Finding number of real roots of an equation

Question

Equation is--> $$ x^{13} + x - 1/e^x - \sin(x) =0 $$

To find number of real roots of the equation.

Context--> I am solving previous years questions of IIT Jam Mathematical Statistics (MS entrance exam) .

My approach--> I took $e^{-x}$ and $\sin(x)$ to other side of the quation and expanded them. The coefficients of $x$ and $x^{13}$ were zero on RHS and if I put $x=0$ or $x=1$ there was no way to equate both sides. So I thought the answer would be no real roots. After that I put that graph on Desmos. Then I saw this graph cuts x axis at one point. So thank you for reading and answering this.

Also my go to approach for finding real roots of a polynomial is to put values of $x$ and check the sign changes. Can you suggest me a different approach. I know about Descartes rule but that doesn't give exact number of real roots.

Example equation:-

$3x^3 - 12x^2 + 11x - 31=0$

For your second question, the rational root theorem gives all possible rational roots of a polynomial. If you find one root of a cubic, you can divide synthetically to get an easily factorable quadratic. — Gnumbertester, Jan 29 '19 at 05:12
If you consider $f(x)=x^{13} + x -e^{-x} - \sin(x) $, then $f(0)=-1$ and $f(1)=2-\frac{1}{e}-\sin (1) >0$ — Claude Leibovici, Jan 29 '19 at 05:13

Robert Lewis · Accepted Answer · 2019-02-07T18:01:58.613

Setting

$f(x) = x^{13} + x - \dfrac{1}{e^x} - \sin x = x^{13} + x - e^{-x} - \sin x, \tag 1$

it is easy to see that there exists $M > 0$ such that

$x \le - M \Longrightarrow f(x) < 0, \tag 2$

and

$x \ge M \Longrightarrow f(x) > 0, \tag 3$

since the term $x^{13}$ dominates the others for large $x > 0$, as does $e^{-x}$ for large $x < 0$; it then follows from the intermediate value theorem that

$\exists x_0 \in (-M, M), f(x_0) = 0; \tag 4$

thus, $f(x)$ has at least one real zero. Next, we have

$f'(x) = 13 x^{12} + 1 + e^{-x} - \cos x > 0, \; \forall x \in \Bbb R, \tag 5$

since

$1 + e^{-x} - \cos x > 0, \; \forall x \in \Bbb R, \tag 6$

and

$13 x^{12} \ge 0, \; \forall x \in \Bbb R, \tag 7$

it follows that

$f'(x) > 0, \; \forall x \in \Bbb R; \tag 8$

therefore $x_0$ is the sole real zero of $f(x)$.

As for polynomials $p(x) \in \Bbb R[x]$, for $\deg p(x) \ge 5$ there is not a lot to be said in general, but there are a few useful rules of thumb, e.g. $p(x)$ has at least one real root if $\deg p(x)$ is odd. Of course, when $\deg p(x) \le 4$ one has well-known procedures and formulas to find solutions from the coefficients of $p(x)$. In the case of

$3x^3 - 12x^2 + 11x - 31 = 0, \tag 9$

we may affirm the existence of a root; but to get much further, we might resort to Cardano's method which, though arithmetic intensive, is guaranteed to yield the desired results.

score 1 · Answer 2 · answered Jan 29 '19 at 06:38

$$y(x)=x^{13}+x-e^{-x}-\sin(x)$$ $y(x)$is continuous (Sum of continuous functions). $$y'(x)=13x^{12}+1+e^{-x}-\cos(x)$$ $13x^{12}+1+e^{-x}>1\quad$ and $\quad-\cos(x)\geq -1\quad$ thus $y'(x)>0\quad$ any $x$.

$y(x)$ is continuous and strictly increasing from $-\infty$ to $+\infty$.

Thus there is a root $f(x_0)=0$ with Bolzano's theorem and this root is unique.

Numerical solving (Newton-Raphson) : $x_0\simeq 0.907676...$

Finding number of real roots of an equation

2 Answers2