In Exercises 41-44, use the error bound to find a value of n for which the given inequality is s

In Exercises 41-44, use the error bound to find a value of n...

Key Concepts

Lagrange Error Bound

The Lagrange error bound provides an estimate on the error made when approximating a function by its Taylor polynomial of a certain degree. It quantifies the difference between the true function value and the Taylor polynomial approximation using the (n+1)-th derivative evaluated at some point between the center of expansion and the point of interest, ensuring that the approximation stays within a specified precision.

Remainder Term Estimation

In the context of Taylor series, the remainder term (or error term) estimates the truncation error incurred by stopping the series after a finite number of terms. By bounding the (n+1)-th derivative on an interval, one can determine how many terms (or what degree of polynomial) are required to achieve a desired level of accuracy in the approximation.

Taylor Series Expansion

A Taylor series represents a function as an infinite sum of terms calculated from its derivatives at a single point. This method is used to approximate functions using a polynomial, which is especially useful when the function itself is complex or difficult to compute directly near a particular point.

Transcript

00:01 All right, we are looking at solving this equation for n, and we want to try to keep our error bound as being less than 10 to the power of negative 6.

00:13 So to do this, it's just a bunch of guess and checking.

00:16 This one's a little bit of a trickier problem, though, because k is not one, and it's not a set value.

00:23 As my derivatives change, my k changes.

00:26 It actually gets larger.

00:27 So you can use k as one as a nice estimate, just to kind of give you a rough idea of whereabouts you'll be, but it won't get you an exact value in this case.

00:41 So if you did solve it as k as one, you're going to get a different answer than if you use different ks for each derivative.

00:48 So if you want to be more precise, then we're just going to have to do a bunch of derivatives for a bunch of k's.

00:59 To test which one actually works.

01:06 And i mean, there is a bit of a pattern here to try to figure out what it is, but it's just easier to write out the derivatives.

01:17 The derivatives actually aren't that bad to find.

01:20 It's kind of a nice pattern.

01:24 We'll just go up to 10, and hopefully by then we've got one that works because the k values for this one also happen to increase.

01:33 So in this case i'm going to write my f of x is x to the power of a half.

01:37 So my first derivative is a half x to the negative half.

01:42 And then we've got negative a quarter x to the negative three halves.

01:47 And then three -eighths x to the power of negative five halves.

01:53 And then negative 15 over 16, x to the power of negative seven halves.

02:02 So you can kind of see here they are smaller than one.

02:07 It's right around the fourth and fifth derivative where we finally actually increase above one for a fraction out front.

02:15 Actually dramatically increase above one.

02:19 And that's just because my top number starts to get quite a bit bigger.

02:25 945 over 64 negative x the power negative 11 over 2.

02:31 And then we've got our seventh one, which is 10 ,395 over 128, x to the power of negative 13 over 2.

02:44 And then our 8th one, we'd have to move the 13 over 2 down, which is 135 ,135, over 256, x to the power of negative 15 over 2 over 256, x to the power of negative 15 over 2.

03:04 That one would be negative.

03:07 And then we'd have to move that down.

03:09 5 .56 times 15 over 2 gives us...

03:18 Doesn't even want me to do this one.

03:20 5 times 15.

03:27 27025 over 500 and 12.

03:36 And then x to the power of...

03:38 Oops.

03:40 12.

03:41 X to the power of negative 17 over 2.

03:43 And for one more.

03:48 3, 4, 4, 5, 9, 4, 25, x the power of negative 19 over 2.

03:58 And i totally forgot my fraction in there.

04:02 Or 1 ,024 x the power of negative 19 over 2.

04:07 All right, so those are our standard derivatives from 1 all the way up to 10.

04:14 Now, finding the k for each of these, because i'm looking on the interval of 1 to 1 .3, and because i've got a root on the bottom, your biggest answer is always going to be at 1.

04:35 These are exponential regression curves.

04:38 So on the plus side, the ks are pretty easy.

04:42 So the k for this one, that would be a half.

04:46 And remember the ks are always the absolutes.

04:51 So the ks are really just the coefficient that's out front as a positive.

05:01 K equals 945 over 64.

05:05 K equals 10395 over 128.

05:14 K equals 135, 135 over 256 over 256.

05:24 K equals 2027 over 512 and k equals 344 -525 over 1024 so those are our case so for each of these cases then we're just going to test our formula so for this first one it'd be a half x minus a so that would be 0 .3 this is my first derivative so that'd be two over two factorial so if i test that one it's not going to be anywhere close but at least it gives us an idea of how to set these up we would end up with an answer of 0 .0 225 so nowhere near the 10 to the negative 6 that i need i'm going to jump down to the fourth one and try the fourth one let's see if that gets us closer...

Please give Ace some feedback