Converting Sigma Notation: Simplifying Binomial Theorem

[Saitama]

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 \int f'(x)dx = x² + C
(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?

 

[I like Serena]

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

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

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

We want to find f(x).

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

So C = 3.

 

[Saitama]

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

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

We want to find f(x).

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

So C = 3.

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

 

[I like Serena]

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

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.

 

[Saitama]

i think the constant is zero.

 

[I like Serena]

i think the constant is zero.

Now I really don't understand you.

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

 

[Saitama]

Now I really don't understand you.

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 \int f'(x)=x^2+C at the place of C.

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

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

Did you get me now?

 

[I like Serena]

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:
f'(x) = x(x+1)^n \color{red}{\textbf{- x}}

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

So you should have:
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

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

 

[Saitama]

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:
f'(x) = x(x+1)^n \color{red}{\textbf{- x}}

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

So you should have:
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

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.

 

[I like Serena]

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

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

 

[Saitama]

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

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

Is it correct..?

 

[I like Serena]

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

 

[Saitama]

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

Ok fixed.
f(x) =-\frac{1^{n+2}}{(n+1)(n+2)}+ C

Is it ok..?

 

[I like Serena]

Yes...

 

[Saitama]

Yes...

But now what's the C?

 

[I like Serena]

But now what's the C?

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

 

[Saitama]

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

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

So therefore our final answer is:-
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)}

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

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

 

[I like Serena]

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

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

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.

 

[Saitama]

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

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.
So now we are done with this question, Right..?

 

[I like Serena]

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

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

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.