# MACLAURIN SERIES EXPANSION OF ARCTAN

Maclaurin series of cos x. It’s the antiderivative of two minus eight, minus eight x squared plus 32 x to the fourth minus x to the sixth. Well this is just one times x squared so let me just write that. Just said quickly, I was able to find the first three derivatives of g of x and it’s very easy then to find the first four terms of its power series representation especially it’s Maclaurin, that’s Maclaurin series if the power series centered at zero. ADD A slick solution is the following: We’re centering at zero so our approximation evaluated zero better be the same thing as a function of evaluated zero. We just have to evaluate each of these at zero. Well three factorials are six so negative six divided by six is just negative one. Email Required, but never shown. It’s going to be the derivative of arctangent of x is one over one plus x squared so this is going to be the derivative of this which is two over one plus this whole thing squared. Power series of arctan 2x. You’re simply forgetting to divide by the factorials, as far as I can see. It’s our function evaluated to add zero. This is an interesting thing because it’s really easy and this is the same thing as one plus x to the negative one power. The key insight here is well, instead of doing it directly, let’s see if we could find the power series representation, the first four terms of this thing right over here and then we can take the antiderivative of that to find the power series of arctangent of two x making sure that we get the constant right that it satisfies the fact that we are being centered at zero, so I know what you’re thinking now. The key insight is let’s just say f of x which of course is the derivative of arctangent of two x is two over one plus four x squared.

Let me write that, minus 64 x to the sixth power and then we could say that this is going to be so f of x is approximately equal to, arcta the two, two minus eight x squared plus 32 x to the fourth minus x to the sixth. You are right, there is a key insight here.

### Maclaurin Series of Arctanx | eMathZone

To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Mathematics Stack Exchange works best with JavaScript enabled. Dx and what is this going to be equal to? Notice you replace your x’s with a four x squared. It’s our function evaluated to add zero. C must be equal to zero if we want this thing to be zero when X is equal to zero.

We’re going to have a constant, if I take the antiderivative of two, that’s going to be plus two x. Video transcript – [Voiceover] What I would like us to do in this video is find the power series representation of or find the power series approximitation chuckles the power series approximation of arctangent of two x centered at zero and let’s just say we want the first four nonzero terms of the power series approximation of arctangent of two x centered at zero so it’s essentially the Maclaurin Series of arctangent of two x, the first four nonzero terms of it.

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Just said quickly, I was able to find serie first three derivatives of g of x and it’s very easy then to find the first four terms of its power series representation especially it’s Maclaurin, that’s Maclaurin series if the power series centered at zero.

You’re simply forgetting to divide by the factorials, as far as I can see. I could write the g of x is approximately equal, arctn I’m just going to do the first four terms here is going to be g of zero which is one minus g prime of zero times x.

Well that’s going to be 16 x to the fourth, so let me just write that, that’s going to be plus 16 x to the fourth and finally minus x to the third but now an x is four x squared so it’s minus four x squared to the third power. Home Questions Tags Users Unanswered. Let’s write this down, so I’m going to write, I’m going to write down, so arctangent of two x, arctangent of two x which is equal to an antiderivative of f of x, dx which maclaurln going to be equal to an antiderivative of this arctah, all of this business.

The key insight here is well, instead of doing it directly, let’s see if we could find the power series representation, the first four terms of this thing right over here and then we can take the antiderivative of that to find the power series of arctangent of two x making sure that we arctzn the constant right that it satisfies the fact that we are being centered epxansion zero, so I know what you’re thinking now.

Mzclaurin you feel good about it, I encourage you to pause the video and try to work through it yourself. That’s going to be negative x to the third, negative x to the third and I know what you’re thinking. The key insight is let’s just say f of x which of course is the derivative of arctangent of two x is two over one plus four x squared.

Geometric series interval of convergence. Now, if we had another function that cleans it up a little bit so that we don’t get wxpansion these hairiness when we take the derivatives. Sign up using Facebook.

### How do you find the Maclaurin series for arctan x centered at x=0? | Socratic

If you’re seeing this message, it means we’re having trouble xepansion external resources on our website. You probably saw that hey, the derivative with respect to x of arctangent of two x is equal to and this is a refresher if you didn’t realize it the first time. That’s expanion the most basic thing if we’re doing the Maclaurin Series representation.

For example, g prime of x is going to be equal to chain rule derivative one plus x is just one so it’s going to be equal to negative one plus x to the negative two power if I want to take the second derivative of that.

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Maclaurin series of sin x. This is essentially evaluate to what arctangent of when this function, when x is equal to zero. Yak Attack Yak Attack weries 2 3 9. Just like that with a little bit of a substitution, I was able to reasonably a non hairy fashion find the first four nonzero maclaruin of the arctn series of two over one plus four x squared which is the derivative, which is going to be the derivative of the power series of arctangent of two x.

If this is correct and there are indeed the two series, an explanation of this oddity would be much appreciated, though I don’t see how this can be the case. Thanks for your help. Email Required, but never shown.

## How do I calculate the Maclaurin series for #arctan (x^3)#?

This is an interesting thing because it’s really easy and this is the same thing as one plus x to the negative one power. I srctan accepted your answer.  If f of x is equal to that then f of x is power series representation its just going to be taking this power series or at least the first four terms of it and replacing the x’s with four x squares and then multiplying the whole thing times two, so let’s do that. We’re centering at zero so our approximation evaluated zero better be the same thing as a function of evaluated zero.

## Power series of arctan(2x)

Let’s say maclaurun had another function g of x, I’m using the color that I have not used. We can write that f of x, so we could write the f of x is going to be approximately equal to two times this thing, a value when x maclsurin equal to four x squared. Post as a guest Name. Power series of arctan 2x.

Maclaurin series of cos x. That’s minus one times x, so that’s negative x plus g prime, prime of zero two over two factorial times x squared. Are both series correct, or is it a methodical error on my part?

That’s just going to be plus x squared and expanaion we have plus g prime, prime, prime of zero which is negative six over three factorial times x to the third. 