# Thread: use f'(x) to find f(x) explicitly

1. ## use f'(x) to find f(x) explicitly

If

$\displaystyle \displaystyle f(x)= \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{2n+1}$,

calculate $\displaystyle f'(x)$ and use it to find $\displaystyle f(x)$ explicitly. Justify each of your steps.

so I have obtained $\displaystyle f'(x)= (-1)^nx^{2n}$, what is the next step?

2. Originally Posted by wopashui
If $\displaystyle \displaystyle f(x)= \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}$, calculate $\displaystyle f'(x)$ and use it to find $\displaystyle f(x)$ explicitly. Justify each of your steps.

so I have obtained $\displaystyle f'(x)= (-1)^nx^{2n}$, what is the next step?
$\displaystyle \displaystyle f'(x) = \sum_{n=0}^{\infty} (-1)^n x^{2n} = 1 - x^2 + x^4 - x^6 + ...$

looks like a geometric series w/ common ratio $\displaystyle -x^2$ ...

3. Originally Posted by skeeter
$\displaystyle \displaystyle f'(x) = \sum_{n=0}^{\infty} (-1)^n x^{2n} = 1 - x^2 + x^4 - x^6 + ...$

looks like a geometric series w/ common ratio $\displaystyle -x^2$ ...
okay, but how is this related to f(x)?

4. Offhand, I'd say we can rewrite f' - which is a geometric series with common ration "-x^2" - using a certain formula.

Then integrate. It looks like arctan to me...

5. Originally Posted by wopashui
okay, but how is this related to f(x)?
For $\displaystyle |x|<1$ use the formula for the sum of a convergent geometric series to find the closed form expression for $\displaystyle f'(x)$. Now integrate this and use the fact that $\displaystyle f(0)=0$ to find the constant of integration and so the closed form for $\displaystyle $$f(x) CB 6. Originally Posted by CaptainBlack For \displaystyle |x|<1 use the formula for the sum of a convergent geometric series to find the closed form expression for \displaystyle f'(x). Now integrate this and use the fact that \displaystyle f(0)=0 to find the constant of integration and so the closed form for \displaystyle$$f(x)$

CB

I dun really understand what the question is asking, why does it mean to find f(x) explicitly when f(x) is given?

7. $\displaystyle f'(x) = 1 - x^2 + x^4 - x^6 + ... = \dfrac{1}{1 - (-x^2)} = \dfrac{1}{1 + x^2}$

integrate the last expression and use the given initial condition to find an explicit expression for $\displaystyle f(x)$

8. Originally Posted by skeeter
$\displaystyle f'(x) = 1 - x^2 + x^3 - x^4 + ... = \dfrac{1}{1 - (-x^2)} = \dfrac{1}{1 + x^2}$

integrate the last expression and use the given initial condition to find an explicit expression for $\displaystyle f(x)$
i dun think $\displaystyle f'(x) = 1 - x^2 + x^3 - x^4 + ...$, it's $\displaystyle 1 - x^2 + x^4 - x^6 + ...$, and what do you mean by the given initial condition, why are we using f'(x) to find f(x) if we already have f(x)? Im very confused here....

9. Originally Posted by wopashui
i dun think $\displaystyle f'(x) = 1 - x^2 + x^3 - x^4 + ...$, it's $\displaystyle 1 - x^2 + x^4 - x^6 + ...$, and what do you mean by the given initial condition, why are we using f'(x) to find f(x) if we already have f(x)? Im very confused here....
sorry about the typo ... $\displaystyle f'(x) = 1 - x^2 + x^4 - x^6 + ... = \dfrac{1}{1 + x^2}$

you were given $\displaystyle f(x)$ as a series ... the question wants you to find the explicit function it represents (that means not a series)

$\displaystyle f(x)$ is the antiderivative of $\displaystyle \dfrac{1}{1+x^2}$

you need to initial condition to find the constant of integration.

10. Originally Posted by wopashui
I dun really understand what the question is asking, why does it mean to find f(x) explicitly when f(x) is given?
For this once I will complete the solution for this question so that you can see what we mean by a closed form solution.

The question means: find a closed form expressions for $\displaystyle $$f(x), by closed form we mean a finite representation with no infinite series etc in it. As has been pointed out elsewhere in this thread we have: \displaystyle f'(x)=1-x^2+x^4-... = \dfrac{1}{1+x^2} Integrating this gives: \displaystyle f(x)=\arctan(x)+c Now we know that \displaystyle f(0)=0, so we have: \displaystyle \arctan(0)+c=0 so \displaystyle$$c=0$ and:

$\displaystyle f(x)=\arctan(x)$

CB

11. Originally Posted by wopashui
okay, but how is this related to f(x)?
Originally Posted by TheChaz
Offhand, I'd say we can rewrite f' - which is a geometric series with common ration "-x^2" - using a certain formula.

Then integrate. It looks like arctan to me...
I don't mind being ignored, but when you ignore three helpful people...

f'(x) = a geometric series.
There is a formula for this: 1/(1-r), where r is the common ratio. So now you can rewrite f'(x).
"How is f' related to f ?" - you integrate f' to get f.
Where was the trouble in this process for you?