Math Notes by Dr. Raghad - PDF

# lec. 1 - Determinants ## Math + Pirstrow a11 a12 a13 + a21 a22 a23 det(A) = a11a22 - a12a21. a31 a32 a33 **Matrices** Jl determinants الاختلاف بين الـ ** Matrices can be solved using determinants but matrices cannot be solved using determinants** sqare no.raws=no. @lons **If matrix is...

# lec. 1 - Determinants ## Math + Pirstrow a11 a12 a13 + a21 a22 a23 det(A) = a11a22 - a12a21. a31 a32 a33 **Matrices** Jl determinants الاختلاف بين الـ ** Matrices can be solved using determinants but matrices cannot be solved using determinants** sqare no.raws=no. @lons **If matrix is square, the number of rows must equal the number of columns** **If matrix is not square, then the number of rows does not necessarily equal the number of columns** **To solve using determinants, you use the first row of the matrix with the following rules:** * You take the first number in the row and its sign. * You then delete the row and column containing the number. * This leaves a smaller determinant with 2 rows and 2 columns. * You solve the smaller determinant and multiply by your original number. * You repeat for the remaining numbers in the row. **Example:** | | 1 | 2 | |---|---|---| | A=| 3 | 0 | 1 | | 4 | 2 | 1 | ``` = [(0x1)_(2x1)] - 2 [3/1)-(1x1)]_[(3x2)-(4x0)] = -2 + 2 - 6 = -6 ``` **(row / colon** ) are known as charges > The **charges** of the first row are represented by (0/0/0) whereas the **charges** of the second row are represented by (01010). ## Solving an Equation for a Variable **Solve the following equation for the variable a:** ``` X X+1 = 7 X-2 ``` ``` x(X-2) + (x+1) = 7 ``` ``` x^2 - 2x + x + 1 = x^2 - x+b = 0 ``` ``` x = 3/ -2 ``` # Computing Determinants of 2x2 and 3x3 Matrices: ## 2x2 Matrices: ``` a11 a12 a13 A= a21 a22 a23 a31 a32 a33 ``` ## 3x3 Matrices: ``` |A| = (a11 a22 a33 + a12 a23 a31 + a13 a21 a32) - (a13 a22 a31 + a11 a23 a32 + a12 a21 a33) ``` **The first part of the equation is the sum of the products of the diagonals from left to right. The second part of the equation is the sum of the products of the diagonals from right to left.** # Cramer’s Rule **Cramers rule is a method for solving systems of linear equations.** * The coefficients of the equations will create our determinants. * We find the determinant by finding the determinant of the coefficients for the whole system (Δ). * We then find the determinant for each coefficient (Δx, Δy). * We can now solve for x and y using the formula ```x = Δx/ Δ``` and ```y = Δy/ Δ```. **The formula for solving the coefficients using determinants:** * We eliminate each variable by replacing its coefficient with the constant. **Example:** **Solve for x and y using the matrix below.** ``` x - y = 4 x + y = 6 ``` **First, find the determinant of the coefficients:** * The equations are set up as ``` Ax + By = C``` * The determinants are found as follows ``` Δ = 1 -1 | | 1 1| ``` **Find the determinant of the coefficients for x and y:** * We eliminate x from the first coefficient by replacing it with 4 and do the same with y for the second coefficient. These will be the determinants. * These are set up as follows: ``` Δx = | 4 -1 | | 6 1 | ``` ``` Δy = | 1 4 | | 1 6 | ``` **Solve for x and y using the formulas:** ``` x = Δx/ Δ = 10/2 = 5 y = Δy/ Δ = 2/2 = 1 ``` # lec. 2 – Matrices ## Matrices Matrices are scalar quantities arranged in a rectangular array containing *m* rows and *n* columns. ``` A = [a11 ..... ..... ain] [a21...... ..... a2n] [am1..... ..... amn] ``` This is called a *m* x *n* matrix, having *m* rows and *n* columns. ## Square Matrices Square matrices are matrices where the number of rows equals the number of columns. **Important Note:** * **AB ≠ BA.** * Square matrices can be solved using det(A) but non-square matrices **cannot** be solved using determinants. ## Operations with Matrices **1. Addition** > **Important Note:** Matrix Addition is possible when both matrices have the same number of rows and columns. **Example:** ``` -1 2 3 A= 4 0 5 2x3 1 2 3 ``` ``` 1 3 4 B = 0 2 6 2x3 1 0 2 ``` **Find A + B** ``` A+B = [-1 2 3] + [1 3 4] = [0 5 7 ] [ 4 0 5 ] + [0 2 6] = [4 2 11] [1 2 3 ] + [1 0 2] = [2 2 5] ``` **2. Subtraction** > **Important Note** Subtraction is the opposite of addition and is only possible with matrices with the same number of rows and columns. **3. Number of Matrices** > **Important Note** The number of matrices is equal to the number of elements in the matrix. **4. Multiplication** > **Important Note:** You must multiply the rows of the first matrix by the columns of the second matrix. You can only multiply matrices if they have the same number of columns in the first and rows in the second. * To do this, Multiply the first element of the row by the first element of the column. You then multiply the second element of the row by the second element of the column. Lastly, you add the two results together. > **Important Note:** The result will be a smaller matrix with the same number of rows as the first matrix and the same number of columns as the second. **Example:** ``` 2 1 A = 3 4 2x2 ``` ``` -1 2 B= 1 3 2x2 ``` **Find AB & BA** First row x first column ``` AB = [2 1] x [-1 2] = (2x-1)+(1x1) = -1 [3 4] x [-1 2] = (3x-1)+(4x1) = 1 ``` First row x second column ``` [2 1] x [1 3] = (2x2)+(-1x3) = 1 [3 4] x [1 3] = (3x2) + (3x4) = 18 ``` ``` AB = [-1 1] [1 18] ``` **Find BA** ``` BA = [-1 2] x [2 1] = (2+6)=8 [1 3] x [2 1] = (1+8) = 9 [-1 2] x [3 4] = (2+9)=11 [1 3] x [3 4] = (1+12) = 13 ``` ``` BA = [8 9] [11 13] ``` **Important Note: AB ≠ BA and multiplication of matrices is not commutative.** ## Matrix Transpose A transposed matrix changes the rows to columns and the columns to rows. > **Important Note:** (AT)T = A. **Example:** ``` [2 -1] A = 3 4 2x3 1 2 ``` ``` [2 3 1] AT= 1 4 3x2 -1 2 ``` ## Symmetric Matrix * A symmetric matrix is when the transpose of the matrix (AT) is equal to the original matrix (A). * If the matrix is not a square matrix, it cannot be symmetric. **Important Note:** (AT)T = A. **Example:** ``` [-1 3] A = 1 5 6 3x3 3 6 4 ``` ``` [-1 1 3] AT = 1 5 6 3x3 3 6 4 ``` **Therefore, the matrix is symmetric.** ## Skew Symmetric Matrix A skew symmetric matrix is when the transpose of the matrix (A) is equal to -1 times the original matrix (A). **Important Note:** (AT)T = A. **Example:** ``` [ 0 -1 2] A = [-1 0 -3] 3x3 [ 2 -3 0] ``` ``` [ 0 -1 2] AT = [-1 0 -3] 3x3 [ 2 -3 0] ``` **Therefore, the matrix is skew symmetric.** ## Inverse Matrix * The inverse of a 2x2 matrix is only possible if the determinant does not equal zero. * The inverse matrix is the matrix that produces the identity matrix when multiplied by the original matrix. **Important Note:** * 1 / any number = ∞(infinity) **To solve, you must find:** * The determinant (|A|) * The adjoint of the matrix adj (A) **The formula for finding the inverse of a matrix:** ``` A^-1 = 1/|A| adj(A) ``` **Steps to solve:** 1. Find the determinant, |A|. 2. Find the adjoint of the matrix, adj(A). 3. Divide adj(A) by the determinant, |A|. **Example:** **Find the inverse of the following matrix:** ``` [3 1] A= 1 5 ``` 1. **Find the determinant.** ``` |A| = 15 + 1 = 16 ``` 2. **Find the adjoint of the matrix.** ``` adj (A) = [5 -1] [-1 3] ``` 3. **Divide adj(A) by the determinant |A|.** ``` A^-1 = 1/16 * [5 -1] = [ -1 3] [5/16 -1/16] [-1/16 3/16] ``` # Lec. 3 - Partial Functions ## Partial Function Partial fractions are when you have multiple terms in the denominator of a fraction. > **Important Note:** It is much easier to work with these fractions if we separate them into multiple fractions with single terms in their respective denominators. **Example:** ``` (x - 1) + 2(x-1) (x-1)(x+1) ``` The goal is to rewrite this as ``` 3x-1 -------------- (x^2-4) ``` There are three cases to consider. **1. First Degree and different:** * A, B and C are constants. **Example:** ``` A ------- (x - a) ``` ``` B ------- + (x +b) ``` ``` C ------- (x - c) ``` **2. First Degree and similar:** * A and B are constants. **Example:** ``` A ------- + (x - a) ``` ``` B ------- (x - a)2 ``` > **Important Note:** The exponent on the denominator becomes a factor in the number of terms. **3. Mix of the previous two:** * A, B and C are constants. **Example:** ``` A ------- + x ``` ``` B ------- + (x - a) ``` ``` C ------ (x-a)2 ``` ## Resolve * The goal is to solve for the variables in the denominators above. * We multiply both sides by the common denominator. * We then solve for the constants A, B and C.. **Example:** * Resolve the following equation for the variables in the denominators. ``` x+1 ---------------- (x+1)(x - 2) ``` ``` x + 1 ------------ = (x+1)(x-2) ``` ``` A --- + (x+1) ``` ``` B ---. (x-2) ``` Multiply both sides by the common denominator. ``` x+1 = A(x-2) + B(x+1) ``` Let's solve for A. The easiest way to do that is to plug in values of x that will make the other denominator go to 0. >> If x = 1, the right-hand side of the equation simplifies to -A. >> If x = 2, the right-hand side of the equation simplifies to 3B. ``` If x=1, then 2 = -A, so A = -2 If x=2, then 3 = 3B, so B = 1 ``` * We can now substitute these values back into the original equation to find the final solution. **Final Solution:** ``` x+1 = (-2) / (x+1) + 1 / (x-2) ``` # lec. 4 - Complex Numbers ## Complex Numbers > **Important Note:** i = √-1 **Example:** ``` z = x+ iy ``` **where:** * z = complex number. * x = real part of the complex number (Re z). * y = imaginary part of the complex number (Im z). **Example:** ``` z = 3 + 4i ``` **where:** * Re z = 3 * Im z = 4 ## Operations with Complex Number **1. Addition** * To add two complex numbers, simply add the real parts and the imaginary parts separately. **Example:** ``` z1=3 + 2i z2 = 4 + 6i ``` **Find z1 + z2:** **Add the real parts together and separately add the imaginary parts together.** ``` z1+z2 = 7 + 8i ``` >**Important Note:** Re(z1 + z2) = 7 & im(z1 + z2) = 8. **2. Subtraction** * To subtract two complex numbers, simply subtract the real parts and the imaginary parts separately. **Example:** ``` z1=3 + 2i z2= 4 + 6i ``` **Find z1 - z2:** **Subtract the real parts together and separately subtract the imaginary parts together.** ``` z1 - z2 = - 1 - 4i ``` **3. Multiplication** * Multiply the two complex numbers together using the FOIL method (First, Outer, Inner, Last) or the distributive property. **Example:** ``` z1 = 3 + 2i z2 = 4 + 6i ``` **Find z1 * z2:** ``` (3 + 2i)(4 + 6i) = 12 + 18i + 8i + 12i^2 ``` >**Important Note:** Remember i^2 = -1. ``` = 12 + 18i + 8i - 12 = 26i ``` > **Important Note:** Re(z1 * z2) = 0 & Im(z1 * z2) = 26. **Therefore, the product of the two complex numbers is 26i.** **4. Division** * To divide two complex numbers, you must rationalize the denominator. * Multiply the numerator and denominator by the conjugate of the denominator. * The conjugate of z = x + iy is z = x - iy. **Example:** ``` z1=3 + 2i z2= 4 + 5i ``` **Find z1/z2:** ``` z1/z2 = (3 + 2i) / (4 + 5i) ``` * **Multiply the top and bottom by the conjugate of the denominator.** ``` (3 + 2i) / (4 + 5i) * (4 - 5i) / (4 - 5i) ``` > **Important Note:** The conjugate of 4 + 5i is 4 - 5i. * **Multiply the numerator and denominator using FOIL or the distributive property.** ``` (3+2i)(4-5i) / (4+5i)(4-5i) ``` ``` (12 -15i + 8i - 10i^2 ) / (16 - 20i + 20i - 25i^2) ``` > **Important Note** Remember: i^2 = -1 . ``` (12 - 7i + 10) / (16 + 25) = (22-7i)/41 ``` **Therefore, the quotient of the two complex numbers is (22 - 7i)/41.** ## Representing a Complex Number **1. Rectangular Form** ``` z = x + iy ``` **2. Polar Form** * The polar form of a complex number is represented by the magnitude and angle of the complex number, r and θ. * **Magnitude**: r = |z| = √(x^2 + y^2) * **Angle**: θ = arg(z) = tan^-1(y/x) **3. Conjugate** The conjugate of a complex number is the number with the opposite sign of the imaginary part. * z = x + iy * The conjugate is z = x - iy. **Example:** z = 1 + i√3 find |z| and arg(z) * **Magnitude:** ``` r = |z| = √(1² + (√3)^2 = 2 ``` * **Angle:** ``` θ = arg(z) = tan^-1(√3/1) = 60° ``` >* **Important Note:** The angle must be expressed in radians. ``` 60 * π/180 = π/3 ``` **The polar form of the complex number is 2(cos(π/3) + i sin(π/3)).** ## Polar Form (Geometric Form) **The polar form (geometric form) of a complex number is:** > **Important Note:** x = rcosθ & y = rsinθ **Example:** * z = 1 - i√3 is written in polar form as: **1. Magnitude:** ``` r = √(1^2 + (√3)^2 = 2 ``` **2. Angle:** * The angle is found by looking at the quadrant in which the complex numbers lies. * The complex number 1 - i√3 is in quadrant 4. * The angle in quadrant 4 is 360° - Θ. * The angle in radians is π - θ. **In this case, the angle is:** ``` tanθ = √3/1 = √3 θ = 60° ``` >* **Important Note:** We must find the angle in the 4th quadrant. ``` θ = 360 - 60 = 300° ``` * **Convert the angle into radians:** ``` θ = 300° * (π/180°) = 5π/3 ``` * **The polar form of the complex number is:** ``` z = 2(cos(5π/3) + i sin(5π/3) ) ``` ## The value of |z| The value of |z| is the magnitude of the complex number. **Example:** ``` |1 + i| ``` * **Multiply everything in the complex number by itself.** ``` (1 + i)(1 + i) = 1 + i + i + i^2 = 1 + i + i - 1 = 2i ``` * **The value of |1+i| is 2** # Lec. 5 - Derivative of Functions ## Derivative of Constants The derivative of a constant is always 0. * f(x) = C * f`(x) = 0 **Example:** * f(x) = 3 * f`(x) = 0 **The power rule:** f(x) = x^n f`(x) = nx^(n-1) **Example:** * f(x) = x^3 * f`(x) = 3x^2 **Derivative of a sum or difference of functions:** > **Important Note:** If u(x) and v(x) are differentiable functions, then the derivative of u(x) ± v(x) is u`(x) ± v`(x). **Example:** * f(x) = x^5 + 3x^2 - 4^3 * f`(x) = 5x^4 + 6x + 0 ## Derivative of Constant times Functions > **Important Note:** If f(X) = c * g(X) > where c is a constant and g(X) is a differentiable function then f`(x) = c * g`(x) **Example:** * f(x) = -3x^2 * f`(x) = -3[1x^2] = -3x - 2 ## Derivative of Products of Functions > **Important Note:** If f(x) = u(x) * v(x) where u(x) and v(x) are differentiable functions, then the derivative of u(x)v(x) is u`(x)v(x) + u(x)v`(x). **Example:** * f(x) = (x^3 + 1) * (x^2+2x) * f`(x) = (x^3 + 1) * (2x+2) + (x^2 + 2x) (3x^2) ## Derivative of a quotient of functions > **Important Note:** If f(x) = u(x)/v(x), where u(x) and v(x) are differentiable functions and v(x) is not equal to 0, then the derivative of [u(x) / v(x)] is [v(x) * u`(x) - u(x) * v`(x)] / [v(x)]^2 **Example:** * f(x)= (x^2 + 1)/(x^3 + 3) * f`(x) = [(x^3 + 3)(2x) - (x^2 + 1) (3x^2)] / (x^3 + 3)^2 ## Derivative of Power of a Function > **Important Note:** If f(x) = [u(x)] ^ n, where u(x) is a differentiable function, then the derivative of [u(x)]^n is n[u(x)]^(n-1) * u`(x). **Example:** * f(x) = (x^3 + 3x + 6)^10 * f`(x) = 10(x^3 + 3x + 6)^9 * (3x^2 + 3) * f(x) = √(x^2+ 4x + 3) >> **Important Note:** The derivative of √(x^2+ 4x + 3) is [1/ (2√(x^2 + 4x + 3)) ]* (2x + 4) **How to solve:** * The root is the same as taking the power of 1/2. * The derivative of a function that is under a root is equal to (1 / (2√(function))). **Therefore, to find the derivative of:** ``` * f(x) = √(x^2+ 4x + 3) ``` **Step 1:** * Rewrite the function as: ``` f(x) = (x^2 + 4x + 3) ^ 1/2 ``` **Step 2:** * Find the derivative of the inside of the function. ``` (x^2 + 4x + 3)’ = 2x +4 ``` **Step 3:** * Use the formula for the derivative of a power of a function. ``` f`(x) = 1/2 * (x^2 + 4x + 3)^ (1/2 - 1) * (2x + 4) ``` **Step 4:** * Simplify the expression. ``` f`(x) = (x^2 + 4x + 3)^-1/2 * (2x +4) ``` **Step 5:** * Rewrite the exponent so it is not negative. ``` f`(x) = 1/( x^2 + 4x +3 ) ^ 1/2 * (2x +4) ``` **Step 6:** * Rewrite the exponent so it is a root. ``` f`(x) = 1 / (√(x^2 + 4x +3) * (2x +4) ``` **Step 7:** * Finally, multiply the derivative of the function by the derivative of the inside of the function. ``` f`(x) = (2x + 4) / (√(x^2 + 4x +3) ``` ## Derivative of Trig Functions **1. Derivative of Sin** > **Important Note** The derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x). **Example:** * f(x) = sin(x) * f´(x) = cos(x) **Example:** * f(x) = x * sin(x) * f´(x) = x * cos(x) + sin(x) * 1 **2. Derivative of Cos** > **Important Note** The derivative of cos(x) is -sin(x). **Example:** * f(x) = cos(x) * f`(x) = -sin(x) **Example:** ``` f(x) = sin(x) / (1 + cos(x) ``` * **Find the derivative** ``` f`(x) = [ (1 + cos(x))( cos(x) ) - (sin(x))(-sin(x)) ] / (1 + cos(x))^2 ``` > **Important Note:** The derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x). **3. Derivative of tan** * **To get the derivative of tan(x), we must remember that tan(x) = sin(x)/cos(x).** * **The derivative of tan(x) = sec^2(x).** # Lec. 6 - Integration ## Integration * Integration is the reverse process of differentiation. > **Important Note:** If you are integrating a function, it is important to remember that the constant of integration (C) MUST be added to the end of the result. **The formula for integrating a power of x, where n is any real number that is not equal to -1 is:** ``` ∫ x^n * dx = x^(n+1)/(n+1) + C ``` **Example:** 1. ∫x^3 * dx = x^4/4 + C 2. ∫ x^1/2 * dx = x^(1/2 +1)/(1/2 + 1) = x^(3/2) / (3/2)= (2/3)x^(3/2) + C **Example:** * Integrate the following equation: ``` x^3 + 5 ``` * **Integrate each term individually.** ``` ∫ x^3dx + ∫ 5dx ``` * **Apply the rule for integration of a power function.** ``` x^4/4 + 5x + C ``` **How to solve integration using a constant:** > **Important Note:** If you are integrating a composite function, you must multiply the integral by the derivative of the internal function. **Example:** ``` ∫ (x^3 + 1)^10 + 3x^2 * dx ``` * The derivative of (x^3 + 1) is 3x^2. **Step 1**: * **Solve for the integral:** ``` ∫ (x^3 + 1)^10 + 3x^2 * dx = 1/11(x^3 + 1)^11 + C ``` * **Multiply by the derivative of the internal function.** ``` 1/11 (x^3 + 1)^11 + C * 3x^2 ``` **Step 2:** **Simplify the equation:** ``` 3/11(x^3 + 1)^11 + C ``` ## Different Rules of Integration * **1. Sum or difference of functions:** ``` ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx ``` **Example:** ``` ∫ (x^2 + 2x^2) dx ``` * **Solve for the integral separately** ``` ∫ x^2 * dx + ∫ 2x^2 * dx ``` * **Apply the formula for the integral of a power of x.** ``` x^3 / 3 + 2x^3 /3 + C ``` * **2. Constant Multiple Rule:** ``` ∫ [cf(x)] dx = c ∫ f(x) dx ``` **Example:** ``` ∫ (x^3 + 2x + 3) dx ``` * **Solve for each integral individually.** ``` ∫ x^3 * dx + ∫ 2x * dx + ∫ 3 * dx ``` **Apply the formula for the integral of a power of x** ``` x^4/ 4 + 2x^2/2 + 3x +C ``` * **3. Integral of a composite function:** ``` ∫ [f(x)]^n * f`(x) dx = [f(x)]^(n+1)/(n+1) + C ``` **Example:** ``` ∫ (x^3 +1)^10 * 3x^2 * dx 1/11 (x^3 + 1)^11 + C ``` >* **Important Note:** If the internal part of the composite function is raised to a power, then the derivative of that internal function is the factor you will multiply by in this step. * **4. Integral of a logarithmic function:** ``` ∫ (f(x)´ / f(x)) * dx = ln|f(x)| + C ``` **Example:** * **Integrate the equation below:** ``` ∫x/(x^2+1) dx ``` * **The denominator is the internal function.** ``` ∫ 1/(x^2 + 1) * 2x * dx ``` * **The derivative of the internal function is 2x, so we need to multiply the inside of the integral by 2.** * **However, this will change the value of the integral, so we must multiply the outside of the integral by the reciprocal.** * **We now have:** ``` 1/2 ∫ 1/(x^2 + 1) * 2x * dx ``` * **Now solve for the integral** ``` 1/2 * ln (x^2 + 1) + C ``` * **The constant of integration (C) is always added in the last step.** * **5. Integration of a constant:** ``` ∫ c * dx = cx + C ``` **Example:** * **Integrate the equation below:** ``` ∫ 2x / (x^2 + 1) * dx ``` * **The integral of 2x / (x^2 + 1) is ln(x^2+ 1) + C.** # Lec 7 - Exponential Model ## Exponential Model **The equation of the exponential model is:** ``` N = N_0 * e^(kt) ``` **where:** * **N:** is the number at time *t*. * **N_0:** the number at time *t* = 0. * **k:** is a constant related to the growth or decay rate. **Conditions:** * **Growth:** k > 0 * **Decay:** k < 0 **Example:** * **The number of bacteria is 1000 after two hours and becomes 2000. Find the relation between the number and time.** **Solution:** * **Find the equation of the exponential model.** ``` N = N_0 * e^(kt) ``` * **The bacteria is 1000 at t = 2 hours.** * **N is 2000 when t = 2.** * **We can substitute these values into the equation to solve for k.** ``` 2000 = 1000 * e^(2k) ``` * **Divide both sides by 1000.** ``` 2 = e^(2k) ``` * **Take the natural log of both sides.** ``` ln(2) = 2k ``` * **Solve for k.** ``` k = ln(2)/2 ``` * **The equation of the exponential model is now:** ``` N = 1000 * e^(ln(2)/2 * t) ``` * **Find the number of bacteria after 4 hours.** * **We substitute t = 4 into the equation.** * **Since we know k = (ln(2))/2, we substitute this value into the equation.** ``` n = 1000 * e^(ln(2)/2 * 4) = 1000 * e^(ln(2) * 2) = 1000 * (e^ln(2))^2 = 1000 * 2 ^