# Error of Taylor Series?

Part of my assignment is to find the third degree Taylor Series of $\tan(x)$ about $\pi/4$ and then estimate the error of this approximation when evaluated at 0.75.

Finding the series was easy enough, but I wasn't sure about the error estimate.

From what I know the error is:

$$E_n \leq \frac{| f ^{(n+1)} (c) |\cdot|x-a|^{(n+1)}}{ (n+1)! } ,$$

where $c$ is between $x$ and $a$.

Now, when I compute this for $\tan{(x)}$ I end up with:

$$E_n \leq \frac{|-4[\cos(2c)-5]\tan(c)\sec^4(c)|\cdot|0.75 - \pi/4|^4} {24}$$

Not only is the fourth derivative of tan(x) really messy, I have no idea what $c$ to choose so that I can simplify things nicely. The whole point was to estimate the error for tan but tan shows up in the error estimate itself! Does this $c$ have to be strictly between the center of the series and the point I'm evaluating at?

The inequality you wrote down is not quite right (or rather, it is slightly misleading). In fact, the error is exactly equal to $$\frac{f^{(n+1)}(c)\cdot(x-a)^{n+1}}{(n+1)!}$$ for some $c$ between $x$ and $a$.

But, as you say, because we don't know what $c$ is, this fact is often useless. Instead, you're supposed to use this fact to find an upper bound on the error. You do this by considering how big $|f^{(n+1)}|$ can be on the interval $(x,a)$. Suppose $|f^{(n+1)}|$ is at most $M$. Then the magnitude of the error will be at most $$\frac{M\cdot|x-a|^{n+1}}{(n+1)!}$$

In your case, because $a=\pi/4>x=0.75$, $c$ is somewhere in the interval $(0.75,\pi/4)$. But on this interval, the fourth derivative of tangent, namely

$$f^{(4)}(x)=-4(\cos(2x)-5)\tan x\sec^4x\,,$$

is positive and increasing. (It is obvious this expression is positive on the specified interval because $\cos(2x)-5$ is certainly negative, while $\tan x$ and $\sec^4x$ are certainly positive. You can see the function is increasing either graphically or by taking another derivative: the fifth derivative is strictly positive, so the fourth derivative is increasing.)

Therefore $$|f^{(4)}(c)|\leq|f^{4}(\pi/4)|=80$$ In other words, in the notation above, we have $M=80$. So the magnitude of the error of the third-degree approximation is at most $$\frac{80\cdot|0.75-\pi/4|^4}{24}\approx 0.0000053$$ Indeed, using the third-order approximation

$$P_3(x)=1+2\left(x-\frac{\pi}{4}\right)+2\left(x-\frac{\pi}{4}\right)^2+\frac{8}{3}\left(x-\frac{\pi}{4}\right)^3$$

we find $$P_3(0.75)\approx 0.9315914...$$ whereas $$\tan(0.75)\approx0.9315964...$$ with error $\approx 0.000005....$.

• @AndréNicolas: I'm not sure what you mean when you say "the third degree approximation is also the fourth degree approximation." Not so for $\tan x$ expanded about $\pi/4$. (But yes, for the expansion about zero.) – symplectomorphic Mar 22 '16 at 4:20
• Sorry, forgot about the $\pi/4$. – André Nicolas Mar 22 '16 at 4:22