Homework Help v1.3

[MuKen]

What level is this class? Would they expect you to make use of time-efficient selection algorithms?

If not, the easiest thing to do is to just sort the array and then choose the element at the appropriate index.

[ahme]

Originally Posted by Ice(v)an

Second Assignment,

finding the kth smallest item in an array. Implement this idea and use it to form the basis for a percentile function. This function should take two arguments, an array or vector of numbers A and a percentile p. It should return the number n in A such that p percent of all items in A are smaller than n.



This specification is intentionally a little imprecise. For example, it will normally be the case that no number in A exactly fits the definition, although one or more comes close. You'll need to decide what to return in that case. Be sure to document your choices and assumptions thoroughly.



Test and demonstrate your percentile function by embedding it in a program that generates sets of random numbers and then finds various percentiles in each set.

I haven't used C++ that much, but here is an algorithm/description of the solution.

Create a class with:
1. One private field that is an array of doubles and let's call it myArray.
2. Constructor that takes an array of doubles as it's argument: The constructor initializes the myArray array to the passed array.
3. Public function that takes a double number as it's argument and returns a double number representing the number n that we want.

pseudo-code/partial code:
class Percentile{

public:
Percentile(double []);
private:
double myArray[];
double getNum(double);

};
Percentile:: Percentile(double nums[]){

myArray = nums;

}
double Percentile::getNum(double p){

double temp[] = myArray;
//1. sort the temp array in ascending order. I think C++ libraries has a sort function or you can create your own sorting function
//2. calculate the index of our target number
// index = (length of the temp array - 1)*p/100 rounded to the nearest integer
// return temp[index]

}

edit: didn't know that someone replied to your post and we have basically said the same thing

[DeathGuiseX]

I just wanna check my work, so can someone do this w/ the work:
∫x³/(x²+1)^(1/2) dx with the limits being 0 --> 1, as in 0 is on the bottom of ∫ representing a and 1 is on the top of ∫ representing b. This is once again, integration by parts.

and also, is ∫sin(3ln x) dx = (xsin(3lnx) - 3xcos(3lnx))/10 . I dont need an explanation on how to do this problem unless I have the wrong answer.

[]MpC[hebe]

Originally Posted by DeathGuiseX

I just wanna check my work, so can someone do this w/ the work:
∫x³/(x²+1)^(1/2) dx with the limits being 0 --> 1, as in 0 is on the bottom of ∫ representing a and 1 is on the top of ∫ representing b. This is once again, integration by parts.

and also, is ∫sin(3ln x) dx = (xsin(3lnx) - 3xcos(3lnx))/10 . I dont need an explanation on how to do this problem unless I have the wrong answer.

The second answer is correct. I don't have time to type out the first question, sorry.

[Atem]

I have to do research on Washington for one of my classes, but I can't find anything about what kind of political issues would affect the state. Can someone help me find something or if you already know something, tell me. Thanks.

[Ice(v)an]

We're not concerning about time efficient as of yet.

And also, I haven't really coded much with arrays and such so...

[MuKen]

Originally Posted by Ice(v)an

We're not concerning about time efficient as of yet.

And also, I haven't really coded much with arrays and such so...

Yeah, then just sort the array and choose an element from it. You can do this by calling qsort.

For example, say you have an array of 10 ints

#include <iostream>
#include <stdlib.h>

int A[] = { 8, 1, 9, 2, 7, 4, 5, 3, 0, 6 };

int compare (const void* x, const void* y)
{
  return *((int*)x) - *((int*)y);
}

int main ()
{
  qsort (A, 10, sizeof(int), compare);
  for (int n=0; n<10; n++) {
    cout << A[n] << endl;
  }
  return 0;
}

Would print out

0
1
2
3
4
5
6
7
8
9

The compare function is something you provide to qsort to tell it how to sort ints.

[ahme]

Originally Posted by DeathGuiseX

I just wanna check my work, so can someone do this w/ the work:
∫x³/(x²+1)^(1/2) dx with the limits being 0 --> 1, as in 0 is on the bottom of ∫ representing a and 1 is on the top of ∫ representing b. This is once again, integration by parts.

and also, is ∫sin(3ln x) dx = (xsin(3lnx) - 3xcos(3lnx))/10 . I dont need an explanation on how to do this problem unless I have the wrong answer.

For the first question, it is not a hard problem if you take a moment before you do the integral by parts. Integral by parts is one of those techniques that you have to have a sharp eye to use.
Back to the question...
∫x³/(x²+1)^(1/2)
When choosing the u and dv in you want to make the problem less complicated and one way to make the problem easier is reducing the power of a polynomial. Here we have the polynomial x^3, so I can take u=x^3, but you need to stop and think for a second here because if u=x^3 then dv = (x²+1)^(1/2)dx which will require you to use trig substitution to get v and will involve very ugly looking function which will make the problem even more complicated. But, if you have sharp eyes and remember the trick of having the derivative of the numerator in the denominator (I guess I've said this too many times) then you know that taking your dv = x/(x²+1)^(1/2)dx would make this very easily integrable.
So, take u=x^2 and dv=x/(x²+1)^(1/2)dx and the integral should be very easy to do.
I don't have time to write down the whole solution, but the final answer is
2^(1/2)-2/3*(2^(3/2)-1) which after doing some simplification turns down to be 2/3 - (2^(1/2))/3

[DeathGuiseX]

^Wait, what would v be then? Cause I tried u = x^3, and dv = 1/(x2 + 1)^.5 dx, and got du/3 = x^2 dx and v= 2(x^2 +1)^1/2
and it became 2x^3 (x^2 + 1)^1/2 - 1/3∫2x^2 (x^2 + 1) ^1/2 dx, but I dont know how to do that second integral. Is my v correct? If not, what is v if you use your dv(x/(x²+1)^(1/2) dx)?
Also, I tried putting the integral in my calculator to get a numerical answer and it doesnt match your answer.

[ahme]

Originally Posted by DeathGuiseX

^Wait, what would v be then? Cause I tried u = x^3, and dv = 1/(x2 + 1)^.5 dx, and got du/3 = x^2 dx and v= 2(x^2 +1)^1/2
and it became 2x^3 (x^2 + 1)^1/2 - 1/3∫2x^2 (x^2 + 1) ^1/2 dx, but I dont know how to do that second integral. Is my v correct? If not, what is v if you use your dv(x/(x²+1)^(1/2) dx)?
Also, I tried putting the integral in my calculator to get a numerical answer and it doesnt match your answer.

Originally Posted by ahme

So, take u=x^2 and dv=x/(x²+1)^(1/2)dx and the integral should be very easy to do.
I don't have time to write down the whole solution, but the final answer is
2^(1/2)-2/3*(2^(3/2)-1) which after doing some simplification turns down to be 2/3 - (2^(1/2))/3

so du = 2x dx and v = (x²+1)^(1/2)

also, I've double checked my answer I think it's correct. It's (2-2^(1/2))/3

[Easybigboi]

Can anyone help me on some english? I need some comparisons between Nicholas Nickleby and Hard Times the books by Charles Dickens preferably the characters i.e Bounderby and Mr Squeers. Any help would be appreciated whether it is your own interpretations or a website link. Thanks

[4thseal]

Here's a resources:

The Life and Adventures of Nicholas Nickleby (Cri...: Information and Much More from Answers.com

^Not the best, though, if you have access to any literary databases, i'm sure you'd have better luck finding what you want. Some questions to ask your self when comparing character: Are there any similar traits between the two? Do they have simulare motives/wants? Do the two characters carry simular allusions in names, clothing, residence, and so on.

[P]

Q: Suppose that f is continuous on the interval [0,1] and its strictly positive there. Show that there exists an epsilon > 0 such that f(x) > epsilon in [0,1]. (Hint: one of the hard theorems will help)

[Nivikran]

edited

[]MpC[hebe]

Originally Posted by P

Q: Suppose that f is continuous on the interval [0,1] and its strictly positive there. Show that there exists an epsilon > 0 such that f(x) > epsilon in [0,1]. (Hint: one of the hard theorems will help)

Seems trivial

The best chance f would have is to have an asymptote y=0, but since f(x) is continuous and positive (> 0), that means y = f(x) has a positive, real value for every x on [0,1]. Any positive real number can be divided by any other positive real number to yield a positive real number (the Reals are a field under addition and multiplication).

So, let f(x) = y and let epsilon = p where y, p > 0

y > y/2 for all y

Therefore, taking the local min of f(x) in [0,1] will offer us a y[sub]min[sub] for which all other y >= y_min.

That in mind, y >= y_min > y_min/2

Let p = y_min/2, and y > p for all y.

Still, that's just a long way of saying something I think is fairly trivial.

[Ice(v)an]

This function should take two arguments, an array or vector of numbers A and a percentile p. It should return the number n in A such that p percent of all items in A are smaller than n.

----

From the code you guys gave me, I don't see any vectors.

So would the function be something like this?

vector<string> vect;

someFunction(vect, p)
{
return n
}

[MuKen]

They said you could use an array or a vector right?

The code I showed above uses an array...

If you are required to use a vector, I'd just go ahead and fill an array from the vector anyway so you can make use of the qsort function; it saves a lot of work.

[Ice(v)an]

I do have a choice for array or vector, I want to choose vector.

So the post above, did I have the outline right?

If I wrote a function that had two arguments, one vector, and the other P?

[MuKen]

Yeah, that structure is right, though the syntax should be something like

someFunction(vector<int>& vect, double p) {
return n;
}

[Ice(v)an]

^
do i still have to put Vector<int>&, doublep)

if it's a class, and I have a vector as a private data member?