Estimating integral with Maclaurin series

[lxman]

Homework Statement

Assume that sin(x) equals its Maclaurin series for all x. Use the first two terms of the Maclaurin series for sin(7x^2) to approximate the integral:

$$\int_{0}^{0.77}sin(7x^{2})\ dx$$

The Attempt at a Solution

If I understand correctly, a Maclaurin series is just a Taylor series centered at 0. Therefore, my first two terms would be:

$$sin(7(0)^{2})$$

and

$$14(0)^{2}cos(7(0)^{2})$$

I feel like I'm getting off on the wrong foot right away.

Help anyone?

 

[armolinasf]

your first derivative is slightly off it should just 14xcos(7x^2), evaluated at zero. which just equals zero.

 

[AlexChandler]

Homework Statement

If I understand correctly, a Maclaurin series is just a Taylor series centered at 0. Therefore, my first two terms would be:

$$sin(7(0)^{2})$$

and

$$14(0)^{2}cos(7(0)^{2})$$

I feel like I'm getting off on the wrong foot right away.

Help anyone?

Yes you are doing this correctly so far except your second term should be

$$14(x) cos(7(x)^{2}) \right _{x=0}$$

which of course gives the same result. But this could have messed you up with your next derivatives.

Now try to take a few more derivatives until you get some non zero terms. Don't forget about the product rule

 

[lxman]

The original problem statement says to use the first two terms of the Maclaurin series. So when I am constructing the series, I just basically throw away any terms with zero values, and the non-zero terms then become my series.

Is this correct?

 

[SammyS]

One way to do this is to take the Maclaurin series for sin(x), then substitute 7x² for x. Then integrate that tern by term.

 

[AlexChandler]

The original problem statement says to use the first two terms of the Maclaurin series. So when I am constructing the series, I just basically throw away any terms with zero values, and the non-zero terms then become my series.

Is this correct?

Yes, just use the first two non zero terms.

 

[lxman]

SammyS, your idea sounds MUCH simpler, but you would have to explain the details of implementation to me.

So, doing it the hard way - and yes, it gets a bit messy, so I won't post all the details.

I end up with the first two non-zero terms being term 2, which evaluates to 14, and term 6, which evaluates to -41160.

Dividing these with their factorials, multiplying by the x-a terms (which is just x, as a is zero) and simplifying, I end up with:

$$7x^{2}-\frac{343\ x^{6}}{6}$$

Plugging in 0.77, I get a result of -7.764512728421

I know the actual value of the integral is about 0.3

I presume the error comes from the fact that I am only using two terms.

Am I correct?

 

[Char. Limit]

SammyS, your idea sounds MUCH simpler, but you would have to explain the details of implementation to me.

So, doing it the hard way - and yes, it gets a bit messy, so I won't post all the details.

I end up with the first two non-zero terms being term 2, which evaluates to 14, and term 5, which evaluates to -41160.

Dividing these with their factorials, multiplying by the x-a terms (which is just x, as a is zero) and simplifying, I end up with:

$$7x^{2}-\frac{343\ x^{6}}{6}$$

Plugging in 0.77, I get a result of -7.764512728421

I know the actual value of the integral is about 0.3

I presume the error comes from the fact that I am only using two terms.

Am I correct?

But you have to integrate it first. You can't just plug in .77 before you integrate.

 

[lxman]

Okay, yep, missed that one completely.

Integrating gives me:

$$\frac{7x^{3}}{3}-\frac{49x^{7}}{6}$$

Now, plugging in 0.77, I get -0.245385 which is much closer.

Again, I presume the error is from only using two terms.

 

[Char. Limit]

Yes it is. If you had tried three terms, you would get a number even close, but above the correct answer.

 

[lxman]

Thank you all for your help.

 

[Char. Limit]

Have a great day!