{Use of Tech} Small argument approximations Consider the follo...

Question

{Use of Tech} Small argument approximations Consider the following common approximations when x is near zero.

b. Estimate f(0.2) and give a bound on the error in the approximation.

f(x) = sin ⁻¹ x ≈ x

Accepted Answer

Hello. In this video we are given the function f x equal to sin inverse. It is often approximated near zero by the linear expression fx approximately equal to x. Using this approximation, we want to estimate the value of f of 0.17, then determine the upper bound on the error of this approximation. We want to express the upper bound in scientific notation and rounded two decimal places. OK, so we are given the function F X equal to sin inverse of X. Now, the approximation that the problem tells us is usually defined as FX is approximately X. And so what this means is that the approximation for F of 0.17 should approximately equal to 0.17 for the function. So this is going to be the approximation that we are looking for. But now what we want to do is you want to determine the upper bound of the error approximation. So we can go ahead and use the error approximation for a Taylor polynomial. The error approximation is defined as rn x is equal to f's derivative, which is defined as m + 1 of some constant value c divided by m + 1 factorial multiplied by x minus a raised to the power of n + 1. Now here we are going to let n equal to 1. And we're going to let A equal to 0. And so with that being said, the error approximation is going to be R1 of X, which is equal to F2 derivative of C divided by 2 multiplied by X 2. And so what we need to do is we need to go ahead and calculate the 2nd derivative for the given function. So, FX by sin inverses derivative is going to be 1 divided by the square root of 1 minus X2. And this is going to simplify to be 1 minus x 2 to raise the power of -1/2. Then the second derivative fx is going to equal to 1/2 multiplied by 1 minus x2 raise the power of 3 divided by 2 multiplied by -2 x, and this second derivative can be rewritten as x divided by 1 minus x 2 raised the power of 3 halves. So, this is going to be the error approximation for a Taylor polynomial. Now for the error bound, the error bound is defined as the absolute value of R1 of X, which is less than or equal to m divided by 2 multiplied by x 2. Now here, M is going to equal to F of 0.17, and by plugging in 0.17 into the second derivative, we will get 0.17 divided by 1 minus 0.17 squared, raised to the 3 halves power. So by plugging this value into the error bound approximation, we get the absolute value of $$r_1 of 0.17$$, and that is going to be less than or equal to 0.17 divided by 1 minus 0.17 quantity squared, all raised to the power of 3 halves. All of this will be divided by 2 and multiplied by x square, which is going to be 0.17 squared. And with the help of a calculator, the scientific notation of the output is going to be 2.57. 2.57 times 10 to 3, which is going to be the error bound for the approximation of fx equal to 0.17, and with that being said, the solution to this problem is going to be a. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

{Use of Tech} Small argument approximations Consider the following common approximations when x is near zero. b. Estimate f(0.2) and give a bound on the error in the approximation.f(x) = sin ⁻¹ x ≈ x

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{Use of Tech} Small argument approximations Consider the following common approximations when x is near zero.

b. Estimate f(0.2) and give a bound on the error in the approximation.

f(x) = sin ⁻¹ x ≈ x