Taylor's formula is
now reevaluate your series as you are missing the terms in your expansion.
Kalyan.
I've just covered Taylor polynomials and series theory, but there are precious few examples so i'm trying to do the following exercise:
Compute Taylor series of f(x) = cos 3x at pi/12
My attempted answer:
f(x)= cos3x = cos pi/4
f'(x) = -3 sin 3x = -3 sin pi/4
f''(x) = -9 cos 3x = -9 cos pi/4
etc
so Tn cos3x = cos3x + (-3 sin 3x) x/2! + (-9 cos pi/4) x^2/ 3!...
I can't figure how I should go about formulating a series formula for this , especially as trig functions are involved. I'm also not even sure I'm doing the Taylor formula right as i've seen some wierd trig series. Any advice on how to start would be much appreciated
Thanks
Thank you, so the corrected taylor polynomial, where a=pi/12, is
Thank you
so Tn cos3x = cos3x + (-3 sin 3x) (x-a)/2! + (-9 cos 3x) (x-a)2/ 3!... = cos pi/4 + (-3 sin pi/4) (x-pi/12)/2! + (-9 cos pi/4) (x-pi/12)2/ 3!... = 1/root(2) -3/root(2) (x-pi/2)/2 - ( 3/2 )(x- pi/12)^2/ root(2)...
But how do i go about rewording this into to a series?
I am sure that the generalization is within the scope of your of your course material but you have to take note of a some facts before you proceed to the generalized term.
Lets consider the values contributed by term in each term
So you see that the pattern is recursive in the sense that you have
We need to find a trigonometric function fitting the pattern so all I did was observe that the pattern is recursive over four terms with signs of the terms occurring as +,-,-,+,.... this is the cos pattern i.e cosine function is +ve in first quadrant, -ve in the second, -ve in the third and again +ve in fourth quadrant so I was sure cos would fit in. And so I use to generate the quadrant nature and add to it.
The powers of occurring due to the derivative can be clubbed with the term to create term. When n=1 this term is in fact 1 (I hope you can observe that) so just leaving you with it fits in nicely.
Voila!.. As everything is in place you can now generalize it as .
That you could have generated a function say where [0.5n] denotes the greatest integer function as also generates the same sign pattern of +,-,-,+,+,-,-,+,.... so I could have said that the generalized nth term is
but I choose the cosine function to give the function a trigonometric touch. As you see the generalization is all about observation and packing things into a nice presentable form and its all about how you want to present your function in the generalized form and there is no one concrete way to say that this is the only way as I have discussed it.
So the Taylor series can be expressed as
or as
which ever way you choose
Hope that helps without confusing much ...
Kalyan
Wow that makes alot more sense now. but the part where you fit in the 3 to the power term doesn't though ie :
1st term is ok, but the 2nd onward not, the Taylor definition uses the and you slotted in the 3 there but when i try calculate the 2nd term from this i get , 3rd term etc
The powers of occurring due to the derivative can be clubbed with the term to create term. When n=1 this term is in fact 1 (I hope you can observe that) so just leaving you with it fits in nicely.
I don't understand this. I would have put but i know that is wrong and also not Taylorish
Ok but I'm still going wrong, sorry please bear with me:
For n=1:
For n=2:
For n=3:
But i can now see that i could get the answer if i take the derivative of what is in brackets to get at least some of what I want eg:
For n=2 it works out:
But for n=3 it doesn't as I still have the x in my result:
But besides, why would i differentiate these terms, if indeed this was the way to go about it? From the Taylor polynomial we only differentiate the function part cos3x as we did already, not the
right?
BTW, tell whoever has set this question that this is a stupid question.
The point of a Taylor series is to use a particular combination of addition, subtraction, multiplication, division and exponentiation to approximate numbers you normally would not be able to evaluate.
There is no point in centering a Taylor series about , because it gives surd coefficients (which you can not easily evaluate) around a point that can not be easily evaluated.