How to solve $3x+\sin x = e^x$

Question

How doese one solve this equation?

$$ 3x+\sin x = e^x $$

I tried graphing it and could only find approximate solutions, not the exact solutions. My friends said to use Newton-Raphson, Lagrange interpolation, etc., but I don't know what these are as they are much beyond the high school syllabus.

Problems like these generally only have numerical answers. Graph it. — Qiaochu Yuan, Jun 28 '12 at 13:03
Have you seen this http://math.stackexchange.com/questions/148129/how-to-solve-axbx2-ecx/148137#148137? — Simon Markett, Jun 28 '12 at 13:05
@QiaochuYuan Sir, I don't know how to graph h(x) = f(x) + g(x) — Bazinga, Jun 28 '12 at 13:12
@Bazinga: graph $y = e^x - 3x - \sin x$ with a calculator (or WolframAlpha I suppose). The points at which it crosses the $x$-axis are your solutions. — Qiaochu Yuan, Jun 28 '12 at 13:13
@QiaochuYuan Sir what if this question came in my exam? Is there any analytic approach to this? Yrogirg Yes I know C++ — Bazinga, Jun 28 '12 at 13:30
Wow, you must have a rough teacher if this was in your exam. — Neal, Jun 28 '12 at 13:31
@Bazinga: you will never get asked this question on an exam. — Qiaochu Yuan, Jun 28 '12 at 13:37
Let Wolframalpha solve for you: http://www.wolframalpha.com/input/?i=3x%2Bsinx%3De%5Ex++ For better insight use the alternative form $sinx=e^x - 3x$ since left hand side will be bounded between -1 and 1. — karakfa, Jun 28 '12 at 15:30
Well we know $x\neq 1$ because $e<3$. we also know $x\neq 2$ because $e^2 > 7$ but $\sin 2 < 1$. And from there, we know that $x\not> 2$ because $e^x$ grows much more quickly as opposed to $3x + \sin x$. Therefore, we know that $x\in(1, 2)$. This may serve as a hint :) — Mr Pie, Feb 27 '18 at 15:11

score 5 · Accepted Answer · answered Jun 28 '12 at 23:14

5

A nice method to find an approximate solution is to successively cut intervals in half, as follows: let's first rewrite this as $$f(x) = 3x + \sin x - e^x = 0$$ Now pick two values, $a$ and $b$, such that $f(a) < 0$ and $f(b) > 0$. (You might have to make a few guesses before finding such values!) In this case, let's choose $a = 0$ and $b = 1$: $$f(a) = 3(0)+\sin(0)-e^{0} = -1 < 0$$ $$f(b) = 3(1)+\sin(1)-e^{1} = 1.12... > 0$$

Now, because our function $f(x)$ is "smooth", there must be a solution somewhere between $a$ and $b$. Find the point halfway in between them, $\frac{0+1}{2} = 0.5$, and check to see whether it makes $f(x)$ positive or negative: $$f(0.5) =3(0.5)+\sin(0.5)-e^{0.5} = 0.33... > 0$$ So, since $f(x)$ is positive here, set $0.5$ as the new value for $b$. Once again calculate the midpoint of $a$ and $b$ (in this case, $\frac{0+0.5}{2} = 0.25$), evaluate it in $f$, and so on. Continue until you have the precision you are looking for.

answered Jun 28 '12 at 23:14

Théophile

24,627

Here is the plot made by maple: http://i.stack.imgur.com/puDet.png – Sigur Jun 28 '12 at 23:30
how does this differ from fixed-point method versus your method of estimation? – User69127 Mar 17 '15 at 19:33
@User69127 What do you mean by "fixed-point method"? – Théophile Mar 17 '15 at 19:46
http://en.wikipedia.org/wiki/Fixed-point_iteration – User69127 Mar 17 '15 at 22:02
@User69127 I see. This method, the bisection method, is a naïve but reliable approach. It doesn't necessarily converge quickly, but it is guaranteed to converge. – Théophile Mar 18 '15 at 02:24

score 0 · Answer 2 · answered May 23 '13 at 12:52

0

sorry

f(b)=3(1)+sin(1)−e1= is Not Equal to ---> 1.12...

f(b)=3(1)+sin(1)−e1= is Equal to ----> 0.299170578

And

f(0.5)=3(0.5)+sin(0.5)−e0.5= is Not Equal to ---> 0.33...>0

f(0.5)=3(0.5)+sin(0.5)−e0.5= is Equal to ----> -0.1399947352

answered May 23 '13 at 12:52

Dereen Jalal Askary

1

2

Welcome to MSE! It helps to format questions using MathJax (see FAQ). Also, it helps to provide details as to how you arrived at answers. Regards – Amzoti May 23 '13 at 13:11

How to solve $3x+\sin x = e^x$

2 Answers2