# FP2 Maclaurin Question

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For question 1, I have integrated to say (sin^-1(t)) between limits x and 0 equals sin^-1(x)

Then I determined the Maclaurins of (1-t^2)^-1/2 = 1+(1/2)t^2 +(3/8)(t^2)^2 ...

Then integrating that you get :

(t+(1/6)(t^3)+(3/40)t^5....) between limits x and 0 gives

Sin^-1(x) = x+(1/6)x^3+(3/40)x^5...

Then since we know that the derivative of cos^-1(x) is -1(the integral of sinx) and remembering that since cos^-1(0) is 1/2pi for the first term we have

Cos^-1(x) = 1/2pi-x-(1/6)x^3-(3/40)x^5 ....

Then by taking x=1/2 in order to find the value of pi we have

Cos^-1(1/2) =1/3pi = (1/2)pi-x-(1/6)x^3...

Then multiplying that result by 3 will give you pi.

I took this to a reasonable number of terms so up to degree 9 I think and my approximation is well off. For these "approximate" questions how on earth can I get to the solution without typing into my calculator an infinite number of terms. Does a graphical calculator do these ...?

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May sound like a stupid question, I don't know, but I am teaching myself this module so wouldn't know..

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For question 1, I have integrated to say (sin^-1(t)) between limits x and 0 equals sin^-1(x)

Then I determined the Maclaurins of (1-t^2)^-1/2 = 1+(1/2)t^2 +(3/8)(t^2)^2 ...

Then integrating that you get :

(t+(1/6)(t^3)+(3/40)t^5....) between limits x and 0 gives

Sin^-1(x) = x+(1/6)x^3+(3/40)x^5...

Then since we know that the derivative of cos^-1(x) is -1(the integral of sinx) and remembering that since cos^-1(0) is 1/2pi for the first term we have

Cos^-1(x) = 1/2pi-x-(1/6)x^3-(3/40)x^5 ....

Then by taking x=1/2 in order to find the value of pi we have

Cos^-1(1/2) =1/3pi = (1/2)pi-x-(1/6)x^3...

Then multiplying that result by 3 will give you pi.

I took this to a reasonable number of terms so up to degree 9 I think and my approximation is well off. For these "approximate" questions how on earth can I get to the solution without typing into my calculator an infinite number of terms. Does a graphical calculator do these ...?

**Mathematicus65**)For question 1, I have integrated to say (sin^-1(t)) between limits x and 0 equals sin^-1(x)

Then I determined the Maclaurins of (1-t^2)^-1/2 = 1+(1/2)t^2 +(3/8)(t^2)^2 ...

Then integrating that you get :

(t+(1/6)(t^3)+(3/40)t^5....) between limits x and 0 gives

Sin^-1(x) = x+(1/6)x^3+(3/40)x^5...

Then since we know that the derivative of cos^-1(x) is -1(the integral of sinx) and remembering that since cos^-1(0) is 1/2pi for the first term we have

Cos^-1(x) = 1/2pi-x-(1/6)x^3-(3/40)x^5 ....

Then by taking x=1/2 in order to find the value of pi we have

Cos^-1(1/2) =1/3pi = (1/2)pi-x-(1/6)x^3...

Then multiplying that result by 3 will give you pi.

I took this to a reasonable number of terms so up to degree 9 I think and my approximation is well off. For these "approximate" questions how on earth can I get to the solution without typing into my calculator an infinite number of terms. Does a graphical calculator do these ...?

[OK, I just did the calculation myself. Using terms up to x^9 should give you 4 decimal places].

Note also that it would have been simpler to directly use .

Also, you say: "Cos^-1(1/2) =1/3pi = (1/2)pi-x-(1/6)x^3..." and that multiplying this by 3 gives you pi.

But simply multiplying this by 3 leaves your "answer" on the RHS still involving a pi term, so you still need to know pi to work out what it is.

What you need to do is subtract pi/2 from both sides, giving -pi/6 on the LHS and an expression without pi on the RHS. (You may have done this and just not said so).

Note that at this point you're really just finding arcsin(1/2), which is probably what the question actually meant for you to do (although it's horribly worded and it's totally understandable you took the approach you did).

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