# Can anyone tell me how to expand the follwing in terms of partial fraction?

• May 19th 2006, 06:47 AM
orange890
Can anyone tell me how to expand the follwing in terms of partial fraction?
Can anyone tell me how to expand the follwing in terms of partial fraction?

.....2a^s
------------
(s^4)-(a^4)

I do it like this:

......a.................b..............c
---------- + --------- + ------
s^2 + a^2.........s+a..........s-a

but after I expand it, it is tooooooooo complicated to solve...
• May 19th 2006, 07:44 AM
CaptainBlack
Quote:

Originally Posted by orange890
Can anyone tell me how to expand the follwing in terms of partial fraction?

.....2a^s
------------
(s^4)-(a^4)

I do it like this:

......a.................b..............c
---------- + --------- + ------
s^2 + a^2.........s+a..........s-a

but after I expand it, it is tooooooooo complicated to solve...

1. You are using "a" to mean two different things in the same expression -
never a good idea.

2. This is not the usual type of expression that partial fractions are used
for. What is the background of your question (that is why do you want to
find a partial fraction expansion for this)?

RonL
• May 19th 2006, 08:32 AM
CaptainBlack
Quote:

Originally Posted by orange890
Can anyone tell me how to expand the follwing in terms of partial fraction?

.....2a^s
------------
(s^4)-(a^4)

I do it like this:

......a.................b..............c
---------- + --------- + ------
s^2 + a^2.........s+a..........s-a

but after I expand it, it is tooooooooo complicated to solve...

You could try:

$
\frac{2 a^s}{s^4-a^4}=2 a^s \left\{ \frac{1}{s^4-a^4} \right\}
$
,

then resolve the only the term in braces into partial fractions.

RonL
• May 19th 2006, 10:22 AM
earboth
Quote:

Originally Posted by orange890
Can anyone tell me how to expand the follwing in terms of partial fraction?
.....2a^s
------------
(s^4)-(a^4)

I do it like this:
......a.................b..............c
---------- + --------- + ------
s^2 + a^2.........s+a..........s-a

but after I expand it, it is tooooooooo complicated to solve...

Hello,

I tried this way:

1rst step:

${2a^2\over{s^4-a^4}}={A\over{a^2+s^2}}+{B\over{a^2-s^2}}+{C\over{s^4-a^4}}$

Then you'll get the results: A = B = 1, C = 0

2nd step (Take the second summand to partial fractions):

${1\over{a^2-s^2}}={A\over{a+s}}+{B\over{a-s}}+{C\over{a^2-s^2}}$

Then you'll get the results: A = B = 0, C = 1

Thus the only partial fraction I got is:

${2a^2\over{s^4-a^4}}={1\over{a^2+s^2}}+{1\over{a^2-s^2}}$

Is this sufficient?

Greetings

EB
• May 19th 2006, 10:31 AM
CaptainBlack
Quote:

Originally Posted by earboth
Hello,

I tried this way:

1rst step:

${2a^2\over{s^4-a^4}}={A\over{a^2+s^2}}+{B\over{a^2-s^2}}+{C\over{s^4-a^4}}$

Then you'll get the results: A = B = 1, C = 0

2nd step (Take the second summand to partial fractions):

${1\over{a^2-s^2}}={A\over{a+s}}+{B\over{a-s}}+{C\over{a^2-s^2}}$

Then you'll get the results: A = B = 0, C = 1

Thus the only partial fraction I got is:

${2a^2\over{s^4-a^4}}={1\over{a^2+s^2}}+{1\over{a^2-s^2}}$

Is this sufficient?

Greetings

EB

A'ha - you think the denominator is $2a^2$ not $2a^s$.

RonL
• May 19th 2006, 11:28 AM
earboth
Quote:

Originally Posted by CaptainBlack
A'ha - you think the denominator is $2a^2$ not $2a^s$.

RonL

Hello,

I thought so - until now. But now I'm convinced that I need a new pair of glasses. :cool:

Greetings

EB
• May 19th 2006, 08:00 PM
earboth
last attempt
Quote:

Originally Posted by orange890
Can anyone tell me how to expand the follwing in terms of partial fraction?...

Hello,

I pick up the suggestion of CaptainBlack and my own incomplete solution (please, read all previous replies so that you understand what I'm talking about)

$\frac{2 a^s}{s^4-a^4}= a^{s-2} \left\{ \frac{2a^2}{s^4-a^4} \right\}$

$\frac{2 a^s}{s^4-a^4}= a^{s-2} \left({1\over{a^2+s^2}}+{1\over{a^2-s^2}} \right)$

Greetings

EB
• May 19th 2006, 08:54 PM
CaptainBlack
Quote:

Originally Posted by earboth
Hello,

I thought so - until now. But now I'm convinced that I need a new pair of glasses. :cool:

Greetings

EB

What I was thinking is not that you had misread the question, but
your interpretation was a plausible explanation of what the Original

Nice Avatar.

RonL
• May 20th 2006, 01:19 AM
earboth
Quote:

Originally Posted by CaptainBlack
What I was thinking is not that you had misread the question, but
your interpretation was a plausible explanation of what the Original

Nice Avatar.

RonL

Hello,

1. I am the worldchampion in reading what I want to read. So I'm pretty sure that you are right.

2. The "s" and the "2" are separated on the keyboard by a rather wide space. So you can't mess them up by accident.

3. If the Original POster had mixed up something I would presume that it is meant $a^5\ \mbox{instead of} \ a^s$. (And now I'd rather prefer to get a reaction from the Original Poster...)

4. Thanks for the compliment about my avatar. I noticed that some members of the forum changed their outfit (Rebesque, ..., you too). So I decided to take to the public my canned brain, only to demonstrate why I've sometimes so much difficulties to get my precious thoughts out of this container: There are only very small holes in it.

Best wishes

EB
• May 20th 2006, 01:22 AM
CaptainBlack
Quote:

Originally Posted by earboth
4. Thanks for the compliment about my avatar. I noticed that some members of the forum changed their outfit (Rebesque, ..., you too). So I decided to take to the public my canned brain, only to demonstrate why I've sometimes so much difficulties to get my precious thoughts out of this container: There are only very small holes in it.

But pleanty of room for BIG thoughts :D

RonL