Taylor Series / Power Series

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March 3rd 2009, 03:55 PM
Fire Mage

Taylor Series / Power Series

Alright, I'm just a little confused on what my work is asking me. The problem says "Use the definition to find the Taylor series, centered at c, for the function."

The problem is:

$\cos(x), c = \pi/4$

So I have an answer book that has the answer for this problem. I understand Taylor series / power series, but I'm just a bit confused on what is going on.

The answer book goes on to do a few derivatives, plugs in the c, etc. Then it forms the sum which is:

$\frac{\sqrt2}{2}\sum{\frac{(-1)^{n(n+1)/2}(x-\frac{\pi}{4})^n}{n!}}$

Now I understand this is in the form of:

$\sum{\frac{f^n(c)(x-c)^n}{n!}}$

Yet... Why does it not follow the "basic list" of power series? Which has:

$\cos(x) = \sum{\frac{(-1)^nx^{2n}}{(2n)!}}$

Why is there only n powers in the answer, and no 2n to be found, etc.? Can someone please explain? Maybe I am mixing two concepts...
March 3rd 2009, 04:09 PM
skeeter

Quote:

Originally Posted by Fire Mage

Alright, I'm just a little confused on what my work is asking me. The problem says "Use the definition to find the Taylor series, centered at c, for the function."

The problem is:

$\cos(x), c = \pi/4$

So I have an answer book that has the answer for this problem. I understand Taylor series / power series, but I'm just a bit confused on what is going on.

The answer book goes on to do a few derivatives, plugs in the c, etc. Then it forms the sum which is:

$\frac{\sqrt2}{2}\sum{\frac{(-1)^{n(n+1)/2}(x-\frac{\pi}{4})^n}{n!}}$

Now I understand this is in the form of:

$\sum{\frac{f^n(c)(x-c)^n}{n!}}$

Yet... Why does it not follow the "basic list" of power series? Which has:

$\cos(x) = \sum{\frac{(-1)^nx^{2n}}{(2n)!}}$

Why is there only n powers in the answer, and no 2n to be found, etc.? Can someone please explain? Maybe I am mixing two concepts...

the one from the "basic list" is centered at c = 0 ... the odd powered terms cancel out because the odd degree derivatives are of the form $\pm \sin(0)$
March 3rd 2009, 04:16 PM
Fire Mage

Yes, I also understand that.

But what I'm trying to get at is: why does it become just n and not 2n? The only thing that changed it was the c. So why does changing the c automatically change the 2n's?
March 3rd 2009, 05:28 PM
matheagle

Quote:

Originally Posted by Fire Mage

Yes, I also understand that.

But what I'm trying to get at is: why does it become just n and not 2n? The only thing that changed it was the c. So why does changing the c automatically change the 2n's?

Because when you center at c=0, you have every other term evaluated at sin 0=0. BUT in this case all the derivatives of cosine are either plus and minus cosine or sine. And these four functions evaluated at $\pi/4$ you'll never get zero. Instead you'll get plus and minus $\sqrt{2}/2$ .
March 3rd 2009, 06:32 PM
Fire Mage

Quote:

Originally Posted by matheagle

Because when you center at c=0, you have every other term evaluated at sin 0=0. BUT in this case all the derivatives of cosine are either plus and minus cosine or sine. And these four functions evaluated at $\pi/4$ you'll never get zero. Instead you'll get plus and minus $\sqrt{2}/2$ .

Ah thank you for explaining it out. Like I comprehended the problem, understood everything, I can work the problem myself... I just didn't understand WHY.

But now that I read it, it's like a slap in the face. I totally ignored that fact. Thanks.
March 3rd 2009, 08:20 PM
matheagle

Sorry that I had to get physical with you.
March 4th 2009, 01:56 PM
Fire Mage

Haha, of course.