Find the Taylor Polynomial of order 1 for the function f(x)=arctan(x/5) in powers of (x-5). Also find the remainder R1(x) as a function of x and c.
To find the Taylor Polynomial of order 1 for the function f(x) = arctan(x/5) in powers of (x-5), we need to first find the first derivative of the function at x = 5.
Step 1: Find the first derivative
f'(x) = d(arctan(x/5))/dx
We can use the chain rule to differentiate arctan(x/5). The derivative of arctan(u) with respect to u is 1/(1+u^2). So, differentiating arctan(x/5) with respect to x, we get:
f'(x) = 1/(1 + (x/5)^2) * (1/5)
Step 2: Evaluate the derivative at x = 5
To find the first derivative at x = 5, substitute x = 5 into f'(x):
f'(5) = 1/(1 + (5/5)^2) * (1/5)
= 1/(1 + 1) * (1/5)
= 1/2 * (1/5)
= 1/10
Step 3: Write the Taylor Polynomial of order 1
The Taylor Polynomial of order 1 for f(x) = arctan(x/5) in powers of (x-5) is given by:
P1(x) = f(5) + f'(5)(x - 5)
Since f(5) = arctan(5/5) = arctan(1) = π/4, we have:
P1(x) = π/4 + (1/10)(x - 5)
Step 4: Find the remainder R1(x) as a function of x and c
The remainder R1(x) for the Taylor Polynomial of order 1 can be expressed using the Lagrange form of the remainder:
R1(x) = (f''(c))/(2!)(x - 5)^2
To find the second derivative of f(x), we take the derivative of f'(x):
f''(x) = d/dx (f'(x))
= d/dx (1/(1 + (x/5)^2) * (1/5))
= -2x/(25(x^2 + 25)^2)
where c is a number between x and 5.
Hence, the remainder R1(x) is:
R1(x) = (-2c/(25(c^2 + 25)^2))/(2!)(x - 5)^2
= -c/(50(c^2 + 25)^2)(x - 5)^2
Therefore, the remainder R1(x) as a function of x and c is -c/(50(c^2 + 25)^2)(x - 5)^2.