Well, we know how this, what this is like. We know what the Maclaurin series for cosine of x is. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. I will give you a hint. Function as a geometric series. This is going to be one minus I encourage you to pause the video and now try to do it. Let me just use a new color.

Plus x to the eighth over eight factorial, and it just keeps going, minus, plus, on and on and on and on. But it’s only going to get more and more painful as we take the second and third and fourth derivative. We want the first five non-zero terms. We say g of x is equal to cosine of x. I find the Maclaurin series representation for g, every place where I saw an x, I replaced it with what I have over here, the B x to the m power, where m is some exponent. I will give you a hint. If I wanted to write it as, I guess you could say as a function, or if I want to construct it using g of x, I could rewrite it as x to the third times The second derivative, this is going to be painful.

Just that was be pretty painful. Unlimited random practice problems and answers with built-in Step-by-step solutions.

### Expansion of $\cos \sqrt x $? – Mathematics Stack Exchange

We might have to take more, because some of these terms might end up being zeros. That actually will be the Maclaurin cox representation of what we started off with. Plus x to the eighth over eight factorial, and it just keeps going, minus, plus, on and on and on and on. I just told you that g of x Remember, the Maclaurin series is just the Taylor series centered at zero. Let’s say our goal here is the first five non-zero terms of the Maclaurin series representation, or Maclaurin series approximation of this.

A Maclaurin series is a Taylor series expansion of a function about 0.

### Power series for Cosine

G of x squared. I will give you a hint. You could just do this. But we just care about the first five terms, the non-zero terms, and we’re just saying this is an approximation, anyway. So x to the third times g, g of x maclauin.

We actually will get the Maclaurin series representation of this thing right over here. The Maclaurin series of a function up to order may be found using Series [ fx0, n ]. Power series of arctan 2x.

## Worked example: power series from cos(x)

Then you’re going to multiply that times x to the third. I’m jaclaurin you had paused the video and you naclaurin attempted to do this.

We say g of x is equal to cosine of x. Maclaurin series of cos x. You might be asking that question. Hints help you try the next step on your own. My answer to xos is, “Yes, you absolutely can. We’ve done that in the previous video. Walk through homework problems step-by-step from beginning to end. Now that I’ve given you a little bit of a reminder on the Maclaurin series representation of cosine of x, my hint to you is, can we use this to find the Maclaurin series representation of this, right up here?

So what do we do here?

So x squared squared is x to the fourth, x to the fourth over two factorial, which is really just two, but I like to put the factorial there because you see the pattern. Let me get myself some space right over here. Geometric series as a function.

This is going to be painful. Times g of B, B x to some other power, where I can fairly easily, without too much computation, maybe I already know, if I know the Maclaurin series representation of g of x, if I know what g of x is going to be, then I can do exactly what I did just in this video. Look malaurin “cosine Taylor series at zero” on Khan Academy, and you’ll find that.

Let me rewrite what I just wrote. I know we wanted the first five terms of this thing, but bear with me.

But it’s only going to get more and more painful as we take the second and third and fourth derivative. There are the Maclaurin series approximation of that is going to be one minus x squared over two factorial, plus x to the fourth over four factorial, minus x to the sixth over six factorial, plus, and I could keep going on and on and on.

Let’s evaluate them each at zero, and then use those for our coefficients, but you’re probably guessing correctly that there’s an easier way to do this. Cyclic Functions under Differentiation William Perry. Well, we know how this, what this is like. There’s a good chance that you might have gotten quite frustrated when you did this, because in order to find a Taylor series, Maclaurin series, we need to find the derivatives of seriss function, and as soon as you start to do that, it starts to get painful.

We know what the Maclaurin series for cosine of x is. And if you multiply that by x to the third, you’re just going to get another polynomial.

Visualizing Taylor series approximations. When you actually see what we got, you realize it would have taken you forever to do this by, I guess you could say, by brute force, because you would have had to find the 19th derivative of all of this madness.

Maclaurin series are a type of series expansion in which all terms are nonnegative integer powers of the variable. If I can rewrite my function so it is equal to maclaruin We’ll see how this thing right over here is going to be swries. Maclaurin series sin 3x.

G of x is just cosine. If you want to see that again, there’s another video. But when you realize that, “Hey, gee, “if I can just re-express this function “as, essentially, “x to a power times “something that I know the Maclaurin series of, “especially