Question

What i Q(1. 5) if ()=co() f(x)=co(x) and Q(x) i centered atπ/2? Round your anwer to 4 decimal, for example, 0. 1234

Answer 1

On solving the provided question we can say that -by Taylor's series Plug 1.5 for x using the estimates, and we get about 0.0707

What is Taylor series?

A creative technique to approximate any function as a polynomial with an unlimited number of terms is to use a Taylor series. The differentiation of the function at a single point is where each term in the Taylor polynomial originates.

The purpose of Taylor series is to use calculus to approximate non-polynomial functions.

So let's use common sense.

We must match the centre value at some Centre an in order to be precise.

Since cos(pi/2) = 0, we know that our initial constant should also be 0.

$q(x) = 0$

The linear approximation must then be accurate.

where x=pi/2

The fact that

$\frac{d}{dx}cosx = -sinx = -sin(\pi /2) = -1$

Additionally, we want to divide this by 1, which is the only value, thus our constant is, Since we want to be universal and utilise this as a linear approximation, and we have a Centre to maintain, we set the value to -1.

our following phrase is

$0-1(x-\pi /2)$

Next, we must share a concave shape. Therefore, we use the cosine's second derivative, which is simply

$-cosx = -cos(\pi /2) = 0$

Even if we divide by 2, we may skip this step as our constant is 0! Cosine's third derivative is our fourth term. What is

$sinx = sin(\pi /2) =1$

Given we want to be universal and utilize this as a cubic approximation, and since we have a centre to maintain, our following phrase is

$-(x-\pi /2) + 1/6(x - \pi /2)^3$

Plug 1.5 for x using the estimates, and we get about 0.0707

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