Mathematicus65
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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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Mathematicus65
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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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Mathematicus65
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Ignore this I just made a typing error in the calculator
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DFranklin
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(Original post by Mathematicus65)
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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 ...?
Without detailed calculation, I would expect the error term to be roughly the same size as the first discarded term. So if you've done this correctly up to degree 9, you should end up with an error of size roughly 1/2^11. (actually a fair bit less as the Maclaurin coefficient is considerably smaller than 1).

[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 \cos^{-1} x = \pi/2 - \sin^{-1} x.

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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