Where am i going wrong? partial fraction question

In summary: Homework Statement Doing a homework question and I ran it through wolfram and I get a different answer to what I'm working it out as...and I can't see where I'm going wrong. Anyone able to give a pointer?my equation is \frac{1}{s(0.641s + 1)}Wolfram gives the answer as\frac{1}{s}-\frac{1}{s+1.56006}Homework EquationsThe Attempt at a SolutionUsing partial fraction decomposition I get (using cover up) -\frac{1}{s(0.641s + 1)} = \frac{A}{s} + \frac{B}{0
  • #1
braceman
30
0

Homework Statement



Doing a homework question and I ran it through wolfram and I get a different answer to what I'm working it out as...and I can't see where I'm going wrong. Anyone able to give a pointer?

my equation is
[tex]\frac{1}{s(0.641s + 1)}[/tex]


Wolfram gives the answer as

[tex]\frac{1}{s}-\frac{1}{s+1.56006}[/tex]


Homework Equations





The Attempt at a Solution



Using partial fraction decomposition I get (using cover up) -

[tex]\frac{1}{s(0.641s + 1)} = \frac{A}{s} + \frac{B}{0.641s+1}[/tex]

[tex]A = \frac{1}{s(0.641s + 1)} ~~~~~~ where~s~ = 0[/tex]

[tex]=\frac{1}{0.641.0 +1}[/tex]

[tex]=\frac{1}{1}[/tex]

[tex]∴A=1[/tex]

[tex]B = \frac{1}{s}[/tex]

[tex](0=0.641s+1[/tex]

[tex]-1=0.641s[/tex]

[tex]s=\frac{-1}{0.641}[/tex]

[tex]s=-1.56)[/tex]

[tex] ∴B=\frac{1}{s}[/tex]

[tex] ∴B=\frac{1}{-1.56}[/tex]

[tex] ∴B=-0.641[/tex]

[tex] Inserting~A~and~B~into~the~decomposition[/tex]

[tex]\frac{1}{s} + \frac{-0.641}{0.641s+1}[/tex]

which I get to be

[tex]\frac{1}{s} - \frac{1}{s+1}[/tex]

Where am I making my schoolboy error? It's my first go at partial fractions and my textbook aint exactly crammed with examples. Anyone able to point out if I've made a glaring error or just something stupid...any help would be appreciated.
 
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  • #2
You might want to look again at setting s = 0. Given it's a factor of the denominator.
 
  • #3
braceman said:

Homework Statement



Doing a homework question and I ran it through wolfram and I get a different answer to what I'm working it out as...and I can't see where I'm going wrong. Anyone able to give a pointer?

my equation is
[tex]\frac{1}{s(0.641s + 1)}[/tex]


Wolfram gives the answer as

[tex]\frac{1}{s}-\frac{1}{s+1.56006}[/tex]

Let's begin by factoring ##.641## out of the denominator:$$
\frac{1}{s(0.641s + 1)}=\frac 1 {.641}\cdot \frac 1 {s(s+\frac 1 {.641})}=
1.56\cdot \frac 1 {s(s+1.56)}=
\frac a {s(s+a)}$$where ##a=1.56##. You want the fraction in that form to do the "cover up" method, from which you get$$
\frac 1 s + \frac {-1}{s+a}$$which agrees with Wolfram. I suggest you take it from there.
 
  • #4
braceman said:
[tex]\cdots + \frac{-0.641}{0.641s+1}[/tex]

which I get to be

[tex]\cdots - \frac{1}{s+1}[/tex]

Where am I making my schoolboy error?

There. If you wanted to divide the top and bottom by 0.641, you forgot one of the terms in the fractiom.

You might want to look again at setting s = 0. Given it's a factor of the denominator.
Ignore that. Maybe PeroK doesn't understand the "cover up rule".
 
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  • #5
AlephZero said:
Ignore that. Maybe PeroK doesn't understand the "cover up rule".

If the cover up rule allows you to divide by 0 and then tweak the distributive law, then I certainly don't understand it.

[tex]A = \frac{1}{s(0.641s + 1)} ~~~~~~ where~s~ = 0[/tex]

[tex]=\frac{1}{0.641.0 +1}[/tex]

[tex]=\frac{1}{1}[/tex]

That's simply two algebraic wrongs making a right!
 
  • #6
If A/x+ B/(x- a)= C/(x(x- a))
Then multiplying by x(x- a) we have A(x- a)+ Bx= C.
There is now no denominator so we can, without difficulty, let x= 0 and x= a getting
x= 0, -aA= C so A= -C/a and
x= a, aB= C so B= C/a.

Notice that
[tex]\frac{(-C/a)}{x}+ \frac{C/a}{x- a}= \frac{(-(C/a)x+ C}{x(x- a)}+ \frac{(C/a)x}{x(x= a)}= \frac{C}{x(x- a)}[/tex]
as desired.
 
  • #7
Thanks for the replies...I realized where I went wrong suddenly, ie- had to make it 'standard form' and needed a 1 on the numerator instead of the 0.641...so manipulated the equation and it all fell into place.

So once again, thankyou all for the help.
 
  • #8
braceman said:

Homework Statement



Doing a homework question and I ran it through wolfram and I get a different answer to what I'm working it out as...and I can't see where I'm going wrong. Anyone able to give a pointer?

my equation is
[tex]\frac{1}{s(0.641s + 1)}[/tex]


Wolfram gives the answer as

[tex]\frac{1}{s}-\frac{1}{s+1.56006}[/tex]


Homework Equations





The Attempt at a Solution



Using partial fraction decomposition I get (using cover up) -

[tex]\frac{1}{s(0.641s + 1)} = \frac{A}{s} + \frac{B}{0.641s+1}[/tex]

[tex]A = \frac{1}{s(0.641s + 1)} ~~~~~~ where~s~ = 0[/tex]

[tex]=\frac{1}{0.641.0 +1}[/tex]

[tex]=\frac{1}{1}[/tex]

[tex]∴A=1[/tex]

[tex]B = \frac{1}{s}[/tex]

[tex](0=0.641s+1[/tex]

[tex]-1=0.641s[/tex]

[tex]s=\frac{-1}{0.641}[/tex]

[tex]s=-1.56)[/tex]

[tex] ∴B=\frac{1}{s}[/tex]

[tex] ∴B=\frac{1}{-1.56}[/tex]

[tex] ∴B=-0.641[/tex]

[tex] Inserting~A~and~B~into~the~decomposition[/tex]

[tex]\frac{1}{s} + \frac{-0.641}{0.641s+1}[/tex]

which I get to be

[tex]\frac{1}{s} - \frac{1}{s+1}[/tex]

Where am I making my schoolboy error? It's my first go at partial fractions and my textbook aint exactly crammed with examples. Anyone able to point out if I've made a glaring error or just something stupid...any help would be appreciated.

If you do it properly, it is more-or-less straightforward. You went at it the wrong way, so made it longer and more complicated than necessary.

Write
[tex] \frac{1}{s(1+as)} = \frac{A}{s} + \frac{B}{as+1}[/tex]
where ##A,B## are constants. Put the above over a common denominator:
[tex] \frac{A}{s} + \frac{B}{as+1} = \frac{A(as+1)+Bs}{s(as+1)}[/tex]
This supposed to equal ##1/[s(as+1)] ## for ALL ##s##, so the numerators much match exactly (because the denominators match). The numerator is ##A(as+1)+ Bs = (Aa+B)s + A##, so we need
[tex] (Aa+B)s + A = 1[/tex]
for ALL ##s##. The polynomial ##P(s) = (Aa+B)s + A## must satisfy ##P(s) = 1## for all ##s##; that means its constant term must = 1 and the coefficient of ##s## must vanish, so ##A = 1## and ##Aa+B = 0##. In other words, we have
[tex] \frac{1}{s(1+as)}= \frac{1}{s} - \frac{a}{as+1}[/tex]
 
  • #9
PeroK said:
If the cover up rule allows you to divide by 0 and then tweak the distributive law, then I certainly don't understand it.

[tex]A = \frac{1}{s(0.641s + 1)} ~~~~~~ \text{where }s~ = 0[/tex]

[tex]=\frac{1}{0.641.0 +1}[/tex]

[tex]=\frac{1}{1}[/tex]

That's simply two algebraic wrongs making a right!
The OP made an algebra error, and the righthand side should have been
$$\frac{1}{0.641 s + 1}$$ which is defined for ##s=0##.

(Of course, to get to that point, you multiply the original equation by ##s## and therefore are assuming that ##s\ne 0##, but that's a minor complication.)
 

FAQ: Where am i going wrong? partial fraction question

1. What are partial fractions and when are they used?

Partial fractions are a method used to simplify and solve complex algebraic expressions, particularly rational functions. They are typically used in integration, solving differential equations, and simplifying equations for further manipulation.

2. How do I know if I need to use partial fractions in my problem?

If you have a rational function with a denominator that can be factored into linear or quadratic terms, then you may need to use partial fractions to simplify it. Additionally, if you are integrating a rational function, you will need to use partial fractions to solve it.

3. How do I set up the partial fraction decomposition?

The general form for partial fraction decomposition is:
- For distinct linear factors: A/(x-a)
- For repeated linear factors: A/(x-a) + B/(x-a)^2
- For distinct quadratic factors: (Ax+B)/(x^2+bx+c)
- For repeated quadratic factors: (Ax+B)/(x^2+bx+c) + (Dx+E)/(x^2+bx+c)^2
To find the coefficients A, B, C, etc. you can use the method of equating coefficients or plug in values for x to solve for the coefficients.

4. What are some common mistakes when solving partial fraction problems?

Some common mistakes when solving partial fraction problems include:
- Forgetting to consider all possible factors when setting up the decomposition
- Making errors when equating coefficients
- Forgetting to include all terms in the partial fraction decomposition
- Not simplifying the fractions before solving
- Making errors when integrating the decomposed fractions

5. Are there any tips for solving partial fraction problems more efficiently?

Yes, some tips for solving partial fraction problems more efficiently include:
- Familiarizing yourself with the general form and patterns for partial fraction decomposition
- Checking your work by plugging in values for x and comparing the original function to the decomposed fractions
- Using algebraic techniques to simplify the fractions before integrating
- Practicing and becoming comfortable with the method of equating coefficients
- Being patient and double-checking your work to avoid common errors.

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