Let’s see if we can find a similar pattern if we try to approximate sine of x using a Maclaurin series. So let’s do that. I didn’t write it here. And that’s when it starts to get really, really mind blowing. The third derivative is going to be the derivative of this. But it’s essentially 0, 2, 4, 6, so on and so forth.
Sine of x, when you take its polynomial representation, is all of the odd powers of x divided by its factorial, and you switch sines. Then I don’t have x to the first power. So I’ll just write a maclaurkn in parentheses there, instead findd doing prime, prime, prime. The second derivative of the sine of x is the derivative of cosine of x, which is negative sine of x. So f prime prime, the second derivative evaluated at 0 is 0.
You have a negative out there. Then the next term is f prime, the second derivative at 0, which we see here is 0.
We approximated it using this polynomial. And whatever power it is, I’m dividing it by that factorial. But it’s essentially 0, 2, 4, 6, so on and so forth. Yeah, 0 would be an even number.
This green should be nice. So f– that’s hard to see, I think So let’s do this other blue color. Function as a geometric series.
This is x to the third over 3 factorial plus x to the fifth over 5 factorial. So let’s find its polynomial representation using the Maclaurin series. The next term is going to be f prime of 0, which is 1, times x. Cosine of 0 is 1. And let’s do the same thing that we did with cosine of x. This third term right here, the third derivative of sine of x evaluated at 0, is negative 1.
Taylor/Maclaurin Series Calculator – Symbolab
And so the first term here, f of 0, that’s just going to be 0. So for this, let me do this in a different color, not that same blue. So there’s something interesting here.
And then the fourth derivative evaluated at 0 is going to be 0 again. And you really do get sinx combination when you involve imaginary numbers. So I’ll pick a new color. The fourth derivative is the derivative of this, which is positive sine of x again.
Negative this is negative 1 in this case– times x to the third over 3 factorial. So there’s something interesting going on here. And f, the first derivative evaluated at 0, is 1.
And we saw this pretty interesting pattern. So this is our new p of x.
How do you find the maclaurin series expansion of #sin2x#?
Let’s just take the different derivatives of sine of x really fast. The third derivative evaluated at 0 is negative 1. So we won’t have the second term.