractions. This is especially useful when dealing with equations involving polynomials.

2. How do you deduce the equation for given fractions?

The process of deducing the equation for given fractions involves breaking down a complex fraction into smaller fractions with known denominators, and then equating the numerators of the smaller fractions to the numerators of the original complex fraction. This allows us to solve for the unknown coefficients in the smaller fractions.

3. What is the purpose of using partial fractions?

The main purpose of using partial fractions is to simplify complex algebraic equations and make them easier to solve. This method is especially useful in integration, where it allows us to transform a rational expression into a sum of simpler fractions that can be integrated separately.

4. Can you explain the process of finding the coefficients in partial fractions?

To find the coefficients in partial fractions, we first break down the complex fraction into smaller fractions with known denominators. Then, we equate the numerators of the smaller fractions to the numerators of the original complex fraction. This creates a system of equations that can be solved using algebraic techniques to find the unknown coefficients.

5. Are there any limitations to using partial fractions?

Partial fractions can only be used when the denominator of the given fraction can be factored into linear and irreducible quadratic factors. In addition, this method is only applicable to rational functions, so it cannot be used for other types of equations such as trigonometric or exponential functions.

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