math1510e & h Week 10-Day 1 Topics covered ? Partial fraction

101373_61510e_week10_1_new.pdf math1510e & h

Week 10-Day 1

Topics covered

Partial fraction decomposition Trigonometric integrals ݐെsubstitution Introduction to Fundamental Theorem of Calculus

Partial fraction decomposition - Long Division

There is

"dictionary" between "rational numbers" and "rational functions". For rational numbers, there are two types, i.e. proper rational numbers and improper rational numbers. Examples are: 3/4 (proper rational number), 4/3 (improper rational number). Now, by long division, one can convert any improper rational number to become a proper rational number (plus) an integer. E.g. ସ ଷ =1+ ଵ ଷ In the same way, one can always use long division to convert an improper rational function (i.e. a rational function ௣(௫) ௤ (௫) satisfying deg݌(ݔ)൒degݍ(ݔ) ) into the sum of a “polynomial" and a “proper rational function".)

Example:

ݔ ଷ +2ݔ+1 ݔ ଶ +1 =ݔ+

ݔ+1

ݔ 6 +1 Here the red-colored term is a polynomial. The yellow-colored term is a “proper" rational function. To compute integral of the form ׬ ௣(௫) ௤(௫)݀ݔ, where ݌(ݔ) and ݍ(ݔ) are polynomials and deg݌(ݔ)Cases for Proper Rational Functions

To integrate rational functions, the above

discussion tells us that we need only to consider "proper rational function" (because integration of polynomial is easy). Now ௣(௫) ௤ (௫) is "proper" and we have the following cases:

1. ݍ(ݔ) has only simple, linear factors.

2. ݍ(ݔ) has only simple irreducible factors.

3. ݍ(ݔ) has repeated linear factors.

4. ݍ(ݔ) has repeated irreducible quadratic factors.

5. Combination of 3 and 4 above.

Now we deal with these 4 cases one by one, via examples: 1. ௫ାଵ (௫ିଵ)(௫ିଷ)(௫ା଻) = ஺ భ ௫ିଵ + ஺ మ ௫ିଷ + ஺ య ௫ା଻ 2. ௫ାଶ ( ௫ మ ା௫ାଵ)(௫ మ ାଵ) = ஺ భ ା஻ భ ௫ ௫ మ ା௫ାଵ + ஺ మ ା஻ మ ௫ ௫ మ ାଵ , 3. ௫ାଵ ( ௫ିଵ) మ (௫ିଷ) య = ஺ భ (௫ିଵ) మ + ஺ మ ௫ିଵ + ஻ భ (௫ିଷ) య + ஻ మ (௫ିଷ) మ + ஻ య ௫ିଷ 4. ௫ାଶ ( ௫ మ ା௫ାଵ) మ (௫ మ ାଵ) య = ஺ భ ା஻ భ ௫ ( ௫ మ ା௫ାଵ) మ + ஺ మ ା஻ మ ௫ ௫ మ ା௫ାଵ + ஼ భ ା஽ భ ௫ ( ௫ మ ାଵ) య + ஼ మ ା஽ మ ௫ ( ௫ మ ାଵ) మ + ஼ య ା஽ య ௫ ௫ మ ାଵ 5. ௫ାଶ ( ௫ మ ା௫ାଵ) మ (௫ିଵ) మ = ஺ భ ା஻ భ ௫ ( ௫ మ ା௫ାଵ) మ + ஺ మ ା஻ మ ௫ ௫ మ ା௫ାଵ + ஼ భ (௫ିଵ) మ + ஼ మ ௫ିଵ

A computational

Example

(A good internet website with an interactive calculator for this is: http://www.emathhelp.net/calculators/algebra-2/partial-fraction-decomposition- calculator/)

Find partial fraction decomposition of

௫ା଻ ( ௫ మ ାଵ)(௫ మ ା௫ାଵ) మ . Proof of the five cases in “Step 1" makes uses of complex numbers - one can show that (if one allows for complex number solutions) any degree

݊ polynomial has exactly ݊ roots. Also,

whenever ܾ+ܽ is a root, then ܽെܾ

the “irreducible quadratic factor" (ݔെ൫ܾ+ܽξെ1൯൫ܽെܾ

assume that ܾ,ܽ Since the denominator has two irreducible quadratic factors, i.e. ݔ ଶ +1 and ݔ ଶ + ݔ+1. The first of them is a "simple" quadratic factor, the other is a "repeated" one. So the form of the partial fraction decomposition is

ݔ+7

( ݔ ଶ +1)(ݔ ଶ +ݔ+1) ଶ

ܣ=ݔ+ܤ

ݔ ଶ +ݔ+1

ܥ +ݔ+ܦ

( ݔ ଶ +ݔ+1) ଶ

ܧ+ݔ+ܨ

ݔ ଶ +1 Forming common denominator on the right-hand side gives:

ݔ+7

( ݔ ଶ +1)(ݔ ଶ +ݔ+1) ଶ =(ݔ ଶ +1)(ݔ ଶ +ݔ+1)(ܣݔ+ܤ ଶ +1)(ܥݔ+ܦ ଶ +ݔ+1) ଶ (ܧݔ+ܨ ଶ +1)(ݔ ଶ +ݔ+1) ଶ (ݔ ଶ +1)(ݔ ଶ +ݔ+1) ଶ Comparing the numerators on the left-hand side and on the right-hand side gives

ݔ+7=(ݔ

ଶ +1)(ݔ ଶ +ݔ+1)(ܣݔ+ܤ ଶ +1)(ܥݔ+ܦ ଶ +ݔ+1) ଶ (ܧ ܨ+

Expand right-hand side:

ݔ+7=ݔ

ହ ܣ ହ ܧ ସ ܣ ସ ܤ ସ ܧ ସ ܨ ଷ ܣ ଷ ܤ ଷ ܥ ଷ ܧ +2ݔ ଷ ܨ ଶ ܣ ଶ ܤ ଶ ܦ ଶ ܧ ଶ

ܨ+ݔܣ+ݔܤ+ݔܥ

+ݔܧ+2ݔܨ+ܦ+ܤ+ܨ Expanding the right-hand side and then collect up like terms (i.e. terms of the form ݔ ଴ ,ݔ ଵ ,ݔ ଶ ڮ,

ݔ+7=ݔ

ହ (ܧ+ܣ ସ (ܤ+ܣ+2ܨ+ܧ ଷ (2ܥ+ܤ+ܣ+3ܧ+2ܨ ଶ (ܣ +2ܦ+ܤ+2ܧ+3ܨ)+ݔ(ܧ+ܥ+ܤ+ܣ+2ܨ)+ܨ+ܦ+ܤ

Coefficients of the ݔ

଴ ,ݔ ଵ ,ݔ ଶ ڄ, should be equal, so we get the following system of equations:

ܧ+ܣ

ܤ+ܣ+2ܨ+ܧ

2ܥ+ܤ+ܣ+3ܧ+2ܨ

ܣ+2ܦ+ܤ+2ܧ+3ܨ

ܧ+ܥ+ܤ+ܣ+2ܨ

ܨ+ܦ+ܤ

Solving it, we get that ܣ=1,ܤ=8,ܥ=7,ܦ=6,ܧ=െ1,ܨ

Therefore,

௫ା଻ ( ௫ . >5)(௫ మ ା௫ାଵ) మ = ௫ା଼ ௫ మ ା௫>5 + ଻

Politique de confidentialité -Privacy policy