Let’s talk about partial fractions and what I’ve found out after programming them. Partial fractions would never occur in real life. Remember the integral is the area under a smooth curve, nothing more from my knowledge, when you graph any original function; it has at least 2 asymptotes. Well that eliminates the smooth curve. Remember an asymptote is a vertical line or sometimes horizontal line where the original function never touches to infinity, so if you choose a range that goes (crosses) over that asymptote there is no computed answer, so no area under a smooth curve. So partial fractions are a Sudoku of math problem, like so many of calculus problems. So let’s get into what I’ve found in programming this area of calculus. The denominator has to be factored to produce partial fractions. Sometimes there are two factors, then three, then two factors with an exponent in between the parenthesis, then even one factor with an exponent outside the parenthesis. The factored denominator is the key as to how to approach the problem.
In my programs, I have a whole separate program to decide if there are two factors or three or any other choices. Most of us are not good at factoring. Maybe simple functions we can, like the difference of squares (x^2-9) = (x-3)(x+3). When it gets a little deeper that this most of us are lost. For instance, what’s the factors of (x^3-x) or (x^4+7x^3+6x^2)? The calculator knows and because of that, so does my programs. But would those appear in your test? In my opinion if they did then most of the class would fail that. No question about it. Let alone complete the partial fractions. So enough about factoring. Remember in algebra when they said you could do anything to one side of an equation as long as you do it to the other side also. They do this in partial fractions. On the left side of the equation in partial fractions they multiply the original function by the denominator, which effectively gets rid of the denominator. Well they do the same thing on the other side (except factored) to get rid of the added or subtracted fractions, and change the A / (x-1) to a product through common denominator. That said, a few trick to remember in partial fractions. If in the factored denominator you have (x^2+6) for the first factor so you want to set that up for partial fractions you would go Ax+B / (x^2+6), if the second factor was (x^2-3) you’d set this up as (Cx+D) / (x^2-3), Do you understand this? Very important and no exceptions!! If the factor turned out to be (x-5)^5(exponent outside of the parenthesis), this is called re-occurring powers, or exponents, and the factors are A/(x-5)^1 + B/(x-5)^2+C/(x-5)^3+E/(x-5)^4+F/(x-5)^5. Now if the power in the denominator is less or equal to the power in the numerator, you have to use short division to find the remainder to the partials.
My programs do all of this for you, as I required my programming to do for me also when I was in class. Hey good luck in your class, I’m always available to help you if you just ask.
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