Science:Math Exam Resources/Courses/MATH104/December 2015/Question 05 (b)

MATH104 December 2015

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Question 05 (b)
We want to estimate a solution to the equation $2^{x}=x+\pi$ . Find a positive integer $n$ such that a solution lies in the interval $[n,n+1]$ .

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Draw (or imagine) the graphs of $f(x)=2^{x}$ and $g(x)=x+\pi$ and estimate their intersection points.

Hint 2
Try using the intermediate value theorem (see the next hint if you need a refresher).

Hint 3
The intermediate value theorem states that if $f$ is a continuous function on $[a,b]$ and $y$ is a number strictly between $f(a)$ and $f(b),$ then there exists a $c\in (a,b)$ such that $f(c)=y.$ What would $y=0$ entail, and how might you use this to answer the question?

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Solution 1

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Sketching the graphs of $f(x)=2^{x}$ and $g(x)=x+\pi ,$ we observe that there are two solutions to the equation $2^{x}=x+\pi ,$ one positive and one negative. Since we are looking for a positive integer $n$ such that the solution lies in $[n,n+1],$ let us estimate the positive solution.

Evaluating $f$ and $g$ at the positive integers $x=1,2,3,\dots$ and gathering our results in a table, we find the following:

${}x{}$	$f(x)$	$g(x)$
$1$	$2^{1}=2$	$4<1+\pi <5$
$2$	$2^{2}=4$	$5{\color {PineGreen}{}<{}}2+\pi <6$
$3$	$2^{3}=8$	$6<3+\pi {\color {Magenta}{}<{}}7$

where we have used the bound $3<\pi <4.$

Intuitively, it seems that $2^{x}=x+\pi$ for some $x$ between $2$ and $3$ , since $f$ is initially less than $g$ but then surpasses it by $x=3.$ We can make this precise as follows: let $h(x)=f(x)-g(x)=2^{x}-(x+\pi ).$ Now $h$ is continuous on the real line, and

{\begin{aligned}h({\color {OliveGreen}2})&=4-(2+\pi ){\color {PineGreen}{}<{}}4-5<{\color {Orange}0},\\h({\color {OrangeRed}3})&=8-(3+\pi ){\color {Magenta}{}>{}}8-7>{\color {Orange}0}.\end{aligned}}

Hence by the intermediate value theorem, there exists a number $c$ in the interval $({\color {OliveGreen}2},{\color {OrangeRed}3})$ (and therefore also in $[2,3]$ ) such that $h(c)={\color {Orange}0},$ i.e., $2^{c}-(c+\pi )=0\implies 2^{c}=c+\pi .$