Taylor Series for following parts,
a) f(x) = 7 - (7/2)x² + (7/24)x⁶ +......
b)f(x) = 21 - 18(x+2) + 25(x + 2)² - 8(x+3)³ +......
c)f(x) = 2 + 2log(2)(x -1) +(log2)²(x-1)² + (1/3 (log2)³(x-1)³ + --------
d)f(x) = x² + ----
The Taylor series giving the expansion of the function f(x) in the neighborhood of the point a. If the function is continuous in the neighborhood, all its derivatives exist, and the series converges, then The form f(x) = f(a)+f′(a)/1!(x−a)+(f′′(a)/2!)(x−a)²+⋯+(fⁿ(a)/n! )(x−a)ⁿ --(1) where fⁿ(a) is the nᵗʰ derivative of f(x) evaluated at a.
a) f ( x) = 7 cos(-x) = 7 cos(x) , a = 0
f(a ) = 7 cos(0°) = 7
f'(x) = - 7 sin(x) , f'(a) = 7 sin(0°) = 0
f"(x) = - 7cos(x) , f"(a) = -7 cos(0°) = -7
f"'(x) = 7 sin(x) , f"'(a) = 7 sin(0° ) = 0
....................................
put all the values in above equation (1) we get,
f(x) = 7 + 0/1! (x -0) - (7/2!)(x-0)² )+ 0 +(7/4!) (x-0)⁴-0 + ----------
f(x) = 7 - (7/2)x² + (7/24)x⁶ +......
b) f(x) = x⁴ + x² + 1, a = -2
f(a= -2) = (-2)⁴ +(-2)² +1 = 21
f'(x) = 4x³ + 2x
f'(a= -2) = 4(-2)³ + 2×(-2) = -36
f"(x) = 12x² + 2
f"( a = -2) = 12(-2)² + 2 = 50
.....
put all the values in above equation (1) we get,
f(x) = 21 - (36/2)(x+2) + (50/2!)(x + 2)² - ( 48/3!)(x+3)³ +......
f(x) = 21 - 18(x+2) + 25(x + 2)² - 8(x+3)³ +......
c) f(x) = 2ˣ ; a = 1
f( a= 1) = 2¹ = 2
f'(x) = 2ˣ log(2)
f'(a= 1) = 2 log(2)
f"(x) = 2ˣ ( log(2) )²
f"( a = 1) = 2 ( log(2) )²
................
put all the values in above formula we get,
f(x) = 2 + 2log(2) +( 2/2!)(log2)² + (2/3!)(log2)³
+--------
f(x) = 2 + 2log(2)(x -1) +(log2)²(x-1)² + (1/3(log2)³(x-1)³ + --------
d) f(x) = x tan⁻¹(x²) ; a= 0
f(a=0) = 0
f'(x) = x (1/(1+x⁴) + tan⁻¹(x²)
f'(a = 0) = 0
f"(x) = ((1+x⁴) - x 4x³)/(1+x⁴)² + (1/(1+x⁴)
= (2/(1+x⁴) - 4x⁴/(1+x⁴)²
f"( a = 0) = 2 - 0 = 2
f"'(x) = - 8x³/(1+x⁴)² - 12x³/ (1+x⁴)² + 16x⁷/(1+x⁴)⁴
f"'(a = 0) = 0
..............
f(x) = 0 +0 + 2/2! (x-0)² + 0 +-------
f(x) = x² + ----
Hence, we get all required Taylor Series for
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