Quiz

The coefficients of a polynomial $p(x)=a_0 + a_1x + a_2x^2+···+a_nx^n$ are given as a python list pol=[a0,a1, a2,...,an], where the index corresponds to the power of x.

(a)The function evaluate(p,a) evaluates a polynomial p(x) at the value x=a and returns this value. What is the missing line of code?

def evaluate(p,a):
    pa=0
    for k in range(len(p)):
        ## MISSING LINE IS HERE
    return pa

pa += p[k] * (a**k)

Consider the function $f(x)=e^{0.4x}−x+sin(4x)$.

In this exercise, we will use various methods to approximate the roots of this function, i.e., solutions of the equation $f(x)=0$.

(Optional, but helpful!) Graph the function f (x) on the interval [0, 1] along with the x-axis. Identify the root visually and "zoom in" closer by modifying your values of a and b.

(a) Using the grid method with a = 0, b =  1, and n =  1000, determine the value of m so that the tolerance $10^{-m}$ gives exactly one possible root.

(b) In (a), what is the approximate root?

(c) Using the grid method with a =  0, b = 1, and TOL=$10^{-5}$, determine the smallest integer m so that when $n=10^m$ you have at least one possible root which satisfies $|f(x)|<10^{-5}$.

(d) The code below is meant to output an interval on which the equation f (x) = 0 has a solution, however it contains an error. Identify the line where the error occurs and type what the correct line should be.

def IVT_bisection(f,a,b,tol):
    while abs(b-a)>=tol:
        mid=(a+b)/2
        if f(mid)==0:
            print("The solution is:", mid)
        if f(a)*f(mid)<0:
            b=mid
        if f(mid)*f(b)<0:
            b=mid
    return a,b

Determine the smallest value of n so that the following sum approximates $e^3$ with an error less than $10^{-1}$.

$$s_n = \sum_{i=0}^{n} \frac{x^i}{i!}$$

(Recall that the above sum approximates $e^x$, for all real numbers x.) In other words, find the smallest n so that $|s_n−e^3|<10{−1}$. Enter this value of n into the answer box below.

Consider a finite subset $S \subseteq R$. Write a function $setA(S,M)$, which returns the set A containing all elements of S which are less than or equal to some value M.

def setA(S,M):  ##(1)
if el<=M:       ##(2)
A=S             ##(3)
A.add(el)       ##(4)
A=S.copy()      ##(5)
return A        ##(6)
for el in A:    ##(7)
if el>M:        ##(8)
A.discard(el)   ##(9)
for el in S:    ##(10)
A=set()         ##(11)

Enter the line numbers in the order that they would appear in your program, separated with commas. For example, an answer might look like 7,3,2,8,10,7,6.

def setA(S,M):  ##(1)
    A=set()         ##(11)
    for el in S:    ##(10)
        if el<=M:       ##(2)
            

Which of the following functions correctly returns True if A, B, and C form a partition of X, i.e., their union is Xand they are pairwise disjoint, and False otherwise?

A.

def partition(X,A,B,C):
    answer=True
    if A.union(B).union(C)!=X:
        answer=False
    if A.intersection(B)!=0:
        answer=False
    if A.intersection(C)!=0:
        answer=False
    if B.intersection(C)!=0:
        answer=False
    return answer

B.

def partition(X,A,B,C):
    answer=False
    if A.union(B).union(C)==X:
        answer=True
    if A.intersection(B)==0:
        answer=True
    if A.intersection(C)==0:
        answer=True
    if B.intersection(C)==0:
        answer=True
    return answer

C.

def partition(X,A,B,C):
    answer=False
    if (A.union(B).union(C)!=X or A.intersection(B)!=set()
        or A.intersection(C)!=set() or B.intersection(C)!=set()):
        answer=True
    return answer

D.

def partition(X,A,B,C):
    answer=True
    if A.union(B).union(C)!=X:
        answer=False
    if A.intersection(B)!=set():
        answer=False
    if A.intersection(C)!=set():
        answer=False
    if B.intersection(C)!=set():
        answer=False
    return answer

E.

def partition(X,A,B,C):
    answer=True
    if (A.union(B).union(C)!=X and A.intersection(B)!=set()
        and A.intersection(C)!=set() and B.intersection(C)!=set()):
        answer=False
    return answer

F.

def partition(X,A,B,C):
    answer=False
    if (A.union(B).union(C)==X and A.intersection(B)==set()
        and A.intersection(C)==set() and B.intersection(C)==set()):
        answer=True
    return answer

G.

def partition(X,A,B,C):
    answer=True
    if (A.union(B).union(C)!=X or A.intersection(B)!=set()
        or A.intersection(C)!=set() or B.intersection(C)!=set()):
        answer=False
    return answer

Enter 1 if you think that the function correctly performs the required task and 2 if you think that it does not. So, a typical answer would look like '1,2,2,1,2,1,1' (without the quotes).

From the options below, choose the part that includes the three cases that most thoroughly (or best) test your code in the previous problem.

Consider the dictionaries

d1={1:"snow",2:"air",3:"cup",4:"toast",7:"avocado"}
d2={1:"flake",2:"plane",3:"cake",5:"jam",6:"juice"}

**(a)**Which of the code blocks below create a list of the values from d1?

**(b)**Type the missing line to create an inverse dictionary d_inverse for the dictionary d2.

d_inverse={}
for i in d2.keys():
    ## MISSING LINE IS HERE

**(c) **Write a program to create a dictionary d3 (do not use ready-made commands) which merges d1 and d2, adding together the values if two keys are the same, i.e., d3[k]=d1[k]+d2[k].

for i in d1:      ## (1)
for i in d2:      ## (2)
d3[i]=d2[i]       ## (3)
d3[i]=d1[i]       ## (4)
d3=dict()         ## (5)
if i not in d1:   ## (6)
if i in d2:       ## (7)
d3[i]=d1[i]+d2[i] ## (8)
else:             ## (9)

Enter the line numbers in the order that they would appear in your program, separated with commas. For example, an answer might look like 3,2,8,9,1,4,5,7,6.

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