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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?

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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 $\approx 0.000005....$.

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  • $\begingroup$ @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.) $\endgroup$ – symplectomorphic Mar 22 '16 at 4:20
  • $\begingroup$ Sorry, forgot about the $\pi/4$. $\endgroup$ – André Nicolas Mar 22 '16 at 4:22

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