I personally feel this sort of method is very artificial, in fact because of this, I've had disagreements with my teacher. I prefer a more natural approach to Partial fractions, instead of writing the fraction as a sum of multiple fractions with unknowns, I try to get their without any hand waving, like 1/(4-x^2) = 1/(2-x)(2+x) = [(2-x)+(2+x)]\4[(2-x)(2+x)]
Here's the proof of this method. Consider our given fraction to be: N(x)/[(x - p)*R(x)] Where: N(x) is the numerator, that is a polynomial p is the pole of interest R(x) is the remaining polynomial in the denominator, after you've factored out (x - p). Note that x=p is not a root of R(x). This is the proof with only a distinct pole at x=p. You could construct an extended version of this proof for a repeated pole at x=p as well. The partial fraction expansion will be: A/(x - p) + Q(x)/R(x) Where: A is the unknown coefficient for the pole of interest Q(x) is a polynomial with unknown coefficients of a degree one less than R(x). Equate: N(x)/[(x - p)*R(x)] = A/(x - p) + Q(x)/R(x) Multiply by (x - p): N(x)/R(x) = A*(x - p)/(x - p) + Q(x)/R(x) * (x - p) Take the limit as x approaches p. This make the Q(x)/R(x) term approach zero. The expression (x-p)/(x-p) evaluates to 1, as long as x isn't exactly equal to p. But in the limit, it will equal 1. The coefficient A approaches what remains of the given fraction, evaluated at x=p: A = N(p)/R(p) And this is exactly what Heaviside Coverup tells us we can do.
Waiting for a blooper video. They are my favorite because they show hard work behind a small video.
He's leaving the bloopers in, but correcting as he goes.
loved the quick way to find A, B, and C. this was the most confusing part for me but you made it easier!
Thank you! Working hard to pass Calculus 2. You remind me of my old math tutor… Very happy to have you!
First!
I like your way for factor trinomial, Great.
Now, I'm waiting for your next one.
Thank you
My school teacher didn't explain it well, so I came here...and guess what...I'm going back with a new method of integration....
Thanks a lot...
Great Video.....
u are doing god´s work
The legendary teacher mr. BPRP is here to teach partial fraction.
4: 31
Did he forget to cut the video?
I also think so.
It's too real, just like you're present in class. I like it.
I personally feel this sort of method is very artificial, in fact because of this, I've had disagreements with my teacher. I prefer a more natural approach to Partial fractions, instead of writing the fraction as a sum of multiple fractions with unknowns, I try to get their without any hand waving, like 1/(4-x^2) = 1/(2-x)(2+x) = [(2-x)+(2+x)]\4[(2-x)(2+x)]
Here's the proof of this method.
Consider our given fraction to be:
N(x)/[(x - p)*R(x)]
Where:
N(x) is the numerator, that is a polynomial
p is the pole of interest
R(x) is the remaining polynomial in the denominator, after you've factored out (x - p). Note that x=p is not a root of R(x).
This is the proof with only a distinct pole at x=p. You could construct an extended version of this proof for a repeated pole at x=p as well.
The partial fraction expansion will be:
A/(x - p) + Q(x)/R(x)
Where:
A is the unknown coefficient for the pole of interest
Q(x) is a polynomial with unknown coefficients of a degree one less than R(x).
Equate:
N(x)/[(x - p)*R(x)] = A/(x - p) + Q(x)/R(x)
Multiply by (x - p):
N(x)/R(x) = A*(x - p)/(x - p) + Q(x)/R(x) * (x - p)
Take the limit as x approaches p. This make the Q(x)/R(x) term approach zero. The expression (x-p)/(x-p) evaluates to 1, as long as x isn't exactly equal to p. But in the limit, it will equal 1. The coefficient A approaches what remains of the given fraction, evaluated at x=p:
A = N(p)/R(p)
And this is exactly what Heaviside Coverup tells us we can do.