Could you find the value of the error using Taylor's theorem...
cos(0.3) ≈ 1 - (0.3^2) / 2! + (0.3^4)/4! Thanks
2007-04-10 19:47:23 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
If f(x) = cos x, f""(x) = cos x, f""'(x) = -sin x, f"""(x) = -cos x.
This approximation contains the fifth-order term (not just the fourth), so the error is given by
f"""(z).(0.3)^6/6! for some z between 0 and 0.3.
So we can bound the error by 1.(0.3)^6/6! = 1.0125×10^-6.
In fact you can verify using your calculator that the error is 1.0109×10^-6 to 5 s.f.
2007-04-10 20:08:19 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
im sorry i tried but i couldnt.
2007-04-11 03:08:37 · answer #2 · answered by prey of viper 3 · 0⤊ 0⤋
sorry I couldn't
2007-04-11 03:13:15 · answer #3 · answered by alex 3 · 0⤊ 0⤋
what is the answer?
2007-04-11 02:50:17 · answer #4 · answered by grace 2 · 0⤊ 0⤋