Solving the Equation cos2x e^(x^2) using Iterative Methods and Complex Analysis

The equation (cos(2x) e^{x^2}) presents a unique challenge in mathematical analysis. This article delves into the methods used to solve this equation, including both iterative techniques and complex analysis.

Iterative Method for Solving cos2x e^(x^2)

To solve (cos(2x) e^{x^2}), we can use an iterative approach. We start with an initial value and refine it until convergence. Let's go through the steps and results of this process.

Initial Value: x_0  0Next Step: cos2x_1  -1, which corresponds to x_1  π / 2cos2x_2  -e^(-π^2/4), which corresponds to x_2  0.82785164045x_3  1.04946356641x_4  0.95482920863x_5  0.99216136972x_6  0.97687901508x_7  0.98304782891x_8  0.9805431788_9  0.98155773484x_10  0.98114637797x_11  0.98131310041x_12  0.98124551744

We observe that the values are converging to a root, approximated numerically as 0.9812…

Graphical Representation

To visualize the solution, we plot the function (f(x) e^{-x^2} cos(2x) 0). This represents the intersection points of the graphs of (f(x)) and the x-axis. The graph shows multiple roots, providing insights into the behavior of the function.

The red line represents the real part (Rez) of the graph, while the blue line represents the imaginary part (Imz) of the graph. These visualizations help in understanding the complex solutions to the equation.

General Solution with GRP-N Method

For a more comprehensive understanding, we can explore the general solution using the GRP-N method. The equations are as follows:

1. Exponential Form

[ e^{-x^2} u [ x i sqrt{log(u)} 2kpi i ]

2. Negative Exponential Form

[ e^{-x^2} u [ x -i sqrt{log(u)} 2kpi i ]

3. Cosine Form

[ 2x u [ x frac{1}{2} - arccos(u) 2pi k ]

4. Positive Cosine Form

[ 2x u [ x frac{1}{2} arccos(u) 2pi k ]

These forms are derived from the initial iterative method and further refined using repetitions and iterative techniques.

roots derived from formfields

From the formfields, the roots are derived as follows:

3rd Field

For (k0) and (k-1):

x0 -0.9812650099953767418979014829686995492733 x-1 -2.354235980933415240494542162185200559476 x-2 -5.497787143782100831021115813973596638339

4th Field

For (k0) and (k-1):

x0 0.9812650099982906703697079884843882349480 x-1 -3.926990917357004028411154023796289875019 x-2 -7.068583470577034786541047497703163820183

Complex Roots

If we delve into the complex solutions:

Field 1

For (k1) and (k-1):

x-1 -1.193687332904474180610802626424254147441 - 2.315959034318176687754663758925472261167i x-2 -2.135757321101391873513664264633442594883 - 3.206424810254914398920530908538994225641i

Field 2

For (k1) and (k-1):

x-1 1.193687332904474180610802626424254147 - 2.315959034318176687754663758925472261i x-2 2.135757321101391873513664264633442594 - 3.206424810254914398920530908538994225i

These complex roots provide a deeper understanding of the equation's solutions, indicating multiple roots across the complex plane.

For a more detailed understanding, the graph above shows the function in the complex plane, with real and imaginary parts clearly labeled. This visualization aids in comprehending the behavior of the solution across the complex plane.

Conclusion

In conclusion, solving the equation (cos(2x) e^{x^2}) involves a rigorous examination of iterative and complex analysis methods. The iterative process reveals a root near 0.9812, while the general solutions and complex roots provide a broader perspective on the equation's behavior.

For those interested in further exploration, working through the iterative process and understanding the complex solutions can offer additional insights and a deeper appreciation of the equation's nature.