I am a little confused with the Taylor Series at the moment, so please forgive me for my very basic questions. If we were to approximate a function, say $cos(x)$, I let $f(x)=cos(x)$

And I have been shown that $f(x)≈f(x_0)+f'(x_0)(x-x_0)$

Just so I understand, when we say we are approximating it, are we simply saying that we can find the original function's $y$ value when given a specific $x$ value by using another, easier polynomial function that we substitute that same $x$ value into?

Anyway we found that $P_2(x)=-\frac{1}{2}x^2+1$ is an approximation of $cos(x)$. This didn't make much sense though as in the first step, we substituted zero in as $x$, and found $cos(0)=1$ But if this was another number, we would have to find the cosine of that number which defeats the whole purpose? Any help would be appreciated


The idea is that you can control the error. Indeed, we have $\cos x \approx 1 - \frac{x^2}{2}$, if $x$ is close to zero. For instance, if I want to know $\cos 0.1$, may be reasonable that $$\cos 0.1 \approx 1 - \frac{0.1^2}{2} = 1 - 0.005 = 0.995$$

We can use Lagrange's formula for the remainder, to estimate how much we're missing. If $$\sum_{k = 0}^{n} \frac{f^{(k)}(x_0)}{k!} (x - x_0)^k $$

is the Taylor expansion of order $n$ of f, around $x_0$, then the error can be estimated by $$\frac{f^{(k+1)}(c)}{(k+1)!} (x - x_0)^{k+1}$$

for some $c$ between $x$ and $x_0$. Ok?

  • $\begingroup$ So if we found the approximation function centered at 0, we can still use that function to find numbers close by to get a rough y value, but if the numbers are not near zero, it will be way off? Also when would we use that same function to approximate y values rather than just move the centre and recalculate an accurate function? I think what threw me was the whole differentiating between when we CENTER it at zero, and when we find the y value when the function's x value it AT zero. $\endgroup$ – user88720 May 21 '14 at 17:44
  • $\begingroup$ Yes. If you want to estimate the value of the function for some $x$ far from the center, you must use a higher order Taylor expansion. How much higher depends of your objective, and given a maximum margin of error, and a center, you can use Lagrange's formula to calculate the order you need, how far you have to go in the expansion. $\endgroup$ – Ivo Terek May 21 '14 at 17:53

The answer can be found in the following graph representing $\color{darkblue}{\cos x}$ and $\color{darkmagenta}{-\tfrac12x^2+1}:$

$\phantom{XXX}$Thanks Mathematica! You're very kind! ;-)

Near $0$, our polynomial approximation is pretty good, but when you start going far, this approximation becomes less accurate as the graph shows.


The whole purpose (well, one purpose) is that you don't have to find the cosine! You know that, for example,

$$\cos\left(\frac{1}{2}\right) \approx -\frac{1}{2} \left( \frac{1}{2} \right)^2 + 1$$

without having to actually compute a cosine at all.

The actual first few decimal places are

  • $\cos\left(\frac{1}{2}\right) = 0.87758\ldots$
  • $-\frac{1}{2} \left( \frac{1}{2} \right)^2 + 1 = 0.875$

so it is a pretty good approximation; for many purposes, using $0.875$ when you were supposed to use $\cos\left(\frac{1}{2}\right)$ is perfectly fine.


Lets say we have a polynomial $P_n(x)≈\cos(x)$ and assume $P_n(x)=a_0+a_1x+a_2x^2+...+a_nx^n$ and my proposal is that as $n$ approaches infinity, $$P_\infty(x)=\cos(x)=a_0+a_1x+a_2x^2+...$$ and now, since the polynomial and the function are equal, that must mean that their values at a specific point, lets say $x=0$,for example, are equal too: $$P_\infty(0)=\cos(0)$$ ant note, that $$P_\infty(0)=a_0=\cos(0)=0$$ Also since $P_\infty(x)=\cos(x)$ their derivatives are equal too, $$P'_\infty(x)=a_1+2a_2x+3a_3x^2...=\cos'(x)=-\sin(x)$$ (this uses the fact that the derivative of a constant is zero, and the power rule: $(x^r)'=rx^{r-1}$) Now lets analize the derivatives at zero $$P'_\infty(0)=a_1=-sin(0)=0$$ Now the double derivative $$P''_\infty(x)=-sin'(x)=-cos(x)=2a_2+2*3a_3x+2*3*4a_4x^2...=2!a_2+3!a_3x+4!a_4x^2...$$ (as you can see this is where the factorial comes in, it is merely a consequance of the power rule) Once again let us analize the double derivative at $0$ $$P''_\infty(0)=2!a_2=-cos(0)=-1$$ $$a_2=-\frac{1}{2!}$$ now, by derriving the funtion ad infinitum and analyzing its value at zero, we may find all of the infinite polynomials coefficients, and by noting that $$\cos(x)'=-\sin(x)$$ $$\cos(x)''=-\sin(x)'=-\cos(x)$$ $$\cos(x)'''=-\sin(x)''=-\cos(x)'=\sin(x)$$ $$\cos(x)''''=-\sin(x)'''=-\cos(x)''=\sin(x)'=\cos(x)$$ the derivatives of the cosine are periodic with a period of four, we may note that $$a_0=1$$ $$a_1=0$$ $$a_2=-1/2!$$ $$a_3=0$$ $$a_4=1/4!$$ $$a_5=0$$ $$a_6=-1/6!$$ or generally, if $n$ is an odd number, $a_n=0$, if $n$ is divisible by $4$, then $a_n=1/n!$, and if $n$ divisible by 2 but not by 4, $a_n=-1/n!$.

Putting that all together, we get that $$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}...$$

Now as you can see, analyzing the function anywhere else other than at$x=0$ would be a pain, because if, for example $$P_2(x)=a_0+a_1x+a_2x^2$$, and we analyze at $x=3$, we get that $$P_2(3)=a_0+3a_1+9a_2≈\cos(3)$$ So as you can see, getting any coefficients out of this expression would be impossible, further more if we have an infinite polynomial.


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