Proof that cosx =2x + 0.5 has only 1 real solution

[borchester]

I am probably missing something but this question is driving me nuts. Any help would be much appreciated.

 

[Deleted member 4993]

I am probably missing something but this question is driving me nuts. Any help would be much appreciated.

Plot y = 2x+0.5 and y = cos(x) and observe the point of intersection.

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[lex]

Or to prove it, prove that [MATH]y=\cos x-2x-\tfrac{1}{2}[/MATH] only cuts the x-axis once. (I.e. [MATH]\cos x-2x-\tfrac{1}{2}=0[/MATH] has only one solution).
Sketch the graph of [MATH]y=\cos x-2x-\tfrac{1}{2}[/MATH]The graph may suggest how to prove it.

 

[pka]

You posted this in the Math odds & Ends forum making it hard to know which tools we can work.
If \(f(x)=\cos(x)-2x-0.5\) we can see that \(f(0)=0.5~\&~f\left(\frac{\pi}{2}\right)\sim -3.64\)...............corrected
Because \(f\) is continuous then it must have a zero in \(\left[0,\frac{\pi}{2}\right]\).
If you have calculus as a tool you know that \(f^{\prime}(x)=-\sin(x)-2\).
Because \(|\sin(x)|\le 1\) if follows that \(f^{\prime}(x)<0,~\forall x\). SEE HERE
The negative derivative tells us the function \(f\) is decreasing everywhere. SEE HERE
All of which says there in only one zero.

 

[pka]

There are several typo in my post.
It should be \(f(0)=0.5\) and \(f\left(\frac{\pi}{2}\right)\sim -3.64\) ............................ corrected above in response #4
By continuity there is a zero of \(f\) in \(\left[0,\frac{\pi}{2}\right]\).

 

[borchester]

Or to prove it, prove that [MATH]y=\cos x-2x-\tfrac{1}{2}[/MATH] only cuts the x-axis once. (I.e. [MATH]\cos x-2x-\tfrac{1}{2}=0[/MATH] has only one solution).
Sketch the graph of [MATH]y=\cos x-2x-\tfrac{1}{2}[/MATH]The graph may suggest how to prove it.

That is the interesting bit.

The question appeared on the Edexcel A Level Paper 1 2019 and it asked for a graphical solution, which was straightforward enough. But I am hoping for a more rigourous proof.

Anyway, many thanks for your comments.

 

[borchester]

You posted this in the Math odds & Ends forum making it hard to know which tools we can work.
If \(f(x)=\cos(x)-2x-0.5\) we can see that \(f(0)=0.5~\&~f\left(\frac{\pi}{2}\right)\sim -3.64\)...............corrected
Because \(f\) is continuous then it must have a zero in \(\left[0,\frac{\pi}{2}\right]\).
If you have calculus as a tool you know that \(f^{\prime}(x)=-\sin(x)-2\).
Because \(|\sin(x)|\le 1\) if follows that \(f^{\prime}(x)<0,~\forall x\). SEE HERE
The negative derivative tells us the function \(f\) is decreasing everywhere. SEE HERE
All of which says there in only one zero.

Thank you pka, I will chew on the above

 

[JeffM]

I think a rigorous proof is quite simple if you know calculus. Essentially, pka and lex have given it.

[MATH]f(x) = 2x + 0.5 \implies f(x) < -1 \text { if } x < - 0.75 \text { and } f(x) > 1 \text { if } x > 0.25.[/MATH]
[MATH]g(x) = cos(x) \implies - 1 \le g(x) \le 1.[/MATH]
[MATH]h(x) = f(x) - g(x).[/MATH]
[MATH]x < - 0.75x \implies h(x) < -1 -(-1) \implies h(x) < 0. \text { And } x > 0.25 \implies h(x) > 1 - 1 \implies h(x) > 0.[/MATH]
[MATH]\text {Both f(x) and g(x) are continuous. Therefore, h(x) is continuous.}[/MATH]
[MATH]\text {Thus, by the Intermediate Value Theorem, } \exists \ a \text { such that } -0.75 \le a \le 0.25 \text { and } h(a) = 0.[/MATH]
[MATH]h(a) = 0 \implies f(a) = g(a).[/MATH]
This proves there is at least one point of intersection. Now we prove it is unique.

[MATH]h’(x) = 2 + sin(x) \implies h’(x) \ge 1.[/MATH]
[MATH]\therefore x > a \implies h(x) > h(a) \implies h(x) > 0 \implies f(x) \ne g(x).[/MATH]
[MATH]\text {And } x < a \implies h(x) < h(a) \implies h(x) < 0 \implies f(x) \ne g(x).[/MATH]
[MATH]x = a \implies f(x) = g(x) \text { and } x \ne a \implies f(x) \ne g(x). \text { Q.E.D.}[/MATH]

 

[borchester]

I think a rigorous proof is quite simple if you know calculus. Essentially, pka and lex have given it.

[MATH]f(x) = 2x + 0.5 \implies f(x) < -1 \text { if } x < - 0.75 \text { and } f(x) > 1 \text { if } x > 0.25.[/MATH]
[MATH]g(x) = cos(x) \implies - 1 \le g(x) \le 1.[/MATH]
[MATH]h(x) = f(x) - g(x).[/MATH]
[MATH]x < - 0.75x \implies h(x) < -1 -(-1) \implies h(x) < 0. \text { And } x > 0.25 \implies h(x) > 1 - 1 \implies h(x) > 0.[/MATH]
[MATH]\text {Both f(x) and g(x) are continuous. Therefore, h(x) is continuous.}[/MATH]
[MATH]\text {Thus, by the Intermediate Value Theorem, } \exists \ a \text { such that } -0.75 \le a \le 0.25 \text { and } h(a) = 0.[/MATH]
[MATH]h(a) = 0 \implies f(a) = g(a).[/MATH]
This proves there is at least one point of intersection. Now we prove it is unique.

[MATH]h’(x) = 2 + sin(x) \implies h’(x) \ge 1.[/MATH]
[MATH]\therefore x > a \implies h(x) > h(a) \implies h(x) > 0 \implies f(x) \ne g(x).[/MATH]
[MATH]\text {And } x < a \implies h(x) < h(a) \implies h(x) < 0 \implies f(x) \ne g(x).[/MATH]
[MATH]x = a \implies f(x) = g(x) \text { and } x \ne a \implies f(x) \ne g(x). \text { Q.E.D.}[/MATH]

Many thanks JeffM, I shall crack on.

 

[lex]

@borchester
You are very thorough to look for a rigorous proof.
To add to the above suggestions:

\(\displaystyle f(x)=\cos x - 2x - \frac{1}{2}\ \)
\(\displaystyle f'(x)=-\sin x -2\ \)

\(\displaystyle f'(x)≤1-2=-1 \)
f'(x) negative \(\displaystyle \rightarrow\) f is a strictly decreasing function.

Now \(\displaystyle f(0)=\tfrac{1}{2}, \hspace2ex f\left(\tfrac{\pi}{2}\right)=-\pi - \tfrac{1}{2} < 0\)

f is continuous, therefore f(x)=0 for some \(\displaystyle x \in \left(0,\tfrac{\pi}{2}\right)\), and the function is strictly decreasing \(\displaystyle \rightarrow\) there is only one root.

(If there were two roots \(\displaystyle x_1≠x_2\): then (wlog) e.g. \(\displaystyle x_1>x_2\) and \(\displaystyle f(x_1)=0=f(x_2)\), contradicting the fact that f is strictly decreasing).

 

[borchester]

@borchester
You are very thorough to look for a rigorous proof.
To add to the above suggestions:

[MATH]f(x)=\cos x - 2x - \tfrac{1}{2}\\ f'(x)=-\sin x -2\\ \text{ }\\ f'(x)≤1-2=-1[/MATH]f'(x) negative [MATH]\rightarrow[/MATH] f is a strictly decreasing function.

Now [MATH]f(0)=\tfrac{1}{2}, \hspace2ex f\left(\tfrac{\pi}{2}\right)=-\pi - \tfrac{1}{2} < 0[/MATH]
f is continuous, therefore f(x)=0 for some [MATH]x \in \left(0,\tfrac{\pi}{2}\right)[/MATH], and the function is strictly decreasing [MATH]\rightarrow[/MATH] there is only one root.

(If there were two roots [MATH]x_1≠x_2[/MATH]: then (wlog) e.g. [MATH]x_1>x_2[/MATH] and [MATH]f(x_1)=0=f(x_2)[/MATH], contradicting the fact that f is strictly decreasing).

Thanks Lex

The thing is that I am retired. I have gotten a few bit of paper stating that I am good at maths and which I have managed to parley into soft jobs.

To my way of thinking maths is fun and about finding simple solutions to most problems. I am not saying that sitting in meetings while someone goes on and on because they really don't want to return to their desks is a bad thing, but maths cuts through a lot of the brown stuff. The only thing is that I feel as though I am in the garden shed. I have lots of useful tools that I can operate, but I would like to know a bit more about them.

Once again, many thanks for your help