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I am trying to find "$\cosh 4$ using the sixth partial sum ($n=5$) of the Maclaurin series" for the function. I am also trying to use "the Lagrange form of the remainder to estimate the number of decimal places to which the partial sum" is accurate.

For the first part, I am getting $\displaystyle\sum_{n=0}^5 \frac{4^{2n}}{(2n)!} \approx 27.2699$, which agrees with the provided answer. However, I am having trouble estimating the error.

Here is what I have done so far.

The Lagrange form of the remainder is $\displaystyle R_n = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$, where here $a=0$ (Maclaurin), $x=4$ (the value for which we want $\cosh$), and $0=a\leq c\leq x=4$.

To find an upper bound for $f^{(n+1)}(c)$, since all derivatives of $\cosh$ are less than or equal to $\cosh$, I see that using $\cosh$ will work, and I also see that $4$ will yield the highest value, so $f^{(n+1)}(c) \leq \cosh 4$. Since we are assuming that we don't know what $e$ is, we have

$$f^{(n+1)}(c) \leq \cosh 4 = \frac{e^4+e^{-4}}{2} < \frac{3^4+2^{-4}}{2} < 41.$$

Using the Lagrange form of the remainder, I get $\displaystyle R_n < \frac{41\cdot 4^{5+1}}{(5+1)!}\approx 233$, which is much greater than the error provided in the answer.

It seems that the solution is using $2(n+1)$ instead of $n+1$ in the denominator of the Lagrange form (with the factorial), and likewise in the exponent for $x$ (for using these yields the provided answer). However, using this seems rather arbitrary as it does not appear in the general formula. How do I reconcile the general form of the remainder with the provided answer?

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I would not agree that $\sum_0^5 \frac{4^{2i}}{(2i)!}$ is the sixth partial sum. The "zero terms" should be counted.

However, your post provides numerical confirmation that the answer was computed by forgetting about zero terms.

The remainder, in the Lagrange sense, is the same as the one we get if we take the terms up to and including the term in $x^{11}$. (Yes, it is $0$).

Thus if you are going to use the Lagrange remainder formula as quoted in Wikipedia, for the "$k$" in that formula, you need to use $11$.

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I am going to elaborate on André’s answer after having thought about the problem some more.

If we let $T_n(x)$ be the $n$th-degree Taylor polynomial of $\cosh$ at $0$ (since Maclaurin), and $R_n(x)$ be the remainder of the Taylor series, then we can write

$$\cosh x = T_n(x) + R_n(x).$$

We can also write

$$\cosh x = 1+ 0 + \frac{x^2}{2!} + 0 + \frac{x^4}{4!} + \cdots + \frac{x^8}{8!} + 0 + \frac{x^{10}}{10!} + 0+ \cdots = \sum_{i=0}^\infty \frac{x^{2i}}{(2i)!}.$$

Now if we want to use the partial sums of $\cosh$, we notice that we don't get $\sum_{i=0}^n \frac{x^{2i}}{(2i)!} = T_n(x)$, but rather $\sum_{i=0}^n \frac{x^{2i}}{(2i)!} = T_{2n}(x)$.

In the original problem, when it asked to find the “sixth partial sum ($n=5$)”, the “$n$” under consideration was the one associated with the partial sum of $\cosh$, not the one in the Taylor polynomial. In other words, the problem wanted one to evaluate $\left .\sum_{i=0}^5 \frac{x^{2i}}{(2i)!}\right|_{x=4}$. However, for this sum, the associated Taylor polynomial is actually $T_{2\cdot5}(4)=T_{10}(4) = T_{11}(4)$ (the last equality just happens to work since every odd-powered term is $0$, and will help with getting the error down even smaller). Therefore we have $\cosh 4 = T_{11}(4) + R_{11}(4)$. To get the error, we want

$$\begin{align*} R_{11}(4) &= \frac{\cosh^{11+1}(c)\cdot 4^{11+1}}{(11+1)!}\\ &< \frac{41\cdot 4^{12}}{12!} \\ &< 2,\end{align*}$$

which agrees with the provided answer.

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