Question Statement

Find the root of the equation $e^x\sin x=1$ in the interval $[0,1]$ using the Regula Falsi method, correct to two decimal places.

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Background and Explanation

The Regula Falsi (or False Position) method is a numerical technique for finding real roots of equations that combines the bisection method with linear interpolation. It requires an initial interval $[a,b]$ where the function changes sign (i.e., $f(a)\cdot f(b)<0$), guaranteeing a root exists within the interval by the Intermediate Value Theorem.

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Solution

First, we rewrite the equation in standard form $f(x)=0$:

$f(x)=e^x\sin x-1=0$

Verification of Initial Interval

We evaluate the function at the endpoints of the interval $[0,1]$ to confirm a root exists:

$f(0)=e^0\sin 0-1=1\cdot 0-1=-1<0$

$f(1)=e^1\sin 1-1=e\sin (1)-1\approx 2.718\cdot 0.8415-1\approx 1.2874>0$

Since $f(0)<0$ and $f(1)>0$, the function changes sign in $[0,1]$, confirming that a root lies between 0 and 1.

Iteration 1

Using the Regula Falsi formula: $x=\frac{af(b)-bf(a)}{f(b)-f(a)}$

With $a=0$, $b=1$, $f(a)=-1$, and $f(b)=1.2874$:

$x_0=\frac{0\cdot f(1)-1\cdot f(0)}{f(1)-f(0)}=\frac{0-(-1)}{1.2874-(-1)}=\frac{1}{2.2874}\approx 0.4372$

Now evaluate $f(x_0)$: $f(0.4372)=e^{0.4372}\sin (0.4372)-1\approx -0.3444<0$

Since $f(x_0)<0$ and $f(1)>0$, the root lies in the updated interval $[0.4372,1]$.

Iteration 2

Using the Regula Falsi formula on $[0.4372,1]$ with $f(0.4372)=-0.3444$ and $f(1)=1.2874$:

$x_1=\frac{0.4372\cdot f(1)-1\cdot f(0.4372)}{f(1)-f(0.4372)}=\frac{0.4372(1.2874)-(-0.3444)}{1.2874-(-0.3444)}=\frac{0.5629+0.3444}{1.6318}\approx 0.5560$

Evaluate $f(x_1)$: $f(0.5560)=e^{0.5560}\sin (0.5560)-1\approx -0.0797<0$

Since $f(x_1)<0$, the updated interval is $[0.5560,1]$.

Iteration 3

Using the Regula Falsi formula on $[0.5560,1]$ with $f(0.5560)=-0.0797$:

$x_2=\frac{0.5560\cdot f(1)-1\cdot f(0.5560)}{f(1)-f(0.5560)}=\frac{0.5560(1.2874)-(-0.0797)}{1.2874-(-0.0797)}=\frac{0.7158+0.0797}{1.3671}\approx 0.5819$

Evaluate $f(x_2)$: $f(0.5819)=e^{0.5819}\sin (0.5819)-1\approx -0.0165<0$

The updated interval is $[0.5819,1]$.

Iteration 4

Using the Regula Falsi formula on $[0.5819,1]$ with $f(0.5819)=-0.0165$:

$x_3=\frac{0.5819\cdot f(1)-1\cdot f(0.5819)}{f(1)-f(0.5819)}=\frac{0.5819(1.2874)-(-0.0165)}{1.2874-(-0.0165)}=\frac{0.7491+0.0165}{1.3039}\approx 0.5871$

Convergence Check

Comparing the results:

• $x_2\approx 0.5819$
• $x_3\approx 0.5871$

Up to two decimal places, both values round to $0.58$. Therefore, the root of the equation $e^x\sin x=1$ in the interval $[0,1]$, correct to two decimal places, is:

$\boxed{0.58}$

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Key Formulas or Methods Used

• Regula Falsi Formula: $x=\frac{af(b)-bf(a)}{f(b)-f(a)}$
• Intermediate Value Theorem: If $f(a)\cdot f(b)<0$, there exists at least one root in $(a,b)$
• Function Definition: $f(x)=e^x\sin x-1$
• Convergence Criterion: Successive approximations agree to the required number of decimal places

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Summary of Steps

1. Define $f(x)=e^x\sin x-1$ and verify $f(0)=-1<0$ and $f(1)\approx 1.2874>0$ to confirm a root exists in $[0,1]$.
2. First iteration: Calculate $x_0=0.4372$ using Regula Falsi; since $f(x_0)<0$, new interval is $[0.4372,1]$.
3. Second iteration: Calculate $x_1=0.5560$; since $f(x_1)<0$, new interval is $[0.5560,1]$.
4. Third iteration: Calculate $x_2=0.5819$; since $f(x_2)<0$, new interval is $[0.5819,1]$.
5. Fourth iteration: Calculate $x_3=0.5871$.
6. Verify convergence: $x_2$ and $x_3$ agree to two decimal places ($0.58$), so the root is $0.58$.