Hyperbolic cosine

Question

I have an A level exam question I'm not too sure how to approach:

a) Show $1+\frac{1}{2}x^2>x, \forall x \in \mathbb{R}$

b) Deduce $ \cosh x > x$

c) Find the point P such that it lies on $y=\cosh x$ and its perpendicular distance from the line $y=x$ is a minimum.

I understand how to show the first statement, by finding the discriminant of $1+\frac{1}{2}x^2-x>0$, but trying to apply this to part b doesn't seem to work: $$\cosh x > x \Rightarrow \frac{1}{2}(e^x+e^{-x}) > x \Rightarrow e^x+e^{-x}>2x \Rightarrow e^{2x}-2xe^x+1>0$$ Applying $\Delta>0$ to this because I know the two functions do not intersect: $$\Delta>0 \Rightarrow 4x^2-4>0 \Rightarrow x^2>1 \Rightarrow |x|>1$$ This tells me that in fact, the two do meet, but only when $|x|>1$, what have I done wrong here?

This last bit I don't know how to approach, I was thinking maybe vectors in $\mathbb{R}^2$ were involved to find perpendicular distances?

Thanks in advance.

Answer 1

First, you cannot apply $\Delta$ as $2x$ is not a constant coefficient.

Second, to show b), notice that $\cosh x>\frac{e^x}{2}>x$. To see the second inequality, you need to consider the derivative of $\frac{e^x}{2}-x$.

Answer 2

a) You'r right, you can do this with the discriminant and it is very natural. But you can also use the well-known inequality: $2ab\leq a^2+b^2$ which follows from the expansion of $(a-b)^2\geq 0$. So you get $$ 2x=2\cdot x\cdot 1\leq x^2+1^2=x^2+1<x^2+2\qquad \forall x\in\mathbb{R}. $$ Then divide by $2$.

b) By definition, $\cosh x$ is the even part of $e^x$. Since $e^x=\sum_{n\geq 0}\frac{x^n}{n!}$ for every $x$, you are left with $$ \cosh x=\sum_{n\geq 0}\frac{x^{2n}}{(2n)!}=1+\frac{x^2}{2}+\frac{x^4}{24}+\ldots\qquad\forall x\in\mathbb{R}. $$ Now, using inequality a) and the fact that every term of the series is nonnegative: $$ \cosh x=1+\frac{x^2}{2}+\frac{x^4}{24}+\ldots\geq 1+\frac{x^2}{2}> x\qquad\forall x\in\mathbb{R}. $$

c) The distance from the point $(x_0,y_0)$ to the line $x-y=0$ is $$ d=\frac{|x_0-y_0|}{\sqrt{1^2+(-1)^2}}=\frac{1}{\sqrt{2}}|x_0-y_0|. $$ So for a point $(x,\cosh x)$ on the graph of the hyperbolic cosine, we get $$ d(x)=\frac{1}{\sqrt{2}}|x-\cosh x|=\frac{1}{\sqrt{2}}(\cosh x-x) $$ as $\cosh x > x$ for every $x$. Now study the variations of $d(x)$ to find its minimum. It will occur at a critical point $d'(x_0)=0$. And actually, there is only one such critical point, solution of $\sinh x=1$. So your minimum occurs at $$ x_0=\mbox{arsinh} 1=\log (1+\sqrt{2})\qquad y_0=\cosh x_0=\sqrt{2}. $$ Click on this link for a drawing of the situation.

Answer 3

For c), this point will occur when the slopes of the two curves are equal, so $\frac{d}{dx} \cosh(x)=\sinh(x)=1$, so $x = \sinh^{-1}(1)$ and $y =\cosh(\sinh^{-1}(1))=\sqrt{2}$ .

Answer 4

For $(a)$ just consider the function

$$ y = 1+\frac{x^2}{2}-x \implies y-\frac{1}{2}= \frac{1}{2}(x-1)^2 $$

which is a parabola with vertex at $(1,\frac{1}{2})$ and implies that

$$ y > 0 \implies 1+\frac{x^2}{2}-x>0 \implies 1+\frac{x^2}{2}>x, \forall x \in \mathbb{R} .$$