### MACLAURIN SERIES COS3X

We want the first five non-zero terms. 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. My answer to you is, “Yes, you absolutely can. Plus x to the eighth over eight factorial, and it just keeps going, minus, plus, on and on and on and on. We actually will get the Maclaurin series representation of this thing right over here. So we could say that f of x is approximately equal to x to the third times x to the third times Plus, if our x is now x squared, x squared to the fourth power is x to the eighth, x to the eighth power over four factorial, minus x squared to the sixth power is x to the 12th over six factorial, and then plus x squared to the eighth is x to the 16th power over eight factorial, and of course, we can keep going on and on and on.

### Free Maclaurin Series Calculator Online | [email protected]

Function as a geometric series. Remember, the Maclaurin series is just the Taylor series centered at zero. But it’s only going to get more and more painful as we take the second and third and fourth derivative. That’s the first five non-zero terms, and we are done.

But we wanted first five terms, so this is a start. I will give you a hint. You could just do this.

Plus, if our x is now x squared, x squared to the fourth power is x to the eighth, x to the eighth power over four factorial, minus x squared to the sixth power is x to the 12th over six factorial, and then plus x squared to the eighth is x to the 16th power over eight factorial, and of course, we can keep going on and on and on.

Well, can’t we just apply this insight to the approximation? F prime of x is going to be, let’s see, product rule, so it’s three x squared times cosine of x squared plus x to the third times the derivative of this thing, which is going to be two x times negative sine of x, negative sine of x squared. If it’s equal to some A x to the n power, times some function Then the third and fourth derivatives are going to be even more painful.

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Look up “cosine Taylor series at zero” on Khan Academy, and you’ll find that.

Visualizing Taylor series approximations. Geometric series as a function. But we already know from that, and this is one of the most famous Maclaurin series, we know that this is coos3x g of x. This is going to cos33x one minus I encourage you to pause the video again and see if you can work through this. 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.

You kind of see where this is going.

Plus x to the serkes over eight factorial, and it just keeps going, minus, plus, on and on and on and on. 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.

We know what the Maclaurin series for cosine of x is.

## Maclaurinâ€™s series

Let’s say our goal here is the first five non-zero terms of the Maclaurin series representation, or Maclaurin series approximation of this. If you want to see that again, there’s another video.

Just that was be pretty painful. We actually will get the Maclaurin series representation of this thing right over here.

### Taylor Series of cos(3x^2) | Physics Forums

So x to the third times g, g of x squared. So every place where we see an x, let’s replace it with an x squared. 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. I encourage you to pause the video and now try to do it.

Maclaurin series of sin x. The second derivative, this is going to be painful. So we can say that this is going to be approximately equal to, we distribute the x to the third, and I’m going to do it in magenta just for fun, so distribute the x to the third, we get x to the third, minus x to the seventh over two factorial, plus x to the 11th over four factorial, maclaurkn x to the 15th over six factorial, plus x to the 19th 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.

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You might be asking that question. So mqclaurin could say that f of x is approximately equal to x to the third times x to the third times Let me say it in a less confusing way. But we just care about the first five terms, the non-zero terms, and we’re just saying this is an approximation, anyway. As I know I have to calculate the first, the second, the third deriavtive of the initial function, but how does it work in this case?

Let me just use a new color. And if you multiply that by x to the third, you’re just going to get another polynomial. 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 By clicking “Post Your Answer”, you acknowledge that you have read our updated terms of serviceprivacy policy and cookie policyand that your continued use of the website is subject to these policies.

Well, we know how this, what this is like. Now that I’ve given msclaurin a maclayrin 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? I know we wanted the first five terms of this thing, but bear with me.

If I can rewrite my function so it is equal to some Zauberkerl Zauberkerl 1, 1 7 Post coos3x a guest Name. 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 We’ll see how this thing right over here is going to be useful.

G of x is just cosine.