00:01
Hello students, let us discuss the solution of this question.
00:03
Here, we have to find the taylor polynomial of order 0123 for f of x, which is equal to e raised to 2x at a is equal to 0.
00:16
The formula for taylor polynomial generated at x equals to a is summation k going from 0 to infinity, kth derivative of f at a divided by k factor.
00:31
Multiplied by x minus a raised to k, which is equals to f of a plus first derivative of a divided by 1 factorial multiplied by x minus a plus second derivative of a divided by its 2 factorial multiplied by x minus a square plus and so on the nth derivative of a divided by n factorial multiplied by x minus a raised to n.
01:06
So this is the formula for tailor polynomial.
01:10
We are already given that f of x is equal to e raised to 2x and we are to construct the first three tailor polynomials centered at a is equal to 0.
01:29
First, let's find the first three derivatives as follows.
01:32
F of x is equal to e raised to 2x first derivative of x let me adjust it first derivative of f of x is equal to a raised to 2x multiplied with 2 that is 2 e raised to 2x second derivative of f of x is 2 multiplied to 2 x multiplied with 2, which is 4 e raised to 2x.
02:16
Now, third derivative of f of x is 8 e raised to 2x.
02:28
Now we can calculate f at 0, which is equal to e raised to 2 multiplied by 0, which is 1...