Question

C++,

Can anybody help me with this recursive function ? https://www.dropbox.com/s/8j48u3gq1wmqtgq/Screenshot%202015-11-28%2015.29.06.png?dl=0

Answer 1

#include <iostream>
using namespace std;
double n;



int display(double n)
{
if (n == 0)
{

return 1;



}

else{
return float(1)/n + float(1)/display(n-1);

}
}


int main()
{



cout << display(5);

}

Answer 2

what I have so far

Answer 3

my version



#include <iostream>
using namespace std;


double inverted_sum(int x){
if(x==1){
return 1;
}
return (float(1)/x + inverted_sum(x-1) );
}

int main(){
cout<<inverted_sum(1);
getchar();
}


see the difference!

and again, wrong section :P

Answer 4

`float(1)/display(n-1)`

that was wrong
should just be
'display(n-1)'

Answer 5

what is getchar(); ?

Answer 6

getchar waits for a character in the input/output terminal

Answer 7

so that the I/O terminal doesn't vanish immediately after displaying

Answer 8

but still how would you get all the values of the generated sequence ?

Answer 9

means?
sequence will generate one output for one value of x ...

Answer 10

For your code, here's what it would return with each iteration:

n=5: 1/5
n=4: 1/5+1/4
n=3: 1/5+1/4+1/3
n=2: 1/5+1/4+1/3+1/2
n=1 1/5+1/4+1/3+1/2+1/1
n=0: 1/5+1/5+1/3+1/2+1/1+1

As you can see, you have the 1 term duplicated. You'd avoid this by stopping when n=1 as @hartnn's code does. The other issue is it should be display(n-1) not float(1)/display(n-1) as @hartnn said.

Answer 11

so 5 should generate 1/5 ?

Answer 12

right ?

Answer 13

display(5)

Answer 14

5 should generate

1 +1/2+1/3+1/4+1/5
= 2.283

Answer 15

also the questions says to take the input as integer,
so you probably should use cin and take an integer and pass it to your function
before submitting the answer

Answer 16

one best way to verify whether you code is working or not,
take an input and run the code in your mind,

when input is 5,
display function will first check, whether 5 == 1?
it will be false, so it goes to
return 1/5+display(4)

before returning, display(4) needs to be calculated,
so
4 ==1 ? false
so
return 1/4 + display(3)

3==1? false
return 1/3 +display(2)

2==1? false
return 1/2 + display(1)

1==1? true
return 1
..........................

then all returns will execute,

return 1/2 +1
return 1/3+1/2+1
return 1/4+1/3+1/2+1
return 1/5+1/4+1/3+1/2+1

which is returned in main

Answer 17

Since we're in math... hehe...

Yeah the idea is you have this function:

\[f(n)=\sum_{k=1}^n \frac{1}{k}\]

But you can pull out the first term of the sum like this:


\[f(n)=\frac{1}{n} + \sum_{k=1}^{n-1} \frac{1}{k}\]
But then we see that summation is just f(n-1) so we plug it in:

\[f(n)=\frac{1}{n}+f(n-1)\]

So now if you just call your f(n) function, it will return 1/n and add that to the call of f(n-1), which then lets you step down until you get to the bottom. You just gotta be sure to tell it that f(1)=1 so that it stops, otherwise you'll find it trying to calculate 1/0, 1/-1, 1/-2, etc... (Not sure how C++ handles infinity, but I am imagining it's not too different from how Java handles floats)

So at least this is the idea of what you're doing here in terms of probably more confusing math notation, or not, I don't know your background in math >_>

Answer 18

hmm so series in math are symbolized with summation ?

Answer 19

It's been a long time since I have done series and I dont remember. Bascally I dont know if he wants me to output each individual element (1/5 + 1/4 +....1/1) or all of them as a sum

Answer 20

he just wants the sum

Answer 21

fibonacci is a Series right ?

Answer 22

so fibonacci(5) = 12 ? or 0 1 1 2 3 5

Answer 23

sum of fibonacci terms till 5 gives 12

all the fibonacci terms till 5 gives 0 1 1 2 3 5

both are different questions.

what you're asked is just the sum.

Answer 24

hmm is a Series different than a Sequence ?

Answer 25

what if I wanted to write a program that outputs each individual element @hartnn ? Do you thing thats possible with recursion ? I am just practicing to see if I can solve it in different ways

Answer 26

I see the confusion on the fibonnaci thing, f(5) doesn't mean 'add up all the fibonacci numbers that have value less than 5'

Instead, it relies on the fact that fibonacci numbers are defined in a recursive way, when you write them out, you rely on the last two numbers to pick the next one, right? That relation looks like this:

\[f(n)=f(n-1)+f(n-2)\]

so I'll go ahead and write out the first few terms in this format and see if yo understand this recursion relation:

f(0)=0
f(1)=1
f(2)=1
f(3)=2
f(4)=3
f(5)=5
f(6)=8
f(7)=13
f(8)=21

Se how f(7)=f(6)+f(5) ? Plug in the 5th, 6th, and 7th Fibonacci numbers to see: 13=8+5

Answer 27

hmmmm i see

Answer 28

but how I go about displaying the elements individually instead of some for the first problem ? for eg display(5) would result in 1/1 , 1/2 , 1/3 , 1/4 , 1/5

Answer 29

Anyways I found it

Answer 30

#include <iostream>
using namespace std;

double inverted_sum(int x){ if(x==1 || x==0){ return 1; }


return (float(1)/x + inverted_sum(x-1) ); }


int main(){
int i=1;
while (i!=6){
cout << inverted_sum(i) - inverted_sum(i-1) << ' ' ;
i++;
}

getchar(); }

Answer 31

#include <iostream>
using namespace std;

double inverted_sum(int x){ if(x==1){ return 1; }

else if (x==0){
return 0;
}


return (float(1)/x + inverted_sum(x-1) ); }


int main(){
int i=1;
while (i!=9){
cout << inverted_sum(i) - inverted_sum(i-1) << ' ' ;
i++;
}

getchar(); }

Answer 32

seriously ?

if you don't want to evaluate the sum, why not simply do :
```
while (i!=6){
cout << "1/"<<i ;
i++;
}
```

Answer 33

I am practicing recursion

Answer 34

even the strating problem would be simpler with loop, and faster

Answer 35

yeah, i believe the question wants you print out the final value of sum after evaluating

not just printing the string : "1 + 1/2 + 1/3 + ... "