• Support PF! Buy your school textbooks, materials and every day products Here!

Integral question

  • Thread starter Miike012
  • Start date
  • #1
1,011
0
If n is a positive integer, prove that ∫ [t] dt = n(n-1)/2 in [0,n]

Solution:

sub intervals: [0,1] , [1,2] , [n-1,1]

each sub interval has base 1 and height 1, 2 , .... n respectively....

Thus: ∫ [t] dt = Ʃ [t](base) from i = 1 to n = n is base(=1)(1 + 2 + .... + n)

= 1 + 2 + .... n =/= n(n-1)/2

some help please...?
 

Answers and Replies

  • #2
33,287
4,990
If n is a positive integer, prove that ∫ [t] dt = n(n-1)/2 in [0,n]
What function does [t] represent? Greatest integer function - [itex]\lfloor t \rfloor[/itex]? Least integer function - [itex]\lceil t \rceil[/itex]?
Solution:

sub intervals: [0,1] , [1,2] , [n-1,1]

each sub interval has base 1 and height 1, 2 , .... n respectively....

Thus: ∫ [t] dt = Ʃ [t](base) from i = 1 to n = n is base(=1)(1 + 2 + .... + n)

= 1 + 2 + .... n =/= n(n-1)/2

some help please...?
 
  • #3
1,011
0
Greatest integer
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,808
933
Your only mistake is in thinking that 1+ 2+ 3+ ...+ n is NOT equal to n(n+1)/2.

If n= 3, 1+ 2+ 3= 6 and 3(3+1)/2= 6.
If n= 4, 1+ 2+ 3+ 4= 10 and 4(4+ 1)/2= 10.


It can be shown that, for all n, 1+ 2+ 3+ ...+ n= n(n+1)/2.

Here is a proof attributed to Gauss when he was a child:

Write
1+ 2+ 3+ ...+ n and below it
n+ n-1+ n-2+ ...+ 1

And add each vertical pair. Each pair adds to n+ 1 and there are, of course, n pairs. That total sum is n(n+ 1). But that clearly adds the original sum twice. The sum itself is half of that, n(n+1)/2.

It is also fairly easy to prove this by induction. When n= 1 we have 1= 1(1+ 1)/2.

Assume that, for some k, 1+ 2+ 3+ ...+ k= k(k+1)/2.

Then 1+ 2+ 3+ ...+ k+ k+1= k(k+1)/2+ (k+1)

Factor k+ 1 out of the right side: (k+1)(k/2+ 1)= (k+1)(k/2+ 2/2)= (k+ 1)(k+ 2)/2 which is just n(n+1)/2 with n= k+1.
 
  • #5
Dick
Science Advisor
Homework Helper
26,258
618
If n is a positive integer, prove that ∫ [t] dt = n(n-1)/2 in [0,n]

Solution:

sub intervals: [0,1] , [1,2] , [n-1,1]

each sub interval has base 1 and height 1, 2 , .... n respectively....

Thus: ∫ [t] dt = Ʃ [t](base) from i = 1 to n = n is base(=1)(1 + 2 + .... + n)

= 1 + 2 + .... n =/= n(n-1)/2

some help please...?
1+2+...+(n-1) is n(n-1)/2. What's the value of [t] for n-1<=t<n? Is [n]=n going to change the integral?
 
  • #6
1,011
0
Your only mistake is in thinking that 1+ 2+ 3+ ...+ n is NOT equal to n(n+1)/2.

If n= 3, 1+ 2+ 3= 6 and 3(3+1)/2= 6.
If n= 4, 1+ 2+ 3+ 4= 10 and 4(4+ 1)/2= 10.


It can be shown that, for all n, 1+ 2+ 3+ ...+ n= n(n+1)/2.

Here is a proof attributed to Gauss when he was a child:

Write
1+ 2+ 3+ ...+ n and below it
n+ n-1+ n-2+ ...+ 1

And add each vertical pair. Each pair adds to n+ 1 and there are, of course, n pairs. That total sum is n(n+ 1). But that clearly adds the original sum twice. The sum itself is half of that, n(n+1)/2.

It is also fairly easy to prove this by induction. When n= 1 we have 1= 1(1+ 1)/2.

Assume that, for some k, 1+ 2+ 3+ ...+ k= k(k+1)/2.

Then 1+ 2+ 3+ ...+ k+ k+1= k(k+1)/2+ (k+1)

Factor k+ 1 out of the right side: (k+1)(k/2+ 1)= (k+1)(k/2+ 2/2)= (k+ 1)(k+ 2)/2 which is just n(n+1)/2 with n= k+1.
I know of that proof....
So i am guessing the author is assuming who ever is reading the book already knows that 1+ 2+ 3+ ...+ n is equal to n(n+1)/2?

I thought that there might be some way to solve [t] and directly get n(n+1)/2, instead of getting 1+ 2+ 3+ ...+ n then adding n+ n-1+ n-2+ ...+ 1 to it...
 
  • #7
1,011
0
I thought of another way of proving it that seems more natural.... please look at it and tell me what you think....

I created a step function then sections it off into a rectangle and a triangle.....

1. area of rectangle = (b)(h) = (n)(1)

2. area of large triangle = (b)(h)/2 = (n)(n)/2

3. area of small triangle's = n/2


Area of ordinate satisfying greatest integer function from [0,n] is the sum of these areas....

(n)(1) + n^2/2 - n/2 = n(n-1)/2..
 

Attachments

  • #8
1,011
0
Never mind I just looked at it and realized it does not equal n(n-1)/2
 
  • #9
Dick
Science Advisor
Homework Helper
26,258
618
I thought of another way of proving it that seems more natural.... please look at it and tell me what you think....

I created a step function then sections it off into a rectangle and a triangle.....

1. area of rectangle = (b)(h) = (n)(1)

2. area of large triangle = (b)(h)/2 = (n)(n)/2

3. area of small triangle's = n/2


Area of ordinate satisfying greatest integer function from [0,n] is the sum of these areas....

(n)(1) + n^2/2 - n/2 = n(n-1)/2..
You did the algebra wrong. n+n^2/2-n/2=n*(n+1)/2. But yes, you work it out like that, but I think your graph is wrong. Isn't [t] the greatest integer less than or equal to t? Shouldn't [1/2] be 0? That's not what your graph shows.
 
  • #10
1,011
0
your correct... I realized that it was equal to n*(n+1)/2., which I replied back to saying that... and yes my graph is wrong... thank you.
 

Related Threads on Integral question

  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
646
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
941
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
9
Views
2K
Top