could you please help me with solving this problem?
#1) Find the Taylor polynomial Tn(x) for the function 'f' at the number 'a'.
f(x) = sqrt(3+x^2) ; a=1; n=2;
my work so far:
f (x) = sqrt(3+x^2) = (3+x^2)^(1/2)
f ' (x) = (1/2)(3+x^2)^(-1/2)
f '' (x) = (-1/4)(3+x^2)^(-3/2)
f ''' (x) = (3/8)(3+x^2)^(-5/2)
f (x) = 2
f ' (x) = 1/4
f '' (x) = -1/32
f ''' (x) = 3/256
T2(x) = (f(1)/0!)(x-1)^0 +(f(1)/1!)(x-1)^1 + (f(1)/2!)(x-1)^2
T2(x) = 2 + 1/4(x-1) + -1/64(x-1)^2
..do i solve for x???
which i tried..
0 = 2 + 1/4(x-1) + -1/64(x-1)^2
2 = 1/4x - 1/4 - 1/64x^2 + 1/32x - 1/64
8 = x - 1 - 1/16x^2 + 1/8x - 1/16
64 = 8x - 8 - 1/2x^2 + 1/8x - 1/16
72.5 = 9x - 1/2x^2
145 = 18x - 1x^2
-145 = -18x + x^2
145/18 = x + x^2
.. and im not sure where to go with this..
"T2(x) = 2 + 1/4(x-1) + -1/64(x-1)^2
..do i solve for x???"
The answer is T2(x) you just calculated.
No, you do not need to solve for x in this case. The expression T2(x) = 2 + 1/4(x-1) + -1/64(x-1)^2 is the Taylor polynomial for the function f(x) = sqrt(3+x^2) centered at a=1 and approximated to the second degree (n=2).
To find the Taylor polynomial, you have correctly calculated the derivatives of f(x) up to the third derivative. Then you evaluate these derivatives at x=a=1.
Next, you use the formulas for the coefficients of the Taylor polynomial. The general formula for the nth degree Taylor polynomial centered at a is:
Tn(x) = f(a)/0! + f'(a)/1!(x-a)^1 + f''(a)/2!(x-a)^2 + f'''(a)/3!(x-a)^3 + ... + f^(n)(a)/n!(x-a)^n
In this case, you have n=2 and a=1. You have already calculated the values of f(1), f'(1), and f''(1) as 2, 1/4, and -1/32, respectively.
So, substituting these values into the Taylor polynomial formula, you get:
T2(x) = f(1)/0! + f'(1)/1!(x-1)^1 + f''(1)/2!(x-1)^2
= 2/1 + (1/4)(x-1) + (-1/32)(x-1)^2
= 2 + 1/4(x-1) - 1/32(x-1)^2
Thus, T2(x) = 2 + 1/4(x-1) - 1/32(x-1)^2 is the Taylor polynomial for the function f(x) = sqrt(3+x^2) centered at a=1 and approximated to the second degree (n=2).