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Use taylor polynomials about 0 to evaluate cos(0.52) to 4 deciamal places, showing all your working.
What I did;
Since the taylor polynomial about 0 for cosine function is
p n (x) = 1-1/2!x^2 + 1/4!x^4- 1/6!x^6+.....1/n!x^n
Hence (did all working)
p2(0.52) = 0.8646
P4(0.52) = 0.867846
p6(0.52) = 0.867819
p8(0.52) = 0.867819
So cos(0.52) = 0.8678
But you have to prove that is accurate to 4 decimal places ;)Quote:
Originally Posted by fair_lady0072002
By Taylor's theorem,
Where, (the remainder term)
for some,
between zero and
But
Thus, what you have is that the remainder term is,
This is correct up to three decimal places.
Thanx very much.
At this point proof of approximations found by evaluating Taylor Polynomials is beyond the scope of my course. I will definitely keep the workings for later ref.
Thanx once more.
Honestly, the biggest challenge in an approximation problem is not approximation but proof that it is accurate.Quote:
Originally Posted by fair_lady0072002