Comment by RiderOfGiraffes | Hacker News Reader

Comment by RiderOfGiraffes | Hacker News Reader

Parent Full thread

RiderOfGiraffes

You've got real problems with brackets. Major problems that are going to cause significant difficulties down the line. I'll bet the square brackets are necessary, and you need more brackets for the numerator.

Your number 1 above you quote as:

  x/(x+1)(x-4)

I'll bet you don't mean that. When you have multiplication and division at the same level, you evaluate left-to-right. The above is the same as:

  [ x / (x+1) ] (x-4)

which is the same as:

  x (x-4) / (x+1)

which I'll bet is not what you intended.

The second - as you quote it - is:

    3x+1/x^2-6x+8

Division binds more closely than addition and subtraction, while addition and substraction are then read from left-to-right. What you quote is therefore the same as:

    3x + 1/x^2 - 6x + 8

which is

    3x + (1/x^2) - 6x + 8

which simplifies to

    (1/x^2) - 3x + 8

I'll bet that's not what you meant.

Now let's turn to the first of the ones you originally asked about:

    3x-1/[(x+2)(1-x+x^2)]

I'll bet you mean:

    (3x-1) / [(x+2)(1-x+x^2)]

and I'll bet the square brackets are necessary. And you are right that all you need to do is solve for A and B after writing it as A/(x+2) + B/(1-x+x^2).

So remembering that a/b + c/d = (ad+bc)/bd, what does

    A/(x+2) + B/(1-x+x^2)

equal?

all right the brackets concept makes sense.

remembering a/b + c/d = (ad+bc)/bd

A/(x+2) + B/(1-x+x^2) = A(1-x+x^2) + B(x+2)/(x+2)(1-x+x^2)

correct??

Yes, except you forgot the brackets again. You should have:

    [ A(1-x+x^2) + B(x+2) ] / [ (x+2)(1-x+x^2) ]

You probably knew that, but this is really, really important. Not getting it right now will confuse you a great deal later.

However, now you have this as your numerator:

    A(1-x+x^2) + B(x+2)  [Eqn *]

Look again at your original problem

- what do you want the numerator to be?

- Expand out Eqn * to remove the brackets.

- What values can you assign to A and B to make Eqn * be the numerator you want?

As a hint, A and B are not simply numbers.

first of all i don't know the significance of replacing the numerator with capital letters when decomposing partial fractions.i'am not aware of the practical implications of these problems.all i know is it'll come in handy when solving laplace transforms in the future which is used in physics/electrical etc.(that's what i'am after). and if you are curious to know why i wanna learn these..well i don't have a proper reason.i simply want to!!

i looked back at the original problem but i'am unable to figure out the values to assign to A and B.

OK, to recap. You want to decompose this:

    [ 3x-1 ] / [ (x+2)(1-x+x^2) ]

into fractions. In other words, you want to find A and B such that:

    A/(x+2) + B/(1-x+x^2) = [ 3x-1 ] / [ (x+2)(1-x+x^2) ]

Apart from getting the brackets wrong, you said (correctly) that the left hand side is equal to this:

    [ A(1-x+x^2) + B(x+2) ] / [ (x+2)(1-x+x^2) ]

So now you need to solve:

[ A(1-x+x^2)+B(x+2) ] / [ (x+2)(1-x+x^2) ] = (3x-1) / [ (x+2)(1-x+x^2) ]

The denominators are equal, so you just need to make the numerators equal.

So tell me - what equation do you have to solve?

i get A(1-x+x^2) + B(x+2) = 3x-1

Excellent.

+ Expand B(x+2)

+ Expand A(1-x+x^2)

What do you get? You need to play with these sorts of equations and see what happens - see what you get.

expand B(x+2)?? but how??

How would you expand 3(x+2)?

3x+6

Or, before simplifying, 3x+3.2 (using a dot for multiplication)

So how do you expand B(x+2) ?

Bx+B2 ??? now it's getting funny!!!!!!!!

So going back, you need to solve: [ A(1-x+x^2)+B(x+2) ] / [ (x+2)(1-x+x^2) ] = (3x-1) / [ (x+2)(1-x+x^2) ]

As I said, the denominators are already equal, so you need to make the numerators equal.

That means you need to make this:

    A(1-x+x^2)+B(x+2)

equal to (3x-1). Expand this - you already know that the second part becomes B.x+2B - and have a look.

Then think - what can you set A and B to be to make it equal to 3x-1?

What if you set A to be x?

What if you set B to be 2?

What if you set A to be -3?

What if you set B to be 2x?

Try those, try more, see if you can work out how you can get 3x-1. Here's a hint: when you expand you'll get one of the terms as Ax^2. How can you get that to cancel out?

Try things. Tell me what you see, tell me what you get.

i've started out with calculus.(it;s been 30 hours now) and i'am enjoying it as i'am able to understand.so my plan is to continue this partial fraction problem when i reach integral calculus involving partial fractions and eventually laplace transforms. and you said my algebra skills sucks.well i partialy agree with you cause i wouldn't be able to solve limits and differentiate functions without a grasp on algebra(precalculus for namesake). as a matter of fact i've never been good at anything .i suck in everything that i do.!!!!!!!!!

Just to add another question - why is that funny? A and B represent polynomials - why should this be funny to you?

the fun part has got nothing to do with polynomials.i'am finding it funny cause i'am taking ages to learn this basic concept .to put in better sentence a DUMB ASS(that's me) taking advice from a knowledgeable person(that's you)

by the way thanks very much for telling me that 'A' and 'B' represents polynomials.never knew it before.

It's not necessarily that you're dumb - I suspect you aren't. The problem is that you're trying to get a grip on these techniques when you haven't really mastered the absolute basics of algebra.

It's not surprising - you're trying to teach yourself, and it's always going to be that there are gaps. You need properly constructed tutorials and exercises.

Have a look at these:

http://www.google.co.uk/#q=algebra+tutorial+with+exercises

Pick one of the tutorials and work through it completely. Then you can do a google search for partial fraction tutorials and do the same.

i've started out with calculus.(it;s been 30 hours now) and i'am enjoying it as i'am able to understand.so my plan is to continue this partial fraction problem when i reach integral calculus involving partial fractions and eventually laplace transforms. and you said my algebra skills sucks.well i partialy agree with you cause i wouldn't be able to solve limits and differentiate functions without a grasp on algebra(precalculus for namesake). as a matter of fact i've never been good at anything .i suck in everything that i do.!!!!!!!!!

by the way i'am not offended.i'am enjoying myself with calculus.i'll get back to you if i have any doubts.if you got to say something you can.

I didn't say your algebra skills suck - I say there are gaps. You don't put brackets where you should, and you don't seem to be able to manipulate equations quickly and easily. But you have some skills, and if you're only just starting out then you're probably doing well. I can't tell from this distance via this medium.

As a final hint, here's something interesting:

  For a polynomial p(x), if p(a)=0 for some value a,
  then (x-a) is a factor of p(x).

  That means, for example, that (x+1) is a factor of
  x^3+1 and (x-1) is a factor of x^3-1.

I have to go. Good luck!

Reply on news.ycombinator.com