Quadratic approximations

b. Find the quadrati...

Question

Quadratic approximations

b. Find the quadratic approximation to f(x) = 1/(1 − x) at x = 0.

Accepted Answer

Hello. In this video, we are going to be determining the quadratic approximation of the given G of X function at X is equal to 0. We want to note that the G of X has a quadratic approximation of the form P X equal to C0 plus C1 multiplied by X minus M, plus C2 multiplied by X minus M2d. At the value X is equal to M. We are also told that P of M is equal to GM, P is equal to G M, and P is equal to G M. So, how can we start approximating the quadratic approximation of G of X? Well, what we want to do is we want to go ahead and solve for each of the coefficients in the quadratic approximation. So the first thing that we can do is we can go ahead and plug in the value X is equal to M into the quadratic approximation. That is going to give us P of M is equal to C0 plus C1 multiplied by M minus M plus C2 multiplied by M minus M squared. Now, every M minus M is going to become zero, because M minus itself will give 0, and that is going to leave us with P of M equal to C subzero. No The problem also tells us that P of M is equal to GM, and that means that P of M is equal to C subzero, where G of M is the value of G with any value plugged into it. So, recall from earlier that we want to evaluate G of X at the value X is equal to 0. So what we are going to do is we are going to plug in the value of 0 into G of X in order to find our C0 coefficient. So G of 0 is going to equal to 2 divided by 2 minus 0, which will simplify the 2 divided by 2, which is equal to 1. And since C subzero is equal to zero or G of 0, C subzero is going to equal to 1, which is the first coefficient we want. Now what we are going to do is we are going to solve for the 2nd coefficient C1. Now we are told that P M is equal to G M. So let's go ahead and take the derivative of PXX with respect to X. That is going to be P X equal to the derivative of Cze, which is 0 because it's a constant, plus C1 multiplied by the derivative of X minus M, which will be 1, plus the derivative of C2 multiplied by X minus M2d. By the chain rule, this will give us 2 multiplied by C2 multiplied by X minus M. Then multiplied by the derivative of X minus M, which is 1. Now, let's go ahead and plug in the value M into the derivative, and that will give us P M equal to C1 + 2 C2 multiplied by M minus M, which is going to be 0. So the whole second term is zero out, leaving us with P M equal to C1. Now again, P M is equal to G M, so C1 is equal to the derivative of our G of X function with 0 plugged into it. By the quotient rule, the derivative G of X will equal to negative, positive, 2 divided. By 2 minus X squared. So, now if we plug in the value of 0 into this derivative, we will get G 0 equal to 2 divided by 2 minus 0 squad, which will simplified to be 2 divided by 4, which is equal to 1/2, and this means that the coefficient C1 will equal to 2. Now we are going to solve for C2 coefficient by using the final statement that P M is equal to G M. So we will go ahead and take the derivative of our PX function. That will give us P X equal to the derivative of C1, which is 0 because it's a constant, multiplied by the derivative of 2C2 multiplied by X minus M. We have 2 C2, since that's a coefficient, and the derivative of X minus M is just going to be 1. So we have P X equal to C2, which is also equal to G of X. To solve for C2, we will divide both sides of this equation by 2, leaving us with C2 equal to G of M divided by 2. Now, what this means is that for the value of 0 plugged into the second derivative of G of X, we are going to take that result and divide it by 2. But in order to do this, we need to find the second derivative of G of X. The second derivative of G of X will just be the derivative of G of X, and the second derivative of G of X is going to be G of X equal. To 4 divided by 2 minus X cubed. Now, if we plug in 0, we get G 0 equal to 4 divided by 2 minus 0 cubed. This will simplify to be 2 cubed, and 2 cubed simplifies to 8. So we have 4 divided by 8, which simplifies to be 1 divided by 2. Now, this is the result for G double prime, but keep in mind that we are solving for C2. We will need to divide this one half by 2 in order to solve for C2. So we get C2 is equal to 1 divided by 2, all divided by 2, which will simplified to be 1 divided by 4. This is going to be the final coefficient for the quadratic approximation. Now that we've solved for C0, C1 and C2, let's go ahead and plug this back into the approximation and reorder the terms with respect to the powers. That will give us p of X is equal to 1/4 multiplied by X minus M2 plus 1/2 multiplied by X minus M plus 1, and this is going to be the quadratic approximation to the given G of X function. So I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

Quadratic approximations

b. Find the quadratic approximation to f(x) = 1/(1 − x) at x = 0.

Key Concepts

Taylor Series