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I am trying to get the sine12º function to approximate using the Taylor polynomial, for this I do the following:

$x = \pi/15$

$a = 0$

$f(x)=senx$ -> $ f(a)=0$

$f'(x)=cos x$ -> $ f(a)=1$

$f''(x) = -senx$ -> $ f(a)=0$

$f'''(x) = -cosx$ -> $ f(a)=-1$

$f''''(x) = senx$ -> $ f(a)=0$

$f(x) = \frac{x}{1!} - \frac{x³}{3!}+ \frac{x⁵}{5!} - \frac{x⁷}{7!} + \dots + \frac{(-1)^{n+1} x^{2n+1}}{(2n+1)!} + \frac{(-1)^{n+2} x^{2n+3} * senc}{(2n+3)!}$

$0 < c < \pi/15$

$|\frac{(-1)^{n+2} x^{2n+3} * cosc}{(2n+3)!}| < |\frac{(-1)^{n+2} x^{2n+3}}{(2n+3)!}| < 0.0001$

It would now give values to n asta to fulfill the condition

The doubt I have is that if this fraction is correct $\frac{(-1)^{n+1} x^{2n+1}}{(2n+1)!} + \frac{(-1)^{n+2} x^{2n+3} * senc}{(2n+3)!} $ since in some literature I have seen that the appropriate would be the following $\frac{(-1)^{n-1} x^{2n-1}}{(2n-1)!} + \frac{(-1)^{n} x^{2n} * senc}{(2n)!}$, and I do not understand why if $n = 0 (2*0-1)! = -1$:

1 Answers 1

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$\sum _\limits{n=0}^{\infty} \frac{(-1)^{n} x^{2n+1}}{(2n+1)!} = \sum _\limits{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n-1}}{(2n-1)!}$

As for the error term.

you could say:

$\sin x = \sum _\limits{n=0}^{k} \frac{(-1)^{n} x^{2n+1}}{(2n+1)!} + \frac{(-1)^{k+1} x^{2k+2}\sin c}{(2n+2)!}$

or $\sin x = \sum _\limits{n=0}^{k} \frac{(-1)^{n} x^{2n+1}}{(2k+1)!} + \frac{(-1)^{k+1} x^{2k+3}\cos c}{(2k+3)!}$

or you could start counting from $1$ and adjust accordingly.

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    $is (-1)^{n}$ in the second form for the error term why not is $is (-1)^{n+1}$ ? – 2017-01-21
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    Should be in terms of $k$ and not in terms of $n.$ And yes, opposite sign from whatever was the last term in summation portion of the calculation. – 2017-01-21