Zero of e^x + sin x by bisection and Taylor

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Let $g(x)=e^{x}+\sin x$. (a) Find the solution of $g(x)=0$ between $x=-1$ and $x=0$ using the bisection method, with error $\le 0.1$. (b) Find the solution with a first-degree polynomial approximation. (c) Find the solution with a second-degree polynomial approximation.

Solution

(a) Bisection. Check the sign change: $g(-1)=e^{-1}+\sin(-1)=0.368-0.841=-0.473<0$, $\;g(0)=1+0=1>0$. Root in $[-1,0]$.

• $g(-0.5)=0.607-0.479=+0.127>0\Rightarrow$ root in $[-1,-0.5]$.
• $g(-0.75)=0.472-0.682=-0.210<0\Rightarrow$ root in $[-0.75,-0.5]$.
• $g(-0.625)=0.535-0.585=-0.050<0\Rightarrow$ root in $[-0.625,-0.5]$.
• $g(-0.5625)=0.570-0.533=+0.037>0\Rightarrow$ root in $[-0.625,-0.5625]$, width $0.0625<0.1$.

So $x_0\approx-0.59$.

(b) First-degree approximation (Taylor at $0$). $g(0)=1$, $g'(x)=e^{x}+\cos x\Rightarrow g'(0)=2$. Line: $1+2x=0\Rightarrow$ $x_1=-\tfrac12=-0.5.$

(c) Second-degree approximation. $g''(x)=e^{x}-\sin x\Rightarrow g''(0)=1$. Polynomial: $1+2x+\tfrac{x^2}{2}=0\Rightarrow x^2+4x+2=0$ $x=\frac{-4+\sqrt{16-8}}{2}=-2+\sqrt2\approx-0.586.$ The quadratic value matches the bisection root well. $\boxed{x_0\approx-0.586,\qquad x_1=-0.5,\qquad x_2=-2+\sqrt2\approx-0.586}$