Hello,
I am learning about Taylor series in school, but they are quite confusing..
Can you please explain how I should do this problem? Thank you very much!
What is the Taylor polynomial of order 5 approximation to sin(1.5) ?
I know the series of sinx = x - ((x^3)/(3!)) + ((x^5)/(5!))-...+...
but where do i go from here?
THANK YOU~
If
"I know the series of sinx = x - ((x^3)/(3!)) + ((x^5)/(5!))-...+..."
this is given, you only have to substitute 1.5(in radians) as x into the above series up to x^5 as requested.
For example:
sin(1.5)=1.5-1.5^3/3!+...
However, in practice, I would not stop at order 5 because the error is excessive for this value of x.
Of course! I can help you with that.
To find the Taylor polynomial of order 5 approximation to sin(1.5), we can use the series expansion of sinx that you mentioned: sinx = x - ((x^3)/(3!)) + ((x^5)/(5!)) - ...
In this case, we want to find the Taylor polynomial of order 5, which means we need to include terms up to the fifth degree. So, we'll take the terms up to x^5 and ignore the terms beyond that.
To form the Taylor polynomial, we'll substitute x = 1.5 into the series expansion.
For the first term, we have x = 1.5, so the first term becomes:
1.5
For the second term, we need to substitute x = 1.5 as well as evaluate the power of x:
((1.5)^3) / (3!) = (1.5 * 1.5 * 1.5) / (3 * 2 * 1) = 1.125 / 6
For the third term, we do the same, but with the power of x:
((1.5)^5) / (5!) = (1.5 * 1.5 * 1.5 * 1.5 * 1.5) / (5 * 4 * 3 * 2 * 1) = 1.875 / 120
To calculate the Taylor polynomial of order 5, just sum up the terms we got so far:
Taylor polynomial of order 5 ≈ 1.5 - (1.125 / 6) + (1.875 / 120)
Simplifying this expression will give you the approximation of sin(1.5) using the Taylor polynomial of order 5.
Remember, the more terms you include in the Taylor polynomial, the more accurate your approximation will be. In this case, the Taylor polynomial of order 5 will give a fairly good approximation of sin(1.5).