Local Maxima and Minima & Regression - PDF

# Local Maxima and Minima ## Introduction * A peak may not be the highest point on the graph, but it is the highest point in its neighbourhood. Similarly, a valley is the lowest point in its neighbourhood. * Informally, for a continuous function, we may think of a relative maximum as occurring on...

# Local Maxima and Minima ## Introduction * A peak may not be the highest point on the graph, but it is the highest point in its neighbourhood. Similarly, a valley is the lowest point in its neighbourhood. * Informally, for a continuous function, we may think of a relative maximum as occurring on a peak on the graph and a relative minimum as occurring on a valley on the graph. For the function, under the entire range, the maximum value of the function is known as absolute maxima and the minimum value is known as absolute minima. ## Turning Points or Stationary Points, Maxima and Minima * The function *f(x)* is said to have attained its maximum value for *x = a* if the function increases as *x* increases and begins to decrease at *x = a*. * The function is said to have attained its minimum value for *x = b* if the function increases as *x* decreases and begins to increase at *x = b*. * In Figure (ii), *P* and *R* are maximum points, and *Q* and *S* are minimum points. The maximum or minimum points are called the turning or stationary points. * *PM* and *RL* are maximum values of *f(x)* at *P* and *R* respectively. Similarly, *QN* and *ST* are minimum values of *f(x)* at *Q* and *S* separately. * According to the above notion of the concept of maximum and minimum values, it is clear that: * Local maxima and local minima * Maximum and minimum values do not mean the 'greatest and the least' value the function can have, but they only signify that the value considered is greater or less than the value on both sides of it in its immediate neighbourhood. ## Conditions for Maxima and Minima * At maximum points, the function *y = f(x)* changes from an increasing to a decreasing state; *dy/dx* changes from positive to negative. * In changing from positive to negative value, *dy/dx* must pass through the value zero. Hence, *dy/dx = 0* at a maximum point. * Similarly, at a minimum point, the function changes from a decreasing to an increasing state; *dy/dx* changes from negative to positive value. In changing from negative to positive value, *dy/dx* must pass through the value zero. Hence, *dy/dx = 0* at a minimum point. * If *dy/dx = 0*, it is not essential that the function may go on increasing or maximum and minimum, for it may happen that in spite of *dy/dx* being zero, the function may go on increasing or decreasing and it may not change from an increasing to a decreasing state or vice versa. This is an example of what is called a point of inflection. * All such points where *dy/dx = 0* are called Turning points or stationary points. * Now we state below the conditions for maximum and minimum points: * (i) *dy/dx = 0*; (ii) *dy/dx* changes sign from + to - for a maximum point. * (i) *dy/dx = 0*; (ii) *dy/dx* changes sign from - to + for a minimum point. ## Points of Inflection * In the graph of *y = f(x)* as in Figure (iii) and (iv), there is a point *B* common to the portions *AB* and *BC* where *A* is one line which is tangent to the curve *A* at *B* and also to the curve *B* at *C*. * In other words, tangent crosses the curve and is parallel to the *x*-axis. * The function is said to have a point of inflection. * The function is neither a maximum nor a minimum but is said to have a point of inflection. * To obtain conditions at a point of inflection, we notice that in Figure (iii) before *B*, *dy/dx* is as the function is increasing, at *B*, *dy/dx = 0*, for the tangent at *B* is parallel to the *x*-axis; after *B*, *dy/dx* is again + as the function continues to increase. * Thus, in the interval from *A* to *B*, *dy/dx* changes from +ve to *0* and is therefore a decreasing function. * Again in the interval from *B* to *C*, *dy/dx* changes from *0* to + and *dy/dx* is an increasing function. * *dy/dx* changes from - to + ... *d²y/dx² = 0* at *B*. * *d²y/dx²* is an increasing function of *x* ... *d³y/dx³ > 0* at *B*. * Similarly at Figure (iv), we get *dy/dx = 0* at *B* and *d²y/dx² ≤ 0* at *B*. * Hence, at a point of inflection, * (i) *dy/dx = 0*; (ii) *d²y/dx² = 0*; (iii) *d³y/dx³ ≠ 0*, i.e., it may be + or -. ## General Condition for Turning Points * If in a particular condition of *dy/dx=0* for a value of *α* obtained from *dy/dx=0* or *d²y/dx²=0*, then the function *y=f(x)* is a maximum or a minimum according as *d³y/dx³* is - or +. ## Working rule to find Local Maxima and Local Minima * **Let f(x) be the given function.** * **Method I: First Derivative Test** * (1) Find *f'(x)* and put it equal to zero. * (2) Solve *f'(x) = 0*. The root could possibly be the points of local maxima or local minima of *f*. * Suppose one of the roots is *'a'*. * (3) Determine the sign of *f'(x)* for values of *x* slightly less than *'a'*, for values slightly greater than *'a'*. * One of these cases may arise: * (i) If *f'(x)* changes sign from + to -, then *x = a* is a point of maxima. * (ii) If *f'(x)* changes sign from - to +, then *x = a* is a point of minima. * (iii) If *f'(x)* does not change sign as *x* increases through *a*, then *x = a* is neither a point of maxima nor minima, such points are points of inflection. * **Method 2: Second Order Derivative Test** * (1) Find *f'(x)* and put it equal to zero. * (2) Find the roots of *f'(x) = 0*. These could possibly be the points of local maxima or minima. Suppose one of the roots is *x = a*. * (3) Find the second derivative *f"(x)* by differentiating again of *f'(x)* w.r.t. *x*. * (4) Check the sign of *f"(x)* at each root value obtained in step 2. Suppose we do it at *x = a*. Then, * (i) If *f"(a) < 0*, then *'a'* is a point of local maxima. * (ii) If *f"(a) > 0*, then *'a'* is a point of local minima. * (iii) If *f"(a) = 0*, then the test fails. * (5) Now find the maximum or minimum value of *x* by substituting for *a* the values of *a* obtained in the given function. ## Application * In Physics * Maxima and minima are fundamental for analyzing potential energy surfaces, stable equilibrium points and interference patterns in optics. Engineers rely on these concepts to optimize designs, minimizing material usage while maximizing structural integrity or performance. In biology, maxima and minima come into play in studying population dynamics, enzyme kinetics, and ecological systems, helping determine sustainable population levels and optimal conditions for growth. * In Economics * Principles revolve around maximizing utility or profit while minimizing cost, with applications ranging from production and utility functions to cost analysis and resource allocation. Across these subjects, understanding and leveraging maxima and minima are essential for optimizing systems, making informed decisions and solving complex problems. ## Aim * Explain and illustrate with suitable examples, the concept of local maxima and local minima using a graph. ## Theory *(i) If *f'(x)* changes from positive to negative, then *x = a* is a point of maxima. *(ii) If *f'(x)* changes from negative to positive, then *x = a* is a point of minima. *(iii) If *f(x)* does not change sign as *x* increases through *a*, then *x = a* is neither a point of maxima nor a point of minima. ## Procedure * On receiving the question, analyze it. * Then, I referred to formulas and theories related to the topic. * Group discussions and references were taken from various sources. * Example questions were practiced. * Graphs were practiced several times. * A rough project was done, which was followed by the creating a fair project. ## Question * Find the local maxima and minima for the function *y = x³ - 3x + 2*. ## Solution * We will need to find the stationary points of this function for which we need to calculate *dy/dx*. * *y = x³ - 3x + 2* * *dy/dx = 3x² - 3* * At stationary points *dy/dx = 0*. Thus, we have: * *3x² - 3 = 0* * *3(x² - 1) = 0* * *(x - 1)(x + 1) = 0* * *x = 1, x = -1* * Now, we have to determine whether any of these stationary points are extremum points. We'll use the second derivative test: * *d²y/dx² = 6x* * *dy/dx = 3x² - 3, d²y/dx² = 6x* * For *x = 1*, *d²y/dx² = 6/1 = 6, which is positive; so it is a local minima*. * For *x = -1*, *d²y/dx² = 6/-1 = -6*, which is negative; so it is a local maxima*. ## Result * Therefore, local maximum occurs at (*x = -1*) and local minimum occurs at (*x = 1*). # Regression ## Introduction * Regression helps us to estimate or predict the value of one variable (dependent variable) from that of the other variable (independent variable). The statistical methods which help us to estimate or predict the unknown value of one variable from the known value of the related variable is called regression. * The word regression means the act of returning or going back. It was first used by Francis Galton towards the end of the 19th century while describing the relationship between the heights of fathers and heights of sons. He observed that while tall fathers have tall sons and short fathers have short sons, the average height of sons of tall fathers is less than that of the fathers, and the average height of sons of short fathers is greater than that of the fathers. This tendency to regress or back towards the average of the population was described by him as regression. The line, or go back towards, the average of the population, was described by him as regression. The line describing this tendency was called by Galton as a regression line. However, in the present days, we use the word regression line to denote that line which gives best possible estimates of the one variable for given values of other variable. This is why it is also called the line of best fit. * The line of regression of *y* on *x* has the form *y = mx + c*, and the line of regression of *x* on *y* has the form *x = my + c*. * Similarly, the quadratic curve regression of *y* on *x* is of the form *y = a + bx + cx²*. ## Method of Least Squares * Given the data (*x1, y1*), (*x2, y2*), . . . (*xn, yn*), we can plot a scatter diagram and then draw a smooth curve 'by eye' approximating the data. This is called a freehand method of curve fitting. This method has the disadvantage that it depends upon individual judgment, and different observers will obtain different curves and equations. * To motivate a possible definition, consider vertical distances *D1, D2, D3, . . . Dn* drawn from each point. Distances for *Dn* are taken from the points (*x1, y1*), (*x2, y2*), (*x3, y3*), . . . (*xn, yn*) to the curve. These distances will be positive or negative according to whether the points are above or below the curve so we work with the squares of these values and consider their sum *D1² + D2² + D3² . . . Dn²*. * A measure of the goodness of fit of the given data is provided by the above quantity. If this is small, the fit is good, if it is large, the fit is bad. We, therefore, make the following definition: * **Definition:** All curves approximating a given set of data points of a curve, having the property that *D1² + D2² + Dn²* is a minimum is called a best fitting curve. * A curve having this property is said to fit the data in the least square sense and is called a least square curve. This is also called a regression curve. Thus, a line having this property is called a least square line; a parabola with this property is called a least square parabola, etc. * **Note:** It is customary to employ the above definition when *x* is the independent variable and *y* is the dependent variable. If *x* is the dependent variable, the definition is modified by considering horizontal instead of vertical deviations, which amounts to an interchange of the *x* and *y* axis. These two definitions, in general, lead to different least square curves. Unless otherwise specified, we shall consider *y* as the dependent and *x* as the independent variable. ## Angle Between Two Lines of Regression * Consider *y - y = r * σy / σx * (x - x)* and *x - x = r * σx / σy * (y - y)*. * The slopes are *m1 = r * σy / σx, m2 = σy / r * σx* respectively. * Let *θ* be the angle between the two different regression lines, then: * *tan θ = ± (m1 - m2) / (1 + m1 * m2) = ± (r * σy / σx - σy / r * σx) / (1 + r * σy / σx * σy / r * σx) = ± (r² * σy² / σx² - σy² / σx²) / (r² * σy² / σx² + σy² / σx²) = ± (1 - r²) * σy² / σx² / (1 + r²) * σy² / σx² = ± (1 - r²) / (1 + r²) * σy² / σx²*. * Since *r² ≤ 1* and *σx, σy* are positive, therefore, +ve sign gives the acute angle between the lines. * Hence, *tan θ = (1 - r²) * σy² / σx² / (1 + r²) * σy² / σx²*. * **Case 1:** When *r = 0*, then *θ = π/2*. The two lines of regression are perpendicular to each other. Hence, the estimated value of *y* is the same for all values of *x* vice-versa. * **Case 2:** When *r = ±1*, *tan θ = 0*, so that *θ = 0* or *π*. Hence, the lines of regression coincide and there is perfect correlation between the two variates *x* and *y*. * Conversely, *tan θ = 0 => (1 - r²) * σy² / σx² = 0 => r² = 1*. Thus, there is only one regression line. ## Application * Regression analysis, a versatile statistical tool, finds application across various domains. Economically, it models intricate relationships, such as supply and demand dynamics. Finance employs regression to devise asset pricing models, manage risks, optimize portfolios and forecast financial outcomes. Meanwhile in marketing, it elucidates the effect of advertising, pricing strategies and promotional campaigns on consumer behavior and aiding in market segmentation. * In healthcare, regression predicts patients' treatment efficacy and outcomes, assesses disease progression. Finally, it delves into surveys to comprehend behavioral determinants and inter-relations between variables like income and education, fostering insights into human behavior. ## Aim * For the given data, find regression equations by the method of least squares. Also, find the angle between regression lines. ## Theory * *y = mx + c* and, * *x = my + c* * *Σy = n*c* + *mΣx* and, * *Σx = n*c* + *mΣy* * *Σxy = *c***Σx* + *mΣx²* and, * *Σxy = *c***Σy* + *mΣy²* (for regression *y* on *x*) * *(for regression *x* on *y*)* * *tan θ = (1 - r²) * σy² / σx² / (1 + r²) * σy² / σx²* * *m = (nΣxy - ΣyΣx) / [nΣx² - (Σx)²]., b = (Σy - mΣx) / n* ## Procedure * I analyzed the question on receiving it. * Theories and formulas related to the topic were referred. * My understanding of the topic was discussed with my friends for more clarification on the topic. * Few example questions were practiced to clarify any more doubts. * A rough solution for the question provided was made. * After conforming with various sources, I moved on with the main project. ## Solutions * **Find regression equation of *y* on *x*** * | *x* | *y* | |---|---| | 1 | 2 | | 2 | 5 | | 3 | 8 | | 4 | 7 | | 5 | 7 | * *Σx = 15* * *Σy = 25* * *Σx² = 55* * *Σy² = 88* * *Σxy = 88* * **Find value of *m* by using the formula** * *m = (nΣxy - ΣyΣx) / [nΣx² - (Σx)²]* * *m = [(5 * 88) - (15 * 25)] / [(5 * 55) - (15)²]* * *m = (440 - 375) / (275 - 225)* * *m = 65 / 50 = 13 / 10* * **Find the value of *c* using formula** * *b = (Σy - mΣx) / n = (25 - 1.3*15)/5* * *b = (25 - 19.5) / 5* * *b = 5.5 / 5* * *b = 1.1* * **Find the angle between regression lines, *x - 2y + 3 = 0, 4x - 5y + 1 = 0*** * **If angle between lines is *θ*, then *tan θ = ± (m2 - m1) / (1 + m1m2)* ** * **m1 = byx, m2 = 1 / byx** * *x - 2y + 3 = 0* * *4x - 5y + 1 = 0* * From (1): * *-2y = -x - 3 => y = (1/2)x + 3/2 => byx = 1/2* * *From (2):* * *-5y = -4x - 1 => x = (5/4)y - 1/4 => byx = 5/4* * *m1 = 1/2, m2 = 4/5* * *tan θ = ± (m2 - m1) / (1 + m1m2) = ± (4/5 - 1/2) / (1 + 1/2 * 4/5) = ± (8/10 - 5/10) / (1 + 2/5) = ± 3/10 / 7/5 = ± 3/14* * *θ = tan⁻¹(± 3/14)* ## Result * The regression equation is *y = 13/10x + 5.5*, and the angle between regression lines is *tan⁻¹(±3/14)*. ## Bibliography * ocw.mit.edu * www.brilliant.org * wikipedia * Selina publications