# Thread: expanding f(x) into the maclaurin series

1. ## expanding f(x) into the maclaurin series

Expand the function $f(x) = sin^6(x) + cos^6(x)$ into Maclaurin series on the whole real line $\mathbb{R}$.

So here is what I have done so far:

From what I have been able to figure out, though I am not sure this is correct, the Maclaurin series is defined as

$f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f''''(0)x^4}{4!} + \frac{f^5(0)x^5}{5!} + ...$.

I solved for $f(0) = sin^6(0) + cos^6(0) = 0 + 1 = 1$.

Similarly, $f'(x) = 6cos(x)sin^5(x) - 6sin(x)cos^5(x)$, giving $f'(0) = 0$.

Once again, $f''(x) = 6[5sin^4(x)cos^2(x) - sin^6(x)] - 6[cos^6(x) - 5cos^4(x)sin^2(x)]$, giving $f''(0) = -6$.

Is the right direction? Should I be finding some pattern here? What is the final answer supposed to look like?

2. ## Re: expanding f(x) into the maclaurin series

Originally Posted by vidomagru
Expand the function $f(x) = sin^6(x) + cos^6(x)$ into Maclaurin series on the whole real line $\mathbb{R}$.

So here is what I have done so far:

From what I have been able to figure out, though I am not sure this is correct, the Maclaurin series is defined as

$f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f''''(0)x^4}{4!} + \frac{f^5(0)x^5}{5!} + ...$.

I solved for $f(0) = sin^6(0) + cos^6(0) = 0 + 1 = 1$.

Similarly, $f'(x) = 6cos(x)sin^5(x) - 6sin(x)cos^5(x)$, giving $f'(0) = 0$.

Once again, $f''(x) = 6[5sin^4(x)cos^2(x) - sin^6(x)] - 6[cos^6(x) - 5cos^4(x)sin^2(x)]$, giving $f''(0) = -6$.

Is the right direction? Should I be finding some pattern here? What is the final answer supposed to look like?
I'm not seeing any particular pattern. Note though, that the odd n terms are 0.

-Dan

3. ## Re: expanding f(x) into the maclaurin series

Originally Posted by vidomagru
Expand the function $f(x) = sin^6(x) + cos^6(x)$ into Maclaurin series on the whole real line $\mathbb{R}$.

So here is what I have done so far:

From what I have been able to figure out, though I am not sure this is correct, the Maclaurin series is defined as

$f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f''''(0)x^4}{4!} + \frac{f^5(0)x^5}{5!} + ...$.

I solved for $f(0) = sin^6(0) + cos^6(0) = 0 + 1 = 1$.

Similarly, $f'(x) = 6cos(x)sin^5(x) - 6sin(x)cos^5(x)$, giving $f'(0) = 0$.

Once again, $f''(x) = 6[5sin^4(x)cos^2(x) - sin^6(x)] - 6[cos^6(x) - 5cos^4(x)sin^2(x)]$, giving $f''(0) = -6$.

Is the right direction? Should I be finding some pattern here? What is the final answer supposed to look like?
I'd start like this...

\displaystyle \begin{align*} \sin^6{(x)} + \cos^6{(x)} &= \left[ \sin^2{(x)} \right] ^3 + \cos^6{(x)} \\ &= \left[ 1 - \cos^2{(x)} \right] ^3 + \cos^6{(x)} \\ &= 1 - 3\cos^2{(x)} + 3\cos^4{(x)} - \cos^6{(x)} + \cos^6{(x)} \\ &= 1 - 3\cos^2{(x)} + 3\cos^4{(x)} \\ &= 1 - 3 \left[ \frac{1}{2} + \frac{1}{2}\cos{(2x)} \right] + 3 \left[ \frac{1}{2} + \frac{1}{2}\cos{(2x)} \right] ^2 \\ &= 1 - \frac{3}{2} - \frac{3}{2}\cos{(2x)} + 3 \left[ \frac{1}{4} + \frac{1}{2}\cos{(2x)} + \frac{1}{4}\cos^2{(2x)} \right] \\ &= -\frac{1}{2} - \frac{3}{2}\cos{(2x)} + \frac{3}{4} + \frac{3}{2}\cos{(2x)} + \frac{3}{4}\cos^2{(2x)} \\ &= \frac{1}{4} + \frac{3}{4}\cos^2{(2x)} \\ &= \frac{1}{4} + \frac{3}{4} \left[ \frac{1}{2} + \frac{1}{2}\cos{(4x)} \right] \\ &= \frac{1}{4} + \frac{3}{8} + \frac{3}{8}\cos{(4x)} \\ &= \frac{5}{8} + \frac{3}{8}\cos{(4x)} \end{align*}

Now if you recall the MacLaurin Series for \displaystyle \begin{align*} \cos{(t)} = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n)!}\,t^{2n} = 1 - \frac{t^2}{2} + \frac{t^4}{4!} - \frac{t^6}{6!} + \dots \end{align*} you should be able to manipulate it to get the series you are after...

4. ## Re: expanding f(x) into the maclaurin series

Originally Posted by Prove It
Now if you recall the MacLaurin Series for \displaystyle \begin{align*} \cos{(t)} = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n)!}\,t^{2n} = 1 - \frac{t^2}{2} + \frac{t^4}{4!} - \frac{t^6}{6!} + \dots \end{align*} you should be able to manipulate it to get the series you are after...
I have $1 - \frac{6x^2}{2!} + \frac{96x^4}{4!} - \frac{1536x^6}{6!} + ...$ but I cannot figure out how to manipulate that into a series similar to the Maclaurin Series Expansion of $cos(x)$.

Thoughts?

5. ## Re: expanding f(x) into the maclaurin series

$\dfrac{7}{8} + \dfrac{3}{8}\sum_{n\ge 0} \dfrac{(-1)^n}{(2n)!}(4x)^{2n}$

6. ## Re: expanding f(x) into the maclaurin series

Originally Posted by SlipEternal
$\dfrac{7}{8} + \dfrac{3}{8}\sum_{n\ge 0} \dfrac{(-1)^n}{(2n)!}(4x)^{2n}$
I don't think that evaluates out to $1 - \frac{6x^2}{2!} + \frac{96x^4}{4!} - \frac{1536x^6}{6!} + ...$.

7. ## Re: expanding f(x) into the maclaurin series

That was supposed to be a 5/8, not a 7/8. Let's calculate the first few terms:

$\dfrac{5}{8} + \dfrac{3}{8} - \dfrac{3\cdot 16x^2}{8\cdot 2!} + \dfrac{3\cdot 256x^4}{8\cdot 4!} - \dfrac{3\cdot 1024x^6}{8\cdot 6!} \pm \cdots$

$= 1 - \dfrac{6x^2}{2!} + \dfrac{96x^4}{4!} - \dfrac{1536x^6}{6!} \pm \cdots$

Looks the same to me.