Write a python function sqr(n) that returns the square of its parameter n for example: Test Result print(sqr(-3)) 9 print(sqr(11)) 121
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Write a python function sqr(n) that returns the square of its parameter n
for example:
| Test | Result |
| print(sqr(-3)) | 9 |
| print(sqr(11)) | 121 |
Step by step
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- Write a recursive Python function that matches the following docstring: '''Function -- sum_fivesSums all positive multiples of five between 0 and some integer, n, inclusive.Parameters:n -- an integer.Returns:The sum of all positive multiples of five up to and including n.''' Examples: sum_fives(1) returns 0 sum_fives(5) returns 5 sum_fives(10) returns 15In C programming Mathematically, given a function f, we recursively define fk(n) as follows: if k = 1, f1(n) = f(n). Otherwise, for k > 1, fk(n) = f(fk-1(n)). Assume that there is an existing function f, which takes in a single integer and returns an integer. Write a recursive function fcomp, which takes in both n and k (k > 0), and returns fk(n). int f(int n);int fcomp(int n, int k){(Program) Write a program that tests the effectiveness of the rand() library function. Start by initializing 10 counters, such as zerocount, onecount, twocount, and so forth, to 0. Then generate a large number of pseudorandom integers between 0 and 9. Each time 0 occurs, increment zerocount; when 1 occurs, increment onecount; and so on. Finally, display the number of 0s, 1s, 2s, and so on that occurred and the percentage of time they occurred.
- Below,enter code to complete implementation of a recursive function sum_all_integers(), which takes an input n and adds all intergers preceding it, up to n: add_all_integers(n):code required for python: For this question, you will be required to use the binary search to find the root of some function f(x)f(x) on the domain x∈[a,b]x∈[a,b] by continuously bisecting the domain. In our case, the root of the function can be defined as the x-values where the function will return 0, i.e. f(x)=0f(x)=0 For example, for the function: f(x)=sin2(x)x2−2f(x)=sin2(x)x2−2 on the domain [0,2][0,2], the root can be found at x≈1.43x≈1.43 Constraints Stopping criteria: ∣∣f(root)∣∣<0.0001|f(root)|<0.0001 or you reach a maximum of 1000 iterations. Round your answer to two decimal places. Function specifications Argument(s): f (function) →→ mathematical expression in the form of a lambda function. domain (tuple) →→ the domain of the function given a set of two integers. MAX (int) →→ the maximum number of iterations that will be performed by the function. Return: root (float) →→ return the root (rounded to two decimals) of the given function. my code below , however as…Write just the function count(String s) and write it in JAVA
- Consider the following function: void fun_with_recursion(int x) { printf("%i\n", x); fun_with_recursion(x + 1); } What will happen when this function is called by passing it the value 0?Implement a recursive C++ function which takes a character (ch) and a positive integer (n) and prints thecharacter ch, n times on the screen. The prototype of your function should be:void printChar (char ch, int n)For example, calling printChar('*',5) should display ***** on screen.Note: There should NOT be any loop in your function.The goal is to rewrite the function, below, such that passes in a different list of parameters, particularly eliminating the need to pass low and high for each recursive call to binary_search. defbinary_search(nums,low,high,item): mid=(low+high)//2iflow>high:returnFalse #The item doesn't exist in the list!elifnums[mid]==item:returnTrue# The item exists in the list!elifitem<nums[mid]:returnbinary_search(nums,low,mid-1,item)else:returnbinary_search(nums,mid+1,high,item) The new function should be prototyped below. The number of changes between the given version, and the one requested is not significant. defbinary_search(nums,item):pass# Remove this and fill in with your code Tip: If you consider that high and low are used to create a smaller version of our problem to be processed recursively, the version requested will do the same thing, just through a different, more Pythonic technique.
- Write a program in Python with a function squareRoot(x, epsilon) that uses bisection search to return a number y that is close enough to the square root of x, so that abs(y**2 - x) < epsilon.python this is connected to the last problem - the second part of the question is added. my attempt on this problem shows that the part b) (approaches N*NH) is not really wokring.. a) (answered) with a function “harmonic(n)” that computes the n-th harmonic number, write a function “harmonic_all(n)” that returns the number of values generated until all values are obtained as a function of the range of possible values n, then write a function “harmonic_sim(n)” that repeats “harmonic_all(n)” a total of n_sim = 100 times. (Attaching the code from the answer for a) d) Show that as n increases (e.g., with a doubling experiment), from n = 2 to n_max = 1,000, the value of “coupon_sim(n)” approaches “n * Hn”.Write a recursive function, reverseDigits, that takes an integer as a parameter and returns the number with the digits reversed. Also, write a program to test your function.
