20050117, 05:44  #1 
Jan 2005
1101_{2} Posts 
Partial fraction of this expression?please!!
hi everyone
i have a proplem with this excercise: f(t)= (t+1)/(t^2 + 8t + 16) Here what i have: (t+1)/(t^2 + 8t + 16) = (t+1)/(t+4)^2 (t+1)/(t+4)^2 = K1/(t+4) + K2/(t+4)^2 i am stuck from here, anyone please help me, thanks first 
20050117, 06:21  #2 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
Hi tinhnho
(t+1)/(t+4)^2 = K1/(t+4) + K2/(t+4)^2 Continue by multiplying both sides by (t+4)^2: t+1 = K1*(t+4) + K2 t+1 = (K1*t) + (4K1 + K2) Equating coefficients for t, we have K1 = 1 Equating constant coefficients, we have: 4K1 + K2 = 1 4 + K2 = 1 K2 = 3 Thus, the partial fraction representation is: f(t) = 1/(t+4)  3/(t+4)^2 Mathematica agrees. 
20050117, 06:43  #3 
Jan 2005
13 Posts 
Excellent job jinydu. BTW, i have searched around on internet to find the site for partial fraction, but cann't find it. I just wonder if you have any site which talk about this thing, because i need to review. Have a good day

20050117, 19:44  #4 
Dec 2003
Hopefully Near M48
2×3×293 Posts 
Note: I'm only allowed 10,000 characters in a post. This post takes over 13,000 characters, which is why I'm splitting it into two posts.
I don't really know of any good site that teaches how to do partial fractions. The most I can say is to just try a Google Search for "Partial Fractions", or something like that. But remember that in most Google searches, the overwhelming majority of results are irrelevant to what you're looking for. Instead, I can give you my own strategy for partial fractions (with some of my comments along the way): The objective of a partial fractions problem is to "split up" a rational function into a sum of simpler rational functions. Remember that a rational function is defined as a quotient of two polynomial, where the polynomial in the denominator is not equal to 0. "Partial fractions decomposition" is the reverse of putting a sum of fractions under a common denominator. However, as with many other problems in math (multiplying natural numbers vs. factoring natural numbers), "splitting up" is much more difficult than "combining". Aside from being an interesting problem in its own right, partial fraction decomposition is useful in calculus because integrating a product is hard, whereas integrating a sum is easy. First, I'll give an outline of the method if you are interested in working with real numbers only: 1) If necessary, use polynomial division so that the degree of the numerator is less than the degree of the denominator. 2) Factor the denominator into a product of linear and quadratic factors. 3) Set the fraction equal to a sum of simpler fractions. For each linear factor of the denominator of the expression to be split up, make a term with that linear factor in the denominator and an unknown constant in the numerator. For each quadratic factor of the denominator of the expression to be split up, make a term with that quadratic factor in the denominator and a linear expression with two unknown constants (one as a coefficient of x, one as a constant coefficient) in the numerator. If there are repeated linear factors, make a seperate term for each power of the linear factor and place an unknown constant in the numerator of each such term. If there are repeated quadratic factors, make a seperate term for each power of the quadratic factor and place a linear expression (with both coefficients as unknown constants) in the numerator of each. 4) Multiply through by the denominator of the expression to be split up. You should now get an expression without fractions, since all the denominators will cancel out. There are two ways you can do the next step 5a) Expand out all the parentheses on the "sum of factors" side and collect terms according to the power of the variable. Equate coefficients with the expression on the other side of the equation. You then have a system of linear equations where the variables are the unknown constants. Solve for those constants 5b) The equation you have is an identity. Therefore, it should hold true for all (real) values of the variable. Choose a value for the variable such that most of the terms cancel out. Then, solve for the constant(s) that didn't get crossed out. Repeat by choosing different values for the variable, or switch to 5a. Keep going until you've solved for all of the unknown constants. 6) Substitute the values of the constants back into the partial fractions expression you had before, and you're done. Now, I'll make some comments on the steps: 1) If you have mastered long division and/or synthetic division, this step is simple and straightforward. Just remember that you can normally expect to end up with a remainder. Stop when you get to this point, and go to step 2. If you don't end up with a remainder, you're already done! As far as I know, the problems assigned by most teachers don't even have this step. That is, in the fraction they give you, the degree of the denominator is already greater than the degree of the numerator. 2) Despite the simple description, this step is by far the most difficult step in the process, at least in a theoretical sense. Factoring a polynomial into linear and quadratic factors is, in principle, as hard as finding the roots of the polynomial. If (x  a) is a factor of a polynomial, then a is a root of the polynomial (since setting x = a makes the factor equal to 0, and 0 times anything equals 0). If (x^2 + ax + b) is a factor of a polynomial, it is also simple to find roots (just set the quadratic factor equal to 0, and use the Quadratic Formula). The reverse process (going from the roots to the quadratic factor) is also easy. For a polynomial with real coefficients, complex roots always come in pairs, where each root in the pair is the conjugate of the other. So just write (x  r(1))*(x  r(2)). When you multiply that out, you get a quadratic with real coefficients. Therefore, if you can factor a polynomial into linear and quadratic factors, from then on, it is easy to find the roots, at least theoretically. Gauss' Fundamental Theorem of Arithmetic assures us that every nth degree polynomial does have n (not necessarily distinct) complex roots. But unfortunately, finding the roots of a polynomial is, in general, extremely difficult, at least if you insist on finding the exact values of the roots. For n = 1 (linear), 2 (quadratic), 3 (cubic) and 4 (quartic), there are exact formulas for the roots in terms of +, , *, /, raising to powers and taking nth roots. But in the 1820s, two mathematicians named Abel and Galois independently proved that there is no such formula for quintics (fifth degree polynomials) and higher. This does not mean that all quintics and higher polynomials are unsolvable (ex. x^5  1 = 0 is easy to solve), but some are. Later in the 19th century, mathematicians like Hermite discovered how to solve the general quintic, but only in terms of "Jacobi theta functions", really complicated functions that are written as infinite sums of exponential expressions. Things are even worse for 6th degree polynomials and higher. But if you are satisfied with an approximate factorization, there are techniques (ex. Newton's Method) that can easily calculate roots to thousands or even millions of decimal places using today's computers. Therefore, if the polynomial in the denominator is fifth degree or higher, you may not be able to factorize the polynomial exactly, at least, not in terms of functions that anyone except the most advanced mathematicians knows about. But it should be noted that most teachers have a habit of giving "artificial" problems, problems that have a special property that makes them far easier to solve than "realistic" problems. Problems are often "rigged" so that they have a particularly simple solution. I've never been assigned to factor an unsolvable polynomial before (at least, not by hand). In fact, in the partial fractions problems given by most teachers, the denominator is almost always a quadratic, and in the rare cases where it is a cubic, one of the factors is almost always obvious. Teachers generally prefer to concentrate most of the effort in the later parts of the problem (usually solving for the unknown coefficients). I'm not saying that this is bad (since it helps emphasize the process as a whole, rather than getting hung up on a single very difficult step), but I'm just pointing out that "real" partial fractions problems are usually not like that. 3) Yes, I know that the description for this step is long and probably confusing. But in fact, it is the second easiest step. It requires no actual calculation. All you're doing is writing down an expression. If you successfully completed step 2, you'll know all the factors of the denominator. Just write down the fractions in the sum according to the instructions. 4) Instead of multiplying by the original form of the denominator, it might be simpler to multiply by its factored form. Of course, the denominator of the original fraction will cancel out completely. For the fractions on the other side of the equation, the denominator will cancel with some of the factors (exactly one, if the denominator of your original fraction had no repeated roots). In any case, once the multiplication is done, there will be no denominators left in your equation. 5a) This method is more general than 5b, in the sense that it always works while 5b might not give you a complete solution, although it can be somewhat more complicated. Still, you ultimately end up solving a system of linear equations in terms of the unknown coefficients. In principle, this step can always be done using nothing harder than +, , * and /. Thus, this step is simple in theory. But in the artificial examples that teachers give, this is where most of the effort ends up. 5b) The main purpose of this procedure is to simplify the system of equations mentioned in 5a. In "real" problems, you will often find it difficult to make all but one of the unknown constants disappear. However, since most classroom problems are easily factorable quadratics, this method usually works splendidly on the problems that teachers give you. Typically, if A and B are your unknown constants, you'll end up with something that looks like: (x + a) = A*(x + b) + B*(x + c) Then, setting x = c allows you to solve for A and setting x = b allows you to solve for B. If your original expression had a quadratic in the denominator, things usually don't get much harder than this. But just keep in mind that when you have a cubic or higher in the denominator, things rarely cancel out so nicely. 6) This step is the simplest of them all. Go back to the equation you got for step 3, and substitute in the values of the constants that you solved for in step 5. Once this is done, you have your partial fractions decomposition! 
20050117, 19:45  #5 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
This is the second part of the post.
Note that the above process applies only when you want to stay in the real numbers. But the original rational function contains complex coefficients (or when you enjoy working with complex numbers, and your teacher allows it), you can actually go a step further. Maybe you're wondering: Why stop at quadratic factors? After all, Gauss proved that an nth degree polynomial always has n roots. On top of that, there's an easy way to find the two roots of any quadratic: Just use the Quadratic Formula. The answer is that the Quadratic Formula sometimes gives roots that involve the square root of a negative number, which is not allowed in the real number system. But this can also happen with higher degree polynomials, so why give the quadratic a special status? The answer is that with higher degree polynomials, its always possible to split up the polynomial further using only real numbers (although it might not be possible using functions you know about, see my comments on step 2). If the polynomial has an oddnumber degree (3, 5, 7, 9, etc.), and all of its coefficients are real, its not hard to show that it will always have at least one real root. We can then factor out this root (neglecting the difficulty of actually finding this root, see comments on step 2), reducing the polynomial to an evendegree polynomial. It can also be shown that an evendegree polynomial can be written as a product of quadratic factors (again, neglecting the difficulty I pointed out for step 2), which may or may not be further factorable into linear factors using only real numbers. But as I've said before, if you use complex numbers, you can go a step further. If you have the quadratic factor (x^2 + ax + b), it is always possible to find the roots of that factor if you use complex numbers. Then, the factorization is just (x  r(1))*(x  r(2)), where r(1) and r(2) are the (possibly complex) roots of the quadratic factor. In some sense then, the problem actually becomes simpler because you don't have to worry about quadratic factors any more. Also, all the numerators of your sum are now just constants, instead of some being linear expressions with two constants each. Just keep in mind that the final expression may contain complex numbers, even if your original problem only had real numbes. Here's a simple example: Find the partial fraction decomposition of 3/(x^2 + 1) In the complex numbers, x^2 + 1 factors into (x + i)(x  i) 3/(x^2 + 1) = A/(x + i) + B/(x  i) Multiplying both sides by (x + i)*(x  i): 3 = A*(x  i) + B*(x + i) Using the method of step 5b, we can let x = i: 3 = A*(i  i) + B*(i + i) 3 = A*(0) + B*(2i) 3 = (2i)*B B = 3/(2i) = 3/2 * 1/i Since 1/i = i: B = 3i/2 Back in our equation: 3 = A*(x  i) + B*(x + i), we can expand the righthand side: 3 = Ax  Ai + Bx + Bi 3 = (A + B)x + (Ai + Bi) Since the x coefficient on the lefthand side is 0 (i.e. its not there): A + B = 0 A  3i/2 = 0 A = 3i/2 Therefore, our partial fraction decomposition is: 3/(x^2 + 1) = (3i/2)/(x + i)  (3i/2)/(x  i) If you want only one fraction bar in each term, you can simplify this further: 3/(x^2 + 1) = 3i/(2x + 2i)  3i/(2x  2i), and you're done! Notice that if you were limited to real numbers, you wouldn't be able to do anything with the original fraction, 3/(x^2 + 1), since x^2 + 1 = 0 has no real number solution. 
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