Given that y=sin^-1 x, find the series expansion of y in ascending powers up to and including the term in x^3?

Question

Given that y=sin^-1 x, find the series expansion of y in ascending powers up to and including the term in x^3.

I have done this part and i got y=x+x^3/6 but you can double check.

Hence, by choosing a suitable value of x, deduce the result of pi to five decimal place.

i sort of used pi/2=y=sin^-1 x
so x=1
then pi= 2 x (1+1/6)
but this answer is like very far from the true value of pi!!

so which x value is the most suitable?

Hence, by choosing a suitable value of x, deduce the result of pi to five decimal place.

so what is the value so called suitable according to the question.

ok thanks a lot for the answer

Answer 1

The term in x^3 implies we need upto the third derivative
y(0) = 0

y' (x) = 1/sqrt (1 - x^2) => y'(0) = 1
y''(0) turns out to be zero and y'''(0) = 1

Thus sin^-1 x = x + x^3 / 6 +...
The reason that your answer that you are getting is quite different from pi is if the series converges for x = 1( which it does not) the first two terms of the series is only giving you an estimate of pi/2.
Also, the number 1 you used is quite far away from the center of the expansion (which is zero)
The series is a good estimate for pi for values of x close to zero.
If you try using sin^-1 (0.5) = pi/6 => pi = 6( 0.5 + 0.5^3/6)
you will get a much better estimate.

Better still, taking a value even closer to zero will give you an even better estimate.

Example : sin^-1 0.382683432 = pi/8 thus
pi = 8( 0.382683432 + 0.382683432^3/6) approximately.
U can see that our estimates of pi are getting better as I am getting closer to the center of the expansion.

Answer 2

Looks like you did it correctly for a series about x=0 ... y =~ x + (1/6)x^3. Using x=1 (pi/2), as you did, you get the approximation of 7/3 = 2.33333. If you chose to use x=1/sqrt(2) (pi/4), you'd get 3.06413. Convergence depends on about what x you develop your series and at which x you are evaluating this. If you do this for x=1/2 (pi/6), you get pi ~ 3 + 1/8 = 3.125. The closer to zero you are, the less significant the O(x^5) terms are and the better the approximation. I can't think of anything smaller than the pi/6 sine off of the top of my head. If you sort of cheated and did this for pi/100 (x=0.031410759), you'd get pi ~ 3.14159 to 5 decimal places.

Answer 3

Your expansion is a Maclaurin series that only holds for value of x whose absolute value is less than 1. Since you chose 1, the approximation was incorrect.

If you choose y = (30°), that is π/6 , then x =.5

Using .5 we get:

π/ 6 ≈ (.5) + (1/6)(.5)^3 = 0.520833

π ≈ 6 (0.520833)

π ≈ 3.12500

If you had continued and also used the 3rd term of the expansion, (3/40)(x^5), then you would have gotten

π/6 ≈ 0.523177083

π ≈ 3.1390625

Answer 4

Arcsin x =x+x^3/6+3x^5/40+15x^7/336++
I xI <=1