Converting Sigma notation

  • Thread starter Saitama
  • Start date
  • #51
3,816
92


You had to subtract the k=0 term, which ended up being just "x".
I don't see it back in your integrated result.




Let me explain with an example.

Suppose you have the function f(x)=x2 + 3
Now you know that f(0)=3 don't you?

Taking the derivative we get f'(x)=2x.
Taking the integral again we get [itex]\int f'(x)dx = x2 + C[/itex]
(This is called an indefinite integral, since the boundaries were not given.)

As you can see the result is not equal to the original function - we lost the '3' in the process.
But since we know that f(0)=3, we can deduce that the integration constant C must be 3.

In which step should i substitute zero to get the constant? :confused:
 
  • #52
I like Serena
Homework Helper
6,579
176


In which step should i substitute zero to get the constant? :confused:

I'm assuming you mean in my example.
(It's slightly different for the current problem.)

In my example we have as input [itex]f(x) = \int f'(x) dx = x^2 + C[/itex] and f(0)=3.

We want to find f(x).

Filling in x=0 gives: [itex]f(0) = 0^2 + C = 3[/itex].

So C = 3.
 
  • #53
3,816
92


I'm assuming you mean in my example.
(It's slightly different for the current problem.)

In my example we have as input [itex]f(x) = \int f'(x) dx = x^2 + C[/itex] and f(0)=3.

We want to find f(x).

Filling in x=0 gives: [itex]f(0) = 0^2 + C = 3[/itex].

So C = 3.

I got what you said. :smile:
But i am asking in which step should i put x=0?
Should i put x=0 during the integration step? :confused:
 
  • #54
I like Serena
Homework Helper
6,579
176


I got what you said. :smile:
But i am asking in which step should i put x=0?
Should i put x=0 during the integration step? :confused:

I don't understand your question.
Afaik I have explicitly stated where you put x=0.

I only could have added as a last step that the conclusion is that f(x)=x2 + 3.
 
  • #55
3,816
92


i think the constant is zero. :confused:
 
  • #56
I like Serena
Homework Helper
6,579
176


i think the constant is zero. :confused:

Now I really don't understand you. :confused:

Which constant?
In which formula?
In which post?
Why would it be zero?
 
  • #57
3,816
92


Now I really don't understand you. :confused:

Which constant?
In which formula?
In which post?
Why would it be zero?

You gave an example that f(x)=x2+3.
And you said that f(0)=3, i.e you substituted the value 0 at the place of x. Right..?
Now that 3 is our integration constant, we can substitute 3 in [itex]\int f'(x)=x^2+C[/itex] at the place of C.

I did the same in my question. I substituted 0 in f(x), i.e i substituted 0 in:-
[tex]f(x)=\sum_{k=1}^n \frac{{}^nC_k}{k+2}x^{k+2}[/tex]

So i got f(0)=0. Therefore integration constant is 0.

Did you get me now? :confused:
 
  • #58
I like Serena
Homework Helper
6,579
176


Yes, I more or less get what you did, but it is not right.


Let's take it a couple of steps back.

You had:
[itex]f'(x) = x(x+1)^n \color{red}{\textbf{- x}}[/itex]

You integrated this wonderfully, but you forgot the "[itex]\color{red}{\textbf{- x}}[/itex]" I just marked.

So you should have:
[tex]f(x) = \int f'(x)dx =\frac{x(x+1)^{n+1}}{n+1}-\frac{(x+1)^{n+2}}{(n+1)(n+2)} - \frac {x^2} 2 + C[/tex]

Now you need to determine the integration constant C using f(0) = 0, but C is not zero!
 
  • #59
3,816
92


Yes, I more or less get what you did, but it is not right.


Let's take it a couple of steps back.

You had:
[itex]f'(x) = x(x+1)^n \color{red}{\textbf{- x}}[/itex]

You integrated this wonderfully, but you forgot the "[itex]\color{red}{\textbf{- x}}[/itex]" I just marked.

So you should have:
[tex]f(x) = \int f'(x)dx =\frac{x(x+1)^{n+1}}{n+1}-\frac{(x+1)^{n+2}}{(n+1)(n+2)} - \frac {x^2} 2 + C[/tex]

Now you need to determine the integration constant C using f(0) = 0, but C is not zero!

So then what's the C, i don't seem to find any other way to find C. :confused:
 
  • #60
I like Serena
Homework Helper
6,579
176


So then what's the C, i don't seem to find any other way to find C. :confused:

What do you get if you fill in x=0 in this formula?
[tex]f(x) = \frac{x(x+1)^{n+1}}{n+1}-\frac{(x+1)^{n+2}}{(n+1)(n+2)} - \frac {x^2} 2 + C[/tex]
 
  • #61
3,816
92


What do you get if you fill in x=0 in this formula?
[tex]f(x) = \frac{x(x+1)^{n+1}}{n+1}-\frac{(x+1)^{n+2}}{(n+1)(n+2)} - \frac {x^2} 2 + C[/tex]

If i would fill x=0 in this formula i get
[tex]f(x) = \frac{0^{n+2}}{n+1}-\frac{0^{n+2}}{(n+1)(n+2)}+ C[/tex]

Is it correct..?
 
  • #62
I like Serena
Homework Helper
6,579
176


Noooo. ;)
You need to fix the second term.
 
  • #63
3,816
92


Noooo. ;)
You need to fix the second term.

Ok fixed. :smile:
[tex]f(x) =-\frac{1^{n+2}}{(n+1)(n+2)}+ C[/tex]

Is it ok..?
 
  • #64
I like Serena
Homework Helper
6,579
176


Yes..... :smile:
 
  • #66
I like Serena
Homework Helper
6,579
176


But now what's the C?

Set the expression equal to zero (since we had f(0)=0) and solve C.
 
  • #67
3,816
92


Set the expression equal to zero (since we had f(0)=0) and solve C.

I set it to 0 and i get
[tex]C=\frac{1^{n+2}}{(n+1)(n+2)}[/tex]

So therefore our final answer is:-
[tex]f(x)=\frac{x(x+1)^{n+1}}{n+1}-\frac{(x+1)^{n+2}}{(n+1)(n+2)} - \frac {x^2} 2 + \frac{1^{n+2}}{(n+1)(n+2)}[/tex]

Substituting x=1, i get
[tex]f(1)=\frac{n.2^{n+2}-n^2-3n-4}{2(n+1)(n+2)}[/tex]

Is it right..?
Can i simplify it further?
 
  • #68
I like Serena
Homework Helper
6,579
176


Is it right..?
Can i simplify it further?

You're expression for f(x) is right! :smile:

However your simplification after substitution of x=1 is wrong. :(

You can check by for instance filling in n=1.
(What should the result be for n=1 and for n=2? And does your expression match it?)

And actually I would recommend simplifying as little as possible.
Simplify the obvious things.
But do not try to simplify too much, since it opens up the door to mistakes.
And especially if the resulting expression is not really simpler, it's not worthwhile anyway.
 
  • #69
3,816
92


You're expression for f(x) is right! :smile:

However your simplification after substitution of x=1 is wrong. :(

You can check by for instance filling in n=1.
(What should the result be for n=1 and for n=2? And does your expression match it?)

And actually I would recommend simplifying as little as possible.
Simplify the obvious things.
But do not try to simplify too much, since it opens up the door to mistakes.
And especially if the resulting expression is not really simpler, it's not worthwhile anyway.

Thanks, i would correct it. :smile:
So now we are done with this question, Right..?
 
  • #70
I like Serena
Homework Helper
6,579
176


Thanks, i would correct it. :smile:
So now we are done with this question, Right..?

Hmm, you still didn't give the right answer.... :uhh:

But if you substitute x=1 without simplifying it, that would be a proper answer, so I guess we're done, with only a technicality left. :smile:
 

Related Threads on Converting Sigma notation

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
21
Views
7K
  • Last Post
Replies
2
Views
6K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
4
Views
4K
Top