The next term is going to be f prime of 0, which is 1, times x. Geoff Pointer Geoff Pointer 1, 6 Let’s just take the different derivatives of sine of x really fast. So let’s take f of x in this situation to be equal to sine of x. Sign up or log in Sign up using Google. So is using the Maclaurin series considered as using the Taylor series expansion? Geometric series as a function. The coefficients alternate between 0, 1, and

So f of 0, in this situation, is 0. A shortened version is:. And what’s really fascinating is that e to the x starts to look like a little bit of a combination here, but not quite. And then the sines keep switching. This green should be nice. So we’re not even going to need to include that.

## How do you find the Maclaurin Series for #Sin(x^2)#?

And if you go to infinity, you’re going to be pretty much at cosine of x. Email Required, but never shown. And you really do get the combination when you involve imaginary numbers.

So this is interesting, especially when you compare to this.

This is x to the third over 3 factorial plus x to the fifth over 5 factorial. A shortened version is:.

seroes And what you’re going to see here– and actually maybe I haven’t done enough terms for you, for you ror feel good about this. Post as a guest Name. By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service.

And if I kept doing this, it would be positive I have to write the 1 as the coefficient– times x to the fifth over 5 factorial.

## How do you find the Maclaurin Series for #sin(x^2)/x#?

Shouldn’t it be the first term? To find the Maclaurin series coefficients, we must evaluate. Anyway, my brain is in a different place right now.

The coefficients alternate between 0, 1, and Just so it becomes clear. You see almost this– they’re kind of filling each other’s gaps over here. Step 4 This step was nothing more than substitution of our formula into the formula for the ratio test. It really makes more sense to point to the form of the general term and describe terms in the series by index values.

And so the first term here, f of 0, that’s just going to be 0. This third term right here, the third derivative of sine of x evaluated at 0, is negative 1.

What do you mean by the zeroth term? And f, the first derivative evaluated at 0, is 1.

### How do you find the Maclaurin Series for Sin(x^2)? | Socratic

Another approach could be to equared a trigonometric identity. This step was nothing more than substitution of our formula into the formula for the ratio test. Step 3 A helpful step to find a compact expression for the n th term in the series, is to write out more explicitly the terms in the series that we have found: I write it here only because I think it plays against this exposition nicely. Some people believe very strongly that you should.

Home Questions Tags Users Unanswered. I think this is a nice and clear post. Because this limit is zero for all real values of xthe radius of convergence of the expansion is the set of all real numbers.

### Maclaurin Expansion of sin(x) | The Infinite Series Module

Maclauin Questions Tags Users Unanswered. And we care– in order to do the Maclaurin, series– we care about evaluating the function, and each of these derivatives at x is equal to 0. Similarly with Abelian, named after Abel. Geometric series as a function. Negative this is negative 1 in this case– times x to the third over seroes factorial. So I’ll pick a new color. So there’s something interesting going on here. The fifth derivative– we’ll do it in that same color, just so it’s consistent– the fifth derivative evaluated at 0 is going to be 1.