Why does the following Maclaurin series equal this catenary curve?

Question

Given this Maclaurin series:

$$f(x)=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$

And the following Catenary curve, assuming that $a=1$:

$$g(x)=\frac{a(e^\frac{x}{a}+e^{-\frac{x}{a}})}{2}$$

Why does $f(x)=g(x)$ seem to hold true (at least when graphed)?

I'm looking for a purely algebraic reason here as to why these two are equal, ideally in terms that are at or around a high-school calculus level (where I'm at currently).

If I am mistaken, and these two are not equal to each other, an explanation of why that is would be great too.

Answer 1

Remember that $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$

Thus $$\frac{1}{2}(e^\frac{x}{a}+e^\frac{-x}{a})=\frac{1}{2}\sum_{n=0}^{\infty}\left [\left (\frac{x}{a}\right )^n\frac{1}{n!}+\left (\frac{-x}{a}\right )^n\frac{1}{n!}\right ]$$

This can be simplified to $$\frac{1}{2}\sum_{n=0}^\infty \left (\frac{1+(-1)^n}{a^n n!}\right )x^n$$

When $n$ is odd, $1+(-1)^n=0$, so we can replace $n$ by $2n$ everywhere to get $$\frac{1}{2}\sum_{n=0}^\infty \left [\frac{1+1}{a^{2n} (2n)!}x^{2n}\right ]=\sum_{n=0}^\infty\frac{x^{2n}}{a^{2n}(2n!)}$$

When $a=1$, you get the series $$\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$

Answer 2

Let consider the function

$$C(x)=\frac{e^x+e^{-x}}{2}$$

and let define also

$$S(x)=C'(x)=\frac{e^x-e^{-x}}{2} \implies S'(x)=C(x)$$

thus since

• $C(0)=1$
• $C'(0)=S(0)=0$
• $C''(0)=S'(0)=C(0)=1$
• ...

by Taylor's expansion at $x=0$, that is Maclaurin's expansion, we have that

$$C(x)=\frac{e^x+e^{-x}}{2}=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\dots+\frac{x^{2k}}{{2k}!}+\dots=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$

which turns out to converge for any $x\in \mathbb{R}$.

Answer 3

The starting point is the series expansion of the exponential function: $$ e^x=\sum_{k=0}^\infty \frac{x^k}{k!} $$ substituting in $g(x)$ with $a=1$ we have: $$ g(x)=\frac{e^x+e^{-x}}{2} = \frac{1}{2}\left(\sum_{k=0}^\infty \frac{x^k}{k!}+\sum_{k=0}^\infty \frac{(-x)^k}{k!} \right)=\frac{1}{2}\left[\sum_{k=0}^\infty \left(\frac{x^k}{k!}+ \frac{(-x)^k}{k!} \right)\right] $$ now note that the for $k$ odd the terms in the series are null and for $k$ even the terms becomes $2\frac{x^k}{k!}$, so in the series we have only the even terms $k=2n$ and the function becomes: $$ g(x)=\frac{1}{2}\sum_{n=0}^\infty\frac{2x^{2n}}{(2n)!}=\sum_{n=0}^\infty\frac{x^{2n}}{(2n)!} $$