Barion Pixel The error term | mathXplain
 

Contents of this Calculus 1 episode:

Compute, without using a calculator, the value of cos1, Taylor polynomial, Two digits precision, Lagrange error term.

Text of slideshow

Let be differentiable any number of times on interval I that contains number a. Then, the Taylor series generated by function at point is:

Here is, for instance

Let's see the fifth-order Taylor polynomial and Taylor series at point .

The Taylor polynomial only approximates the original function;

the Taylor series, on the other hand, is identical to the function itself at every single point.

The Taylor polynomial generated by function at provides a good approximation of the function around number .

The more terms of the Taylor polynomial we generate, the longer section of the function takes shape.

If we get into generating the Taylor polynomial so much that we forget to stop...

well, then we get an infinitely large number of terms, and we called that Taylor series.

The Taylor polynomial only approximates the original function; the Taylor series, on the other hand, is identical to the function itself at every single point.

Let's see the Taylor series of a few functions.

Let’s start with , for instance.

Let’s generate its Taylor series at .

Taylor series where are often called Maclaurin series, too.

We can figure out the remaining terms based on this.

TAYLOR SERIES OF A FEW SPLENDID FUNCTIONS

Let's see one more.

Let’s generate the Taylor series of at .

Let's try to figure out the kth term.

We got it.

.

Now we can get down to the Taylor series.

In similarly exciting fashion we can create the Taylor series of many other functions.

Here is, for example, the Taylor series of and at zero.

And now, if we want to know the Taylor series of – let’s say – , all we have to do here is to replace x with 2x.

And Voila! – it is done.

Getting the Taylor series of is also similarly pleasant.

Here is a more interesting case:

There are some cases that are even more exciting:

But enough excitement for today.

Now we are launching a very interesting endeavor:

we will compute, without using a calculator, the value of .

We will use the Taylor polynomial of .

Let’s generate the fourth-order Taylor polynomial.

If x is near zero, then , therefore

The question is how accurate this result is.

Well, according to the calculator

It seems we only get the value of cos1 with two digits precision.

Of course, it is easy now.

If we already know the exact value of cos1, it is not a big deal to figure out our error afterwards.

It would be great to know the extent of our error, even without knowing the exact value of cos1.

In other words, we need to state the size of our error while we have no idea what the exact result is.

This sounds impossible, but we will still do it.

This is what the Lagrange remainder term is for:

We need one more derivative.

We need to substitute not zero, but a number c.

This c is always a number between some a and x.

The special thing about the Lagrange remainder term is that the Taylor polynomial only approximates the function...

but by adding the remainder term to it, it will be the same.

So, the remainder term tells us the amount of our error.

We don’t know the exact value of c, so we don’t know the exact remainder either, but we can estimate it.

Based on this, our error is less than 0.00139:

Well, this is why the Lagrange remainder term is useful.

Let's see another case.

Let’s compute, with an error of less than 0.05, the value of

We will use the Taylor polynomial of at .

The error must be less than 0.05.

Let’s see where we are now.

If we use the second-order Taylor polynomial, then we calculate the error term from this:

We want to compute , therefore .

We will try to find an upper estimate of the remainder term.

All we know about this c is that it is a number between 1 and 2.

The error is less than 0.0625, but that is not enough for us.

We want the error to be less than 0.05, so we need to suffer some more.

Well, isn’t this enough yet?

Finally, the error is below the specified 0.05, so we will compute the value of using a third-order Taylor polynomial.

Here comes:

Enter the world of simple math.
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