Master's level mathematics can be daunting, filled with complex concepts that require a deep understanding and a strategic approach. For students navigating these challenges, having a reliable math assignment helper can make all the difference. Websites like MathsAssignmentHelp.com offer the expert assistance you need to conquer these problems. In this blog, we'll explore two advanced math problems, complete with detailed solutions, to guide you through the process. Let's dive in!

Problem 1: Solving a System of Non-Linear Equations

Question:

Solve the following system of non-linear equations:

x2+y2=13x^2 + y^2 = 13x2+y2=13 x2−y2=5x^2 - y^2 = 5x2−y2=5

Solution:

This system of non-linear equations presents a challenge because both equations involve quadratic terms. The goal is to find the values of $x$ and $y$ that satisfy both equations simultaneously.

1. Add the Equations:

   To simplify the system, let's add the two equations together:

   (x2+y2)+(x2−y2)=13+5(x^2 + y^2) + (x^2 - y^2) = 13 + 5(x2+y2)+(x2−y2)=13+5

   This simplifies to:

   2x2=18⇒x2=92x^2 = 18 \quad \Rightarrow \quad x^2 = 92x2=18⇒x2=9

   Taking the square root of both sides:

   $$x=\pm 3$$

2. Substitute $x$ into One of the Original Equations:

   Now that we have the possible values for $x$, let's substitute them into one of the original equations to find the corresponding values of $y$. We'll use the first equation x2+y2=13x^2 + y^2 = 13x2+y2=13:

   • For $x=3$:

     32+y2=13⇒9+y2=13⇒y2=4⇒y=±23^2 + y^2 = 13 \quad \Rightarrow \quad 9 + y^2 = 13 \quad \Rightarrow \quad y^2 = 4 \quad \Rightarrow \quad y = \pm 232+y2=13⇒9+y2=13⇒y2=4⇒y=±2

   • For $x=-3$:

     (−3)2+y2=13⇒9+y2=13⇒y2=4⇒y=±2(-3)^2 + y^2 = 13 \quad \Rightarrow \quad 9 + y^2 = 13 \quad \Rightarrow \quad y^2 = 4 \quad \Rightarrow \quad y = \pm 2(−3)2+y2=13⇒9+y2=13⇒y2=4⇒y=±2

3. Solution Pairs:

   The solutions to the system are the pairs $(x,y)$:

   $$(3,2),(3,-2),(-3,2),(-3,-2)$$

   These pairs represent the points where the two curves (a circle and a hyperbola) intersect.

4. Verification:

   It's important to verify that these solutions satisfy both original equations. Substituting each pair into both equations will confirm their correctness.

   For example, substituting $x=3$ and $y=2$ into the second equation x2−y2=5x^2 - y^2 = 5x2−y2=5:

   32−22=9−4=53^2 - 2^2 = 9 - 4 = 532−22=9−4=5

   Since this holds true, the solution pair $(3,2)$ is correct. Similarly, the other pairs also satisfy both equations.

5. Conclusion:

   The solutions to the system of non-linear equations are the four pairs $(3,2)$, $(3,-2)$, $(-3,2)$, and $(-3,-2)$. If you're working on similar complex systems and need expert guidance, consider reaching out to a math assignment helper at MathsAssignmentHelp.com for personalized assistance.

Problem 2: Finding the Fourier Series of a Function

Question:

Determine the Fourier series of the function $f(x)$ defined as:

f(x)={1if −π<x<0−1if 0<x<πf(x) = \begin{cases} 1 & \text{if } -\pi < x < 0 \\ -1 & \text{if } 0 < x < \pi \end{cases}f(x)={1−1​if −π<x<0if 0<x<π​

Solution:

Fourier series are essential in various fields, from signal processing to solving partial differential equations. The goal here is to express the given piecewise function as a sum of sines and cosines.

1. Determine the Coefficients:

   The Fourier series of a function $f(x)$ with period $2\pi$ is given by:

   f(x)=a02+∑n=1∞(ancos⁡(nx)+bnsin⁡(nx))f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) ight)f(x)=2a0​​+n=1∑∞​(an​cos(nx)+bn​sin(nx))

   where the coefficients ana_nan​ and bnb_nbn​ are calculated as follows:

   an=1π∫−ππf(x)cos⁡(nx) dxa_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dxan​=π1​∫−ππ​f(x)cos(nx)dx bn=1π∫−ππf(x)sin⁡(nx) dxb_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dxbn​=π1​∫−ππ​f(x)sin(nx)dx

   For the given function $f(x)$, it's an odd function, meaning an=0a_n = 0an​=0 for all $n$. We need to compute only the bnb_nbn​ coefficients.

2. Calculate the bnb_nbn​ Coefficients:

   Since the function is odd, we only compute bnb_nbn​:

   bn=1π∫−ππf(x)sin⁡(nx) dxb_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dxbn​=π1​∫−ππ​f(x)sin(nx)dx

   This integral can be split into two parts:

   bn=1π(∫−π01⋅sin⁡(nx) dx+∫0π(−1)⋅sin⁡(nx) dx)b_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} 1 \cdot \sin(nx) \, dx + \int_{0}^{\pi} (-1) \cdot \sin(nx) \, dx ight)bn​=π1​(∫−π0​1⋅sin(nx)dx+∫0π​(−1)⋅sin(nx)dx)

   Solving the integral:

   bn=1π([−1ncos⁡(nx)]−π0+[1ncos⁡(nx)]0π)b_n = \frac{1}{\pi} \left( \left[ -\frac{1}{n} \cos(nx) ight]_{-\pi}^{0} + \left[ \frac{1}{n} \cos(nx) ight]_{0}^{\pi} ight)bn​=π1​([−n1​cos(nx)]−π0​+[n1​cos(nx)]0π​)

   Simplifying:

   bn=2nπ(1−(−1)n)b_n = \frac{2}{n\pi} \left(1 - (-1)^night)bn​=nπ2​(1−(−1)n