MAT 21B Lecture Notes - Lecture 23: Partial Fraction Decomposition, Constant Function

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31 May 2018

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MAT 21B – Lecture 23 – Partial Fraction Decomposition

• Partial fraction decomposition is the procedure of combining ratios of

polynomials in reverse.

• In other words, we will learn to split a rational function such as 

 into its

fractional components, which in this example 





.

• A rational function is a function that is a ratio of two polynomials.

• Examples of rational functions

o . Polynomials are rational functions because when divided by a

constant function 1, it is a polynomial of degree 0.

o 





• Examples of functions that are not rational

o . This is neither a rational nor a polynomial. Polynomials have integer

powers only. Fractional powers are not included.

o 

 . Numerator is not a polynomial; it is a trigonometric function.

o . This is an exponential function

• Example: Perform partial fraction decomposition on 



Factor the denominator, if possible. In this case, 



. Split the

function into two separate fractions, 















 







. Thus, 





. By plugging in the

numbers -1 and 3 (in this case) into this equality of a function, we are eliminating

one of the variables, which make it easier to solve for the constants A and B.

Alternatively, you could use the method of atchig coefficiets such that

when  is rearranged we obtain 

 

 and then solve for A and B.

• Example 2: Evaluate 



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MAT 21B Full Course Notes

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Mat 21b lecture 23 partial fraction decomposition: partial fraction decomposition is the procedure of combining ratios of constant function 1, it is a polynomial of degree 0. powers only. Numerator is not a polynomial; it is a trigonometric function. polynomials in reverse: examples of functions that are not rational, a rational function is a function that is a ratio of two polynomials, examples of rational functions. In other words, we will learn to split a rational function such as (cid:2873) (cid:2871) (cid:3118) (cid:2870) (cid:2871) into its (cid:3118) (cid:2870) (cid:2871)= (cid:2870)+(cid:2869)+ (cid:2871) (cid:2871). fractional components, which in this example (cid:2873) (cid:2871: (cid:2871)+(cid:884). Polynomials are rational functions because when divided by a. This is neither a rational nor a polynomial. Polynomials have integer: sin (cid:4666)(cid:3118)+(cid:2870)(cid:4667) (cid:3118)+(cid:2869, (cid:1857)(cid:2873). This is an exponential function: example: perform partial fraction decomposition on (cid:2873) (cid:2871) (cid:3118) (cid:2870) (cid:2871) (cid:3118) (cid:2870) (cid:2871)= (cid:2873) (cid:2871)

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What are the answers for all of these questions? I am confused on how to solve for these.

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olivezebra225

must understand functions to be able linearize. First let's review what graphs of certain functions looks like. Sketch the shape of each type of y vs. x function below. k is listed as a generic constant of proportionally. Linear y = kx Inverse y = k/x Inverse Square y = k/x^2 Power y = kx^2 You will notice that only the linear function is a straight line. We can easily find the slope of our line by measuring the rise and dividing it by the run of the graph or calculating it using two points. The value of the slope should equal the constant, k, from the equation. Finding k is a bit more challenging in the last three graphs because the slope isn't constant. This should make sense since your graphs aren't linear. So how do you calculate the constant, k? You need to transform the non-linear graph into a linear graph in order to calculate a constant slope. You can accomplish this by transforming one or both of the axes for the graph. The hardest part is figuring out which axes to change and how to change them. The easiest way to accomplish this task is to solve your equation for the constant. Note in the examples from the last page there is only one constant, but this process could be done for other equations with multiple constants. Instead of solving for a single constant, put all of the constants on one side of the equation. When you solve for the constant, the other side of the equation should be in fraction form. This fraction gives the rise and run of the linear graph. Whatever is in the numerator is the vertical axis and the denominator is the horizontal axis. If the equation is not in fraction form, you will need to inverse one or more of the variables to make a fraction. First let's solve each equation to figure out what we should graph. Then look below at the example and complete the last one, a sample AP question, on your own. State what should be graphed in order to produce a linear graph to solve for k. Inverse Graph Vertical Axis: _____ Horizontal Axis: _____ Inverse Square Graph Vertical Axis: _____ Horizontal Axis: _____ Power (Square) Graph Vertical Axis: _____ Horizontal Axis: _____

1.Write a function encipher(s, n) that takes as inputs an arbitrary string s and a non-negative integer n between 0 and 25, and that returns a new string in which the letters in s have been "rotated" by n characters forward in the alphabet, wrapping around as needed. For example:

>>> encipher('hello', 1) result: 'ifmmp'

>>> encipher('hello', 2) result: 'jgnnq'

>>> encipher('hello', 4) result: 'lipps'

Upper-case letters should be "rotated" to upper-case letters, even if you need to wrap around. For example:

>>> encipher('XYZ', 3)

result: 'ABC'

Lower-case letters should be "rotated" to lower-case letters:

>>> encipher('xyz', 3)

result: 'abc'

Non-alphabetic characters should be left unchanged:

>>> encipher('#caesar!', 2)

result: '#ecguct!'

Hints/reminders:

You can use the built-in functions ord and chr convert from single-character strings to integers and back:

>>> ord('a')

result: 97

>>> chr(97)

result: 'a'

You can use the following test to determine if a character is between 'a' and 'z' in the alphabet:

if 'a' <= c <= 'z':

A similar test will work for upper-case letters.

We recommend writing a helper function rot(c, n) that rotates a single character c forward by n spots in the alphabet. We have given you a template for this helper function in ps3pr3.py that checks to ensure that c is a single-character string. We wrote rot13(c) in lecture; rot(c, n) will be very close to rot13(c)! You can test your rot(c, n) as follows:

>>> rot('a', 1)

result: 'b'

>>> rot('y', 2)

result: 'a'

>>> rot('A', 3)

result: 'D'

>>> rot('Y', 3)

result: 'B'

>>> rot('!', 4)

result: '!'

Once you have rot(c, n), you can write a recursive encipher function.

Once you think you have everything working, here are three more examples to try:

>>> encipher('xyza', 1)

result: 'yzab'

>>> encipher('Z A', 2)

result: 'B C'

>>> encipher('Caesar cipher? I prefer Caesar salad.', 25)

result: 'Bzdrzq bhogdq? H oqdedq Bzdrzq rzkzc.'

2.Write a function decipher(s) that takes as input an arbitrary string s that has already been enciphered by having its characters "rotated" by some amount (possibly 0). decipher should return, to the best of its ability, the original English string, which will be some rotation (possibly 0) of the input string s. For example:

>>> decipher('Bzdrzq bhogdq? H oqdedq Bzdrzq rzkzc.')

result: 'Caesar cipher? I prefer Caesar salad.'

Here are two more examples:

>>> decipher('Hu lkbjhapvu pz doha ylthpuz hmaly dl mvynla lclyfaopun dl ohcl slhyulk.')

result: 'An education is what remains after we forget everything we have learned.'

>>> decipher('python')

result: 'eniwdc'

------------------------------------

def rot(c, n):
""" your docstring goes here """
# check to ensure that c is a single character
assert(type(c) == str and len(c) == 1)

def letter_prob(c):
""" if c is the space character (' ') or an alphabetic character,
returns c's monogram probability (for English);
returns 1.0 for any other character.
adapted from:

"""
# check to ensure that c is a single character
assert(type(c) == str and len(c) == 1)

if c == ' ': return 0.1904
if c == 'e' or c == 'E': return 0.1017
if c == 't' or c == 'T': return 0.0737
if c == 'a' or c == 'A': return 0.0661
if c == 'o' or c == 'O': return 0.0610
if c == 'i' or c == 'I': return 0.0562
if c == 'n' or c == 'N': return 0.0557
if c == 'h' or c == 'H': return 0.0542
if c == 's' or c == 'S': return 0.0508
if c == 'r' or c == 'R': return 0.0458
if c == 'd' or c == 'D': return 0.0369
if c == 'l' or c == 'L': return 0.0325
if c == 'u' or c == 'U': return 0.0228
if c == 'm' or c == 'M': return 0.0205
if c == 'c' or c == 'C': return 0.0192
if c == 'w' or c == 'W': return 0.0190
if c == 'f' or c == 'F': return 0.0175
if c == 'y' or c == 'Y': return 0.0165
if c == 'g' or c == 'G': return 0.0161
if c == 'p' or c == 'P': return 0.0131
if c == 'b' or c == 'B': return 0.0115
if c == 'v' or c == 'V': return 0.0088
if c == 'k' or c == 'K': return 0.0066
if c == 'x' or c == 'X': return 0.0014
if c == 'j' or c == 'J': return 0.0008
if c == 'q' or c == 'Q': return 0.0008
if c == 'z' or c == 'Z': return 0.0005
return 1.0

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MAT 21B Lecture Notes - Lecture 23: Partial Fraction Decomposition, Constant Function

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