00:01
Imagine that you need to compute exponential of 0 .2, but you have no calculator or other ed to enable you to compute it exactly only paper and pencil.
00:16
You decide to use a second degree tailored polynomial expanded around x equals 0.
00:24
Use the fact that exponential of 0 .2 is less than e or exponential of 1, which is also less than 1, which is also less, than 3 and the error bound for the tailored polynomials to find an upper bound for the error in your approximation.
00:41
Good so the function we got to consider here is the exponential function because we want to find exponential of 0 .2 that means f at 0 .2 so what we want to estimate is f at 0 .2.
01:07
In fact we want to estimate is f at 0 .2.
01:10
In fact we want to estimate that and find a bound for the error and that's the number you get to enter in the text box for the answer that is the bound for the error in the approximation so we want to use taylor polynomial of degree 2 and develop at zero that is expanded around x equal 0 good so one of the main properties of this exponential function is that any derivative of any order of this function is the same function exponential of x for all or for any k a natural number or even zero because for k equals here we were talking about the function which is exponential of x also we have that and so the derivative of any order of this function at zero is one so the second degree table polynomial can be found easily in this case so we get that p2 of x which is i emphasize here taylor polynomial of degree two expanded around zero that is also called maclory polynomial so this is f at zero plus the first derivative at 0 times x plus f derivative at 0 times x because it's times x divided by 1 factorial so we get this plus the second derivative at 0 over 2 factorial that is 2 times x square and because all derivatives at 0 are equal to 1 here i forgot to put for okay a natural number then we get that p2 of x is 1 plus x plus x squared over 2 then exponential of 0 .2 which is approximated by the second degree to prognonial at 0 .2 that is evaluate this polynomial here at x equal 0 .2 so we get 1 plus x is 0 .2 plus x square over 2 that is 0 .2 square over 2 that is 1 .2 plus 0 .0 .2 is 0 .04 divided by 2 that is 1 .2 plus 0 .02 and that is 1 .22 and that is 1 .22.
05:15
So the exponential of 0 .2 is approximately equal to 1 .02 and that is 1 .22.
05:19
So the exponential of 0 .2 is approximately equal to 1 .2 point 22.
05:27
We have calculated this number easily without the calculator and now we need to have an idea of how large is the error in this approximation here.
05:41
So if we use the taylor's theorem we can say that in this case having all derivatives continuous this function in the real numbers.
06:01
We can apply for any order of table of polynomial, particular degree 2.
06:06
So we get that f of x is equal to p2 of x plus error associated to this second degree table perennomial at x, where p2 of course we know is this, but e2 of x, this is this is the 2 of x, this is this, error is given by the third derivative of f not at zero but at a number which depends on which value of x we are chosen here over three factorial times x minus zero to the third where for some number cx which depends on x that is if we change x c of x in general is different and that number is between the value of x for around which we expanded the term polynomial that is zero and x the value where we are evaluating the function and the terminal and the error so this error in 2 of x is third derivative is exponential function, so we get exponential of x x over 6, which is 3 factorial times x cubed.
08:12
And in particular, we are interested in x equals 0 .2, which is the value we are approximating here.
08:21
So for x equals 0 .2, we get, now we have e2 of 0 .2.
08:31
It's a particular value of x.
08:33
So we get now this is exponential of c, but now i don't write the dependency with x because we have fixed now the value of x times 0 .2, which is the value of x we put here, q over 6.
08:53
4 sum, z between 0 and 0 .2.
09:10
But now we know 0 .2 is greater than 0.
09:17
And so we can write this expression here.
09:19
C between 0 and 0 .2 means that 0 less than 0 .2.
09:29
Before, here, i could not say which inequality i can write because it depends on the value of x.
09:38
If x is negative, we have one way of writing these using inequalities, and if x is positive, there is another.
09:45
Different way...