Using Taylor theorem prove that x-((x^3)/6)<sinx<x-((x^3)/6)+((x^5)/120) for x>0

Answer 1

For all x >= 2.85, x - x^3/6 <= -1; for all x >= 3.69, x- x^3/6 + x^5/120 >= 1.Thus, we only must show the inequalities hold for smaller values.

These results may be shown by noting that the derivate of x - x^3/6 = 1 - x^2/2 < 0 for all x >= √2, and f(2.85) = -1.00819. The derivative of x - x^3/6 + x^5/120 = 1 - x^2/2 + x^4/24, and this is positive for x >

√(6 + 2√3)) = 3.07, and f(3.69) = 1.017093

We know that ∑_n=0^∞ (-1)ⁿx²ⁿ⁺¹/(2n+1)! = sin(x)

Thus, if we can show that ∑_n=2^∞ (-1)ⁿx²ⁿ⁺¹/(2n+1)! > 0 for x <= 2.85, we have shown that x- x³/3! < sin(x)

This is an alternating series. The magnitude of the ratios between successive terms is x²/((2n+2)(2n+3)) beginning with n = 2. At n = 2, 2.85²= 8.1225, (2n+2)(2n+3) = 6*7 = 42 for n= 2 and the product increases, so x²/((2n+2)(2n+3)) < 1 for all x in 0 <= x <= 2.85, and, as the first term is positive ∑_n=2^∞ (-1)ⁿx²ⁿ⁺¹/(2n+1)! > 0

Thus, the result is true for 0 < x <= 2.85, and as it is already true for x >= 2.85, for all x > 0, x- x³/3! < sin(x)

Next, show that ∑_n=3^∞ (-1)ⁿx²ⁿ⁺¹/(2n+1)! < 0 for 0 <= x <= 3.69, so x- x³/3! + x⁵/5! > sin(x)

This is an alternating series.

The magnitude of the ratios between successive terms is x²/((2n+2)(2n+3)) beginning with n = 3. At n = 3, 3.69²= 13.6161, (2n+2)(2n+3) = 8*9 = 72 for n= 3 and the product increases, so x²/((2n+2)(2n+3)) < 1 for all x in 0 <= x <= 3.69, and, as the first term is negative ∑_n=3^∞ (-1)ⁿx²ⁿ⁺¹/(2n+1)! < 0

Thus, for all x > 0, sin(x) < x- x³/3! + x⁵/5!

For all x >= 0, x- x³/3! < sin(x) < x- x³/3! + x⁵/5!