Question

{Use of Tech} Newton's derivation of the sine and arcsine series Newton discovered the binomial series and then used it ingeniously to obtain many more results. Here is a case in point.
a. Referring to the figure, show that x = sin s or s = sin ⁻¹ x.
b. The area of a circular sector of radius r subtended by an angle θ is 1/2r²θ. Show that the area of the circular sector APE is s/2, which implies that
s = 2 ∫₀ˣ √(1 − t²) dt − x √(1 −x²)
c. Use the binomial series for f(x) = √(1 − x²) to obtain the first few terms of the Taylor series for s=sin ⁻¹ x.
d. Newton next inverted the series in part (c) to obtain the Taylor series for x=sin s. He did this by assuming sin s = ∑ aₖ sᵏ and solving x = sin(sin ⁻¹ x) for the coefficients aₖ. Find the first few terms of the Taylor series for sin s using this idea (a computer algebra system might be helpful as well).

Accepted Answer

Step 1: Understand the relationship between the variables in the figure. Since s is the angle subtended by the arc on the unit circle and x is the length of the vertical segment corresponding to sin s, we have the fundamental trigonometric identity: $$x = \sin s. $$Consequently, this implies $$s = \sin$$^{-1}$$ x$$ because the arcsine function is the inverse of the sine function on the appropriate domain. Step 2: Use the formula for the area of a circular sector. The area of a sector with radius $$r$$ and angle $$	heta is $$given by $$\frac{1}{2} r^2$ 	heta. $$For a unit circle ($$r=1$$), the area of sector APE is $$\frac{s}{2}. $$Next, express this area in terms of an integral involving $$x$$ and $$t by $$considering the area under the curve $$y = \sqrt{1 - t^2$}$$ from $$0 to$$ $$x. $$This leads to the equation: $$s = 2 \$$int_0$$^x \sqrt{1 - t^2$} \, dt - x \sqrt{1 - x^2$}. $$Step 3: Expand the function $$f(x) = \sqrt{1 - x^2$}$$ using the binomial series. Recall the binomial series expansion for $$(1 + u)^p is$$ $$\sum_{n=0}^\infty \binom{p}{n} u^n$$, where $$\binom{p}{n} = \frac{p(p-1)\cdots(p-n+1)}{n!}. $$Here, set $$u = -x^2$ and p$$ = \frac{1}{2}$$. Write out the first few terms explicitly to approximate $$\sqrt{1 - $$x^2$$}$$. Step 4: Substitute the binomial series expansion of $$\sqrt{1 - $$t^2$$}$$ into the integral expression for s$. Integrate term-by-term from 0$ to x$ to find the series expansion for $$\$$int_0^x$$ \sqrt{1 - $$t^2$$} \, dt$$. Then, use this result along with the term x$$ \sqrt{1 - $$x^2$$}$$ to write the Taylor series for s$$ = \$$sin^{-1} x$ up to the desired number of terms. Step 5: To find the Taylor series for x$$ = \sin s$$, assume a power series form $$\sin s = \sum_{k=1}^\infty a_k s^k$$. Substitute s$$ = \$$sin^{-1} x$ from the series found in Step 4 into this assumed series and equate it to x$. By matching coefficients of powers of x$, solve for the coefficients a_k$. This inversion process recovers the Taylor series for $$\sin s$$.

Verified video answer for a similar problem:

Key Concepts

Inverse Trigonometric Functions

Area of a Circular Sector and Integral Representation

Binomial Series and Taylor Series Expansion