Leibniz criterion and Maclaurin approximations

[estro]

I'm trying to calculate the following function (at x=1) with accuracy of 10^(-3).

f(x)= \int^x_{0} \frac{1-cost}{t}

What I've tried:
f(1)=f(0)+f'(c)=1-\sum_{k=0}^\infty \frac{(-1)^nc^{2k}}{(2k)!}
But now I don't know how to calculate this expression. [I know that this series is convergent thanks to Leibniz criterion]

Have I done something wrong?

 

[Mark44]

I'm trying to calculate the following function (at x=1) with accuracy of 10^(-3).

f(x)= \int^x_{0} \frac{1-cost}{t}

What I've tried:
f(1)=f(0)+f'(c)=1-\sum_{k=0}^\infty \frac{(-1)^nc^{2k}}{(2k)!}
But now I don't know how to calculate this expression. [I know that this series is convergent thanks to Leibniz criterion]

Have I done something wrong?

I think what you need to do is
1. Write the Maclaurin series for 1 - cos(t).
2. Divide each term in the Maclaurin series by t.
3. Integrate, and evaluate the resulting series at 1 and at 0.
4. Use as many terms as needed to get the desired precision.

 

[estro]

g(x)=1-cosx
g'(x)=sinx
g''(x)=cosx
g^{(3)}(x)=-sinx
g^{(4)}(x)=-cosx
g^{(5)}(x)=sinx
g^{(6)}(x)=cosx
g^{(7)}(x)=-sinx

|R_6(t)| = |\frac {f^{(7)}(c)t^7} {7!}| \leq 10^{-3}

<br /> f(1)=\int_0^{1} \frac {g(0)+g&#039;(0)x+ \frac {g&#039;&#039;(0)t^2} {2!t} + \frac {g^{(3)}(0)t^3} {3!t} + \frac {g^{(4)}(0)t^4} {4!t} + \frac {g^{(5)}(0)t^5} {5!t} + {g^{(6)}(0)t^6} {6!t}} {t}= \int_0^{1} 0+0+\frac {t} {2} - \frac {t^3} {24} + \frac {t^5}{720} +\frac {R_6(t)} {t}

But how can I integrate the Reminder without corrupting my approximation, and even if I will integrate also the reminder how can I know that if f(x)=g(x)+c then also \int_a^{b} f(x) = c(b-a)+\int_a^{b} g(x).

 

[Mark44]

Go back to what I said in post #2.
1. Write the Maclaurin series for 1 - cos(t).
2. Divide each term in the Maclaurin series by t.
3. Integrate, and evaluate the resulting series at 1 and at 0.
4. Use as many terms as needed to get the desired precision.
The Maclaurin series for 1 - cos(t) is 1 - {1 - t²/2! + t⁴/4! - t⁶/6! +- (some more terms) ...}

The result from step 3 will still be a series, and will be an alternating series. There is a simple way to determine the error in truncating an alternating series that pertains to the first unused term.

 

[estro]

Thanks, got it!