Use differentiation to find the taylor series about 0 for the function
g(x)=(1/16+x)^-5/4
giving all terms up to one in x^2, and calculating each coefficient as an integer or a fraction
so, what do you get for g'(x)?
i have no clue
well, assuming you meant
g(x) = (1/16 + x)^(-5/4)
g'(x) = (-5/4)(1/16 + x)^(-9/4)
using the chain rule for g(x) = u^n and u = (1/16 + x)
do that again for g", and evaluate the fractions.
Thanks
To find the Taylor series for the function g(x) = (1/16 + x)^(-5/4) about the point x = 0, we can use differentiation.
First, let's find the derivatives of g(x) at x = 0. We will need to find the first and second derivative.
First derivative:
g'(x) = d/dx (1/16 + x)^(-5/4)
To find the derivative, we can use the chain rule. Let's define u = 1/16 + x.
g'(x) = d/dx (u^(-5/4))
= (-5/4) * u^(-5/4 - 1) * du/dx
= (-5/4) * (1/16 + x)^(-5/4 - 1) * 1
= (-5/4) * (1/16 + x)^(-9/4)
Now, let's find the second derivative:
g''(x) = d/dx (-5/4) * (1/16 + x)^(-9/4)
Applying the chain rule again, we get:
g''(x) = (-5/4) * (-9/4) * (1/16 + x)^(-9/4 - 1) * 1
= (135/16) * (1/16 + x)^(-13/4)
Next, let's expand g(x) into its Taylor series about x = 0 using the derivatives we just found:
g(x) = g(0) + g'(0) * x + (1/2) * g''(0) * x^2 + higher-order terms
Since we want to find all terms up to one in x^2, we only need the first two terms:
g(x) = g(0) + g'(0) * x + (1/2) * g''(0) * x^2
To find g(0), we substitute x = 0 into the original function:
g(0) = (1/16 + 0)^(-5/4)
= (1/16)^(-5/4)
= (16)^(-5/4)
= 2^(-5)
= 1/32
Now, let's find g'(0) by substituting x = 0 into g'(x):
g'(0) = (-5/4) * (1/16 + 0)^(-9/4)
= (-5/4) * (1/16)^(-9/4)
= (-5/4) * (16)^(-9/4)
= (-5/4) * 2^(-9)
= (-5/4) * (1/512)
= -5/2048
Finally, let's find g''(0) by substituting x = 0 into g''(x):
g''(0) = (135/16) * (1/16 + 0)^(-13/4)
= (135/16) * (1/16)^(-13/4)
= (135/16) * (16)^(-13/4)
= (135/16) * 2^(-13)
= (135/16) * (1/(8192))
= 135/131,072
Now, we can substitute these values into the Taylor series formula:
g(x) ≈ (1/32) - (5/2048) * x + (1/2) * (135/131,072) * x^2
Simplifying, we have:
g(x) ≈ 1/32 - 5x/2048 + 135x^2/262,144
Therefore, the Taylor series for g(x) about x = 0, up to one in x^2 with coefficients as an integer or a fraction, is:
g(x) ≈ 1/32 - 5x/2048 + 135x^2/262,144