# Absolute Maximum And Minimum Calculator On Interval

20 at x = –0. > @ 2 f x x x 6 2 on 1,27 11. Occurence of absolute maxima: If f(x) is continuous in a closed interval I, then the absolute maximum of f(x) in I is the maximum value of f(x) on all local maxima and endpoints on I. These values correspond to the probability of observing such an extreme value by chance. If we ﬁnd all possible local extrema, then the global maximum, if it exists, must be the value of f(x) on the interval 1 ≤ x ≤ 4, and the maximum is f(4) = 9. Find more Mathematics widgets in Wolfram|Alpha. F INDING a maximum or a minimum has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. Sample size calculation for trials for superiority, non-inferiority, and equivalence. It is a measure of your capacity for aerobic work and can be a predictor of your potential as an endurance athlete. 80 at x = –1. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval. The absolute maximum value of f x x x( ) 3 12 M 32 on the closed interval > @M2, 4 occurs at x = A) 4 B) 2 C) 1 D) 0 E) M2 2. C) Absolute maximum only. For example, you could say,“The pulse rates are between 56 and 92 bpm. The closed interval method is a way to solve a problem within a specific interval of a function. Inequalities and ‘Maximum-Minimum’ Problems Henry Liu, 26 February 2007 There are many olympiad level problems in mathematics which belong to areas that are not covered well at all at schools. It is important to understand the difference between the two types of minimum/maximum (collectively called extrema) values for many of the applications in this chapter and so we use a variety of examples to help with this. Absolute zero is defined as the point where no more heat can be removed from a system, according to the absolute or thermodynamic temperature scale. Four Function and. Left bounds go on left side of min/max and right bounds go on right side. (3) Because the system has been carefully planned using modern top-down programming techniques, it is relatively easy to modify and extend. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. We take the derivative using the quotient rule: f0(x) =. The MAX function is a built-in function in Excel that is categorized as a Statistical Function. Relative Minimum - The lowest point on an interval of a curve. Once you've found the max-imum and minimum on this line (as well as on the other two lines that make up the boundary), compare all the values you've checked to ﬁnd out that the absolute maximum and the absolute minimum are f(1 4,2) = 673 64 (←absolute maximum) f(1 2, 4. An absolute maximum occurs at c if for all x in the domain of f. The absolute is measured in liters of oxygen per minute. f x x() 3 d. Meaning answers obtained by looking at the graph in the calculator will not earn V = - 4x +3; [1, 3] BIUA. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. Find the absolute maximum and the absolute minimum values of the function shown below, on the given interval. • Find the values of f at the endpoints of the interval. Absolute Maximum/Minimum Values of Multivariable Functions - Part 1 of 2 To find absolute max/min values of a continuous function g on a closed bounded set D: 1. The graph of y = cos x. Absolute Maximum and Absolute Minimum. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. Find the absolute maximum and absolute minimum values of f(x) = x2 −4 x2 +4 on the interval [−4,4]. Subtract 6 6 from 1 1. It is important to understand the difference between the two types of minimum/maximum (collectively called extrema) values for many of the applications in this chapter and so we use a variety of examples to help with this. Find the absolute maximum and absolute minimum values of f on the given interval. MIN([DISTINCT] expr) Minimum value returned by expr MOD(x,y) Remainder of x divided by y MONTHS_BETWEEN(end_date, start_date) Number of months between the 2 dates (integer) NEW_TIME(date, zone1, zone2) Convert between GMT and US time zones (but not CET) NEXT_DAY(date,day_of_week) '12-OCT-01','Monday' will return the next Mon after 12 Oct NLS. 4 is the lower limit. (d) Find the absolute minimum value of f x over the closed interval 5 ddx 5. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. For, sin (x + ) = cos x. denbal87 New member. Therefore, let's consider the function over the closed interval If the maximum value occurs at an interior point, then we have found the value in the open interval that maximizes the area of the garden. f(x)=x+ 9 x on [0. On the graph above of the function f on the closed interval [a, e], the point (a, f (a)) represents the absolute minimum, and the point (d, f (d)) represents the absolute maximum. Question 203087: Find the absolute maximum and absolute minimum values of the function below. An absolute minimum is the lowest y value or output value a. Thread starter denbal87; Start date Nov 4, 2014; D. The largest value found in steps 2 and 3 above will be the absolute maximum and the smallest value will be the absolute minimum. This lesson will focus on the maximum and minimum points. These points are sometimes referred to as max, min, extreme values, or extrema. (d) Find the absolute minimum value of f x over the closed interval 5 ddx 5. Let's find, for example, the absolute extrema of h (x)=2x^3+3x^2-12x h(x) = 2x3 +3x2 −12x. Sure, there are other, more precise, definitions, but that will work for what we want to do. 67, and a relative maximum of 4. Use the Stefan Boltzmann relationship (I = σT 4 where σ = 5. Q: Determine the absolute maximum and minimum values of the function on the given interval. summary_interval # How often to write checkpoints (rounds up to the nearest statistics # interval). I have an assignment where I have to write a program that accepts a series of integers from the keyboard from 1 to 100 using a single sentinel controlled loop, meaning I will need to also assign a sentinel number that when entered, will stop the loop and display the results. This is defined everywhere and is zero at $\ds x=\pm \sqrt{3}/3$. By using this website, you agree to our Cookie Policy. For each x value: Determine the value of f '(x) for values a little smaller and a little larger than the x value. The absolute max occurs at S = The absolute min occurs at S =. Decide whether you have a minimum or a maximum. )Given the function 𝑓(𝑥= 𝑥2+ 𝑥+ , chose values for a, b, and c in that could work for the graph shown. Which method do you prefer? f (x) = 1 + 3x^2 - 2x^3. the absolute (global) maximum 3. Meaning answers obtained by looking at the graph in the calculator will not earn V = - 4x +3; [1, 3] BIUA. VO2 max in men is approximately 40-60% higher in males than females. On the interval, fnmin then finds all local extrema of the function as left and right limits at a jump and as zeros of the function's first derivative. If you're behind a web filter, please make sure that the domains *. Question 203087: Find the absolute maximum and absolute minimum values of the function below. We take the derivative using the quotient rule: f0(x) =. The debt to equity ratio is a financial, liquidity ratio that compares a company’s total debt to total equity. To find the extreme values of a function (the highest or lowest points on the interval where the function is defined), first calculate the derivative of the function and make a study of sign. The absolute maximum on the interval is 138 at x=-2. For example, the following are all equivalent confidence intervals: 20. Find the absolute maximum and absolute minimum values of f on the given interval. Approximating Relative Extrema. Also the lowest value of either the X of the Y is placed first in the set. the absolute (global) maximum 3. Theorem (Extreme Value Theorem) If f is continuous on a closed interval [a,b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some. Example 1: Consider the three curves shown below. In worst case, if all intervals are from ‘min’ to ‘max’, then time complexity becomes O((max-min+1)*n) where n is number of intervals. Find the extreme values of f on the boundary of D. Occurence of absolute minima: If f(x) is continuous in a closed interval I, then the absolute minimum of f(x) in I is the minimum value of f(x) on all local. Three major examples are geometry, number theory, and functional equations. Go to window and set your X minimum to -1 and your X maximum to 5. Extreme Values of Functions Definitions An extreme value of a function is the largest or smallest value of the function in some interval. We usually distinguish between local and global (or absolute) extreme values. An extreme value, or extremum (plural extrema), is the smallest (minimum) or largest (maximum) value of a function, either in an arbitrarily small neighborhood of a point in the function's domain — in which case it is called a relative or local extremum — or on a given set contained in the domain (perhaps all of it) — in which case it is called an absolute or global extremum (the latter. Calculus I: Candidates Test for Global Extrema 1) If a continuous function f is defined on a finite, closed interval, such as −1≤x≤4 or [−1,4], or, more generally, a≤x≤b or [a,b], then f always has a global minimum value and a global maximum value on that interval. At t =0 the position of the object is 5. Examples: Input: v = {{1, 2}, {2, 4}, {3, 6}} first_page Given an absolute sorted array and a number K, find the pair whose sum is K. Use a graphing calculator to approximate the intervals where each function is increasing and Increase and Decrease Absolute minimum:. Answer: First, ﬁnd the critical points by ﬁnding where the derivative equals zero: f0(x) = (x2 +4)(2x)−(x2 −4. If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0. For a strictly unimodal function with an extremum inside the interval, it will find that extremum, while for an interval containing multiple extrema (possibly including the interval boundaries), it will converge to one of them. f(x) = x4 – A: Plot the graph for f(x) in the interval for [-2, 0]. Find the absolute maximum and absolute minimum values of f on the given interval. pow(x, 2) + math. Visual Magnitude Calculator: Computes the visual magnitude of a star from its absolute magnitude and distance. If f has a local maximum or minimum at c and f'(c. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. Extreme Values A Global Maximum A function f has a global (absolute) maximum at x =c if f (x)≤ f (c) for all x∈Df. Absolute Maximum - The highest point on a curve. Note: From our definition of absolute maxima and minima, if $(a, f(a))$ is an absolute max/min, then it is also a local max/min too. Let's Practice:. between -30 to 20 function is decreasing because there are no local minima and maxima in between them. f(x)= 490x x2 +49 on [0,10] 2 Fall 2016, Maya. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Here again we are giving definitions that appeal to your geometric intuition. f(x)=xln(2x)on[0. Local Extreme Values of a Function Let c be an interior point of the domain of the. The absolute is measured in liters of oxygen per minute. ) Find the absolute maximum and minimum values of the function on the given interval. Q: Determine the absolute maximum and minimum values of the function on the given interval. Extreme value theorem tells us that a continuous function must obtain absolute minimum and maximum values on a closed interval. VO2max stands for maximal oxygen uptake and refers to the amount of oxygen your body is capable of utilizing in one minute. so minimum value of f(x) in interval [0. For example, consider the functions shown in Figure(d), (e), and (f). Find more Mathematics widgets in Wolfram|Alpha. It's easy to ﬁnd one with neither absolute extrema. Identify the location and value of the absolute maximum and absolute minimum of a function over the domain of the function graphically or by using a graphing utility. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. Find the absolute maximum and absolute minimum values of f on the given interval. The maximum value of a function that has a derivative at all points in an For a function f(x) that has a derivative at every point in an interval [a, b], the maximum or minimum values can be found by using the following procedure: 1. By using this website, you agree to our Cookie Policy. Similar topics can also be found in the Calculus section of the site. So the absolute max value is 19 and the absolute min value is 1. The minimum and maximum describe the spread of the data. By using this website, you agree to our Cookie Policy. When calculating Intervals the X values are placed "on top of" the Y values. Enter DNE if the absolute maximum or minimum does not exist. Pick the largest and smallest. Find the absolute maximum and minimum values on theinterval: f(x) = x - 2cosx [-pi, pi] f ' (x) = 1+2sinx f (-pi)= -pi - 2cos(-pi) = -pi - 2. Thus if one has a sample {, …,}, and one picks another observation +, then this has / (+) probability of being the largest value seen so far. At t =0 the position of the object is 5. MIN([DISTINCT] expr) Minimum value returned by expr MOD(x,y) Remainder of x divided by y MONTHS_BETWEEN(end_date, start_date) Number of months between the 2 dates (integer) NEW_TIME(date, zone1, zone2) Convert between GMT and US time zones (but not CET) NEXT_DAY(date,day_of_week) '12-OCT-01','Monday' will return the next Mon after 12 Oct NLS. A higher debt to equity ratio indicates that more creditor financing (bank loans) is used than investor financing (shareholders). The range spread then uses the range to find a percentage that the maximum is greater than the minimum, using the minimum as a base. The absolute minimum on the interval is -237 at x=3. f x x( ) cos 2 b. Explain your reasoning. maximum" functions. Similarly, the global minimum is located at the lowest point. A relative minimum is a point that is lower than all the other points around it. The absolute max occurs at S = The absolute min occurs at S =. question_answer. Find the maximum and minimum values of the function f(x) = ln x/x on the interval [1, 3]. Hit graph and then hit 2nd Trace I think to bring up a menu that has minimum and maximum in there. Typical values for are 0. Absolute Maximum And Minimum Calculator. So, f(b) is a relative maximum of f. The maximum will occur at the highest value and the minimum will occur at the lowest value. If you're seeing this message, it means we're having trouble loading external resources on our website. To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. Find the the critical points of f on D. Fermat's Theorem. From this list of values we see that the absolute maximum is 8 and will occur at $$t = 2$$ and the absolute minimum is -3 which occurs at $$t = 1$$. You can take notes in the margins or on the flip-side of each sheet. Finding Extrema on a closed interval: 1. Subtract 6 6 from 1 1. In the single-variable case, it is known, by the Extreme Value Theorem, that if f is continuous on a closed interval [a;b], then it has has an absolute maximum and an absolute minimum on [a;b]. Find more Mathematics widgets in Wolfram|Alpha. Since the function is concave down at x=1 and has a critical point at x=1 (zero slope) then the function has a local maximum at x=1. find the absolute maximum and absolute minimum values of the fuction f(x)=2x-13ln(3x) on interval [1,8] 2. ^2 = 825 Thus function has absolute minimum value at x = 2 and absolute maximum value at x = 5 in the interval [-1, 5]. Wolfram alpha paved a completely new way to get knowledge and information. Enter the equation in the Y= section for Y1. minimum" or "4. f(x)=x^3-3x+1; [0,3]. Local Extreme Values of a Function Let c be an interior point of the domain of the. Let f be a function defined and. Endpoint Discontinuities: only one of the one-sided limits exists. f(x) = x + (4/x) on the interval [1,5]. A relative (or local) maximum occurs at c if for all x in an open interval containing c. 11) A) Absolute minimum only. But there is one very important condition that guarantees both an absolute minimum and an absolute maximum. If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0. Absolute Maximum and Absolute Minimum. So the absolute max value is 19 and the absolute min value is 1. org are unblocked. Absolute Maximum/Minimum Values of Multivariable Functions - Part 1 of 2 To find absolute max/min values of a continuous function g on a closed bounded set D: 1. Get the free "Function Extrema - Math 101" widget for your website, blog, Wordpress, Blogger, or iGoogle. so minimum value of f(x) in interval [0. Sometimes it's important to consider points which are only largest or smallest in small parts of a graph. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Thus, to find the absolute maximum (absolute minimum) value of the function, we choose the largest and smallest amongst the numbers f(a), f(c 1 ), f(c 2. have both an absolute maximum and an absolute minimum. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. Finding Extrema on a closed interval: 1. f(5) = (5)^4 + 8*(5)^3 -32*(5)^2 = 825 Thus function has absolute minimum value at x = 2 and absolute maximum value at x = 5 in the interval [-1, 5]. Local minima and maxima (First Derivative Test) by Paul Garrett is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4. In a blank cell, enter this formula =Max(ABS(A1:D10)), see screenshot: 2. The student familiar with the sum formula can easily prove that. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. False There is a local maximum at x = 0. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. Fold Unfold. Examples: Input: v = {{1, 2}, {2, 4}, {3, 6}} first_page Given an absolute sorted array and a number K, find the pair whose sum is K. This lesson will focus on the maximum and minimum points. Find more Mathematics widgets in Wolfram|Alpha. Relative Minimum - The lowest point on an interval of a curve. # The text file for eval. f(x) = @ 10. absolute minimum value at x = 2 is -48. The first derivative test: Let f (x) be a function and x = c a critical point of f. f(t) = 4t + 4 cot(t/2), [π/4, 7π/4] I'm stuck after I take the first derivative. Occurence of absolute minima: If f(x) is continuous in a closed interval I, then the absolute minimum of f(x) in I is the minimum value of f(x) on all local. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. Occurence of absolute maxima: If f(x) is continuous in a closed interval I, then the absolute maximum of f(x) in I is the maximum value of f(x) on all local maxima and endpoints on I. Mean number of days ≥ 30, 35 or 40 °C The average number of days in the period when the daily maximum air temperature was equal to, or exceeded 30, 35 or 40 °C. Look at the graph of f (x) = x 3 + 4x 2 - 12x over the interval [0, 3], Figure 1a. Free functions extreme points calculator - find functions extreme and saddle points step-by-step This website uses cookies to ensure you get the best experience. We take the derivative using the quotient rule: f0(x) =. A higher debt to equity ratio indicates that more creditor financing (bank loans) is used than investor financing (shareholders). f x x3 2 on 3,1 > @ 15. Question 203087: Find the absolute maximum and absolute minimum values of the function below. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. The minimum value for this range is the mean subtracted by the confidence interval and the maximum value is calculated by the mean added by the confidence interval. f(x) = x4 – A: Plot the graph for f(x) in the interval for [-2, 0]. f(x) = (x^2 - 1)^3, [-1, 5] 14. 2 Maximum and Minimum on an Interval. Use a graphing calculator to approximate the intervals where each function is increasing and Increase and Decrease Absolute minimum:. Find the absolute maximum and absolute minimum values of f on the given interval. In that case, the point right on the border might be the maximum or minimum of the curve. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. f(c) is another relative minimum of f. It could very well continue to increase or decrease once we leave the interval. Example: Find the absolute maximum and minimum of:. Note: From our definition of absolute maxima and minima, if $(a, f(a))$ is an absolute max/min, then it is also a local max/min too. Question 203087: Find the absolute maximum and absolute minimum values of the function below. Occurence of absolute maxima: If f(x) is continuous in a closed interval I, then the absolute maximum of f(x) in I is the maximum value of f(x) on all local maxima and endpoints on I. 1 absolute maximum or minimum. The maximum will occur at the highest f (x) f (x) value and the minimum will occur at the lowest f (x) f (x) value. Find the the critical points of f on D. We say that the function f(x) has a global maximum at x=x 0 on the interval I, if for all. Absolute Maximum and Absolute Minimum. In this case, “absolute extrema” is just a fancy way of saying the single highest point and single lowest point in the interval. D) Absolute minimum and. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. The restrictions stated or implied for such functions will determine the domain from which you must work. 587 Get more help from Chegg Get 1:1 help nowfrom expert CalculustutorsSolve itwith our calculusproblem solver and calculator. For example, consider the functions shown in Figure(d), (e), and (f). Compare the f (x) f ( x) values found for each value of x x in order to determine the absolute maximum and minimum over the given interval. Evaluate the function to find the y -values at all critical numbers and at each endpoint. (To make the distinction clear, sometimes the ‘plain’ maximum and minimum are called absolute maximum and minimum. Absolute minimum definition is - the smallest value that a mathematical function can have over its entire curve. f ( 1) = − 5 f ( 1) = - 5. Sal finds the absolute maximum value of f(x)=8ln(x)-x² over the interval [1,4]. earn credit. Python List max() Method - Python list method max returns the elements from the list with maximum value. True You are given that the function f ( x ) = 2 ( x + 3 ) x 2 + x − 2 has an absolute maximum on the interval − 2 < x < 1. Find the absolute maximum and absolute minimum values of f on the given interval. We usually distinguish between local and global (or absolute) extreme values. The smallest y -value is the absolute minimum and the largest y -value is the. On the interval, fnmin then finds all local extrema of the function as left and right limits at a jump and as zeros of the function's first derivative. The maximum will occur at the highest value and the minimum will occur at the lowest value. Roll your mouse over the Extreme Value Theorem to check your answers. A closed interval like [2, 5] includes the endpoints 2 and 5. For this data set, the minimum (lowest) value is 56 and the maximum (highest) value is 92. It could very well continue to increase or decrease once we leave the interval. Endpoint Discontinuities: only one of the one-sided limits exists. A point at which a function attains its minimum value among all points where it is defined is a global (or absolute) minimum. maximum: ( (62 -√3763)/9, (-461347 +7526√3763)/243)) ≈ (. If an absolute maximum or minimum does not exist, enter NONE. F INDING a maximum or a minimum has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. Every function that’s continuous on a closed interval has an absolute maximum value and an absolute minimum value (the absolute extrema) in that interval — in other words, a highest and lowest point — though there can be a tie for the highest or lowest value. In that case, the point right on the border might be the maximum or minimum of the curve. Absolute maximum is highest of and. f(x) = 4x^3 - 6x^2 - 144x + 9, [-4, 5] 13. Find the extreme values of f on the boundary of D. Before Using this Calculator. The following small array formulas can help you to find out the largest absolute value and the smallest absolute value. Explain the meaning of the result. Get the free "Max/Min Finder" widget for your website, blog, Wordpress, Blogger, or iGoogle. f(x) = x + (4/x) on the interval [1,5]. Learn more about population standard deviation, or explore other statistical calculators, as well as hundreds of other calculators addressing math, finance, health, fitness, and more. Calculates the root of the equation f(x)=0 from the given function f(x) and its derivative f'(x) using Newton method. There are two kinds of extrema (a word meaning maximum or minimum): global and local, sometimes referred to as "absolute" and "relative", respectively. Example: Find the absolute maximum and minimum of:. f(c) is another relative minimum of f. The student familiar with the sum formula can easily prove that. Return to Contents. It is a greatest value in a set of points but not highest when compared to all values in a set. ) Find the absolute max/min values of f(x) = x2 4 x2+4 on the interval [ 4;4]. We say that the function f(x) has a global maximum at x=x 0 on the interval I, if for all. Wolfram alpha paved a completely new way to get knowledge and information. exp(exp) * math. Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value. For, sin (x + ) = cos x. You can take notes in the margins or on the flip-side of each sheet. Maxima and minima are points where a function reaches a highest or lowest value, respectively. Find the local maximum and minimum values of using both the First and Second Derivative Tests. Generating random numbers Problem. Let's find, for example, the absolute extrema of h (x)=2x^3+3x^2-12x h(x) = 2x3 +3x2 −12x. Four Function and. (b) Use calculus to find the exact maximum and minimum values. This important theorem can guide our investigations when we search for absolute extreme values of a function. 4 is the lower limit. Mean number of days ≥ 30, 35 or 40 °C The average number of days in the period when the daily maximum air temperature was equal to, or exceeded 30, 35 or 40 °C. in some open interval containing c. The Organic Chemistry Tutor 200,049 views 1:10:05. Over the long term about one day in ten can be expected to have a (maximum or minimum) temperature exceed the decile 9 value. Sample size calculation for trials for superiority, non-inferiority, and equivalence. To define these terms more formally: a function f has an absolute maximum at x = b if f ( b )≥ f ( x ) for all x in the domain of f. Therefore, let’s consider the function over the closed interval If the maximum value occurs at an interior point, then we have found the value in the open interval that maximizes the area of the garden. You have to use a graphing calculator to find that out. Calculate Field honors the Transfer Field Domain Descriptions environment. Therefore, we are trying to determine the maximum value of A(x) for x over the open interval $$(0,50)$$. The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. Enter the equation in the Y= section for Y1. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Step 6: As mentioned earlier, A (x) is a continuous function over the closed, bounded. If f has a local maximum or minimum at c and f'(c. (c) For any. Some problems may have two or more constraint equations. On a closed interval these points are referred to as absolute or global minimum/maximum points. In worst case, if all intervals are from 'min' to 'max', then time complexity becomes O((max-min+1)*n) where n is number of intervals. The “V” = volume per time. Local Extreme Values of a Function Let c be an interior point of the domain of the. Also time complexity of above solution depends on lengths of intervals. Find the absolute maximum and the absolute minimum values of the function shown below, on the given interval. 2 Maximum and Minimum on an Interval. The golden-section search is a technique for finding an extremum (minimum or maximum) of a function inside a specified interval. Evaluate the function to find the y -values at all critical numbers and at each endpoint. The smallest y -value is the absolute minimum and the largest y -value is the. Hill Sphere Calculator: Computes the Hill Sphere of an object: Sail Calculator: Computes the maximum velocity possible from acceleration caused by light. The calculators will allow you to convert any heart rate between 63% and 102% of your maximum heart rate to a percentage of your VO2max , or any percentage of VO2max. Look at the graph of f (x) = x 3 + 4x 2 - 12x over the interval [0, 3], Figure 1a. save_path = FLAGS. Example 2: Locate the value(s) where the function attains an absolute maximum and the value(s) where the function attains an absolute minimum, if they exist. f(x)=x+ 9 x on [0. f(t) = 4t + 4 cot(t/2), [π/4, 7π/4] I'm stuck after I take the first derivative. Explain your reasoning. Sal finds the absolute maximum value of f(x)=8ln(x)-x² over the interval [1,4]. question_answer. Calculus I: Candidates Test for Global Extrema 1) If a continuous function f is defined on a finite, closed interval, such as −1≤x≤4 or [−1,4], or, more generally, a≤x≤b or [a,b], then f always has a global minimum value and a global maximum value on that interval. The debt to equity ratio is a financial, liquidity ratio that compares a company’s total debt to total equity. 47, an absolute minimum of –5. 2 Maximum and Minimum on an Interval. You should substitute those and pick the greatest for the maximum value. No Local Extrema Compare the f (x) f (x) values found for each value of x x in order to determine the absolute maximum and minimum over the given interval. Find the absolute maximum and absolute minimum values of f on the given interval. It also has its application to commercial problems, such as finding the least dimensions of a carton that is to contain a given volume. (1 point) Find the absolute maximum and absolute minimum values of the function f (x) = x3 + 6x2 — 63x + 8 over each of the indicated intervals. maximum" functions. f (c) is called the global (absolute) maximum value. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. This is achieved with a minimum of manual intervention. A relative minimum is a point that is lower than all the other points around it. The graph of y = cos x is the graph of y = sin x shifted, or translated, units to the left. Learn more about population standard deviation, or explore other statistical calculators, as well as hundreds of other calculators addressing math, finance, health, fitness, and more. f(x)= 490x x2 +49 on [0,10] 2 Fall 2016, Maya. An open interval like (2, 5) excludes the endpoints. X Values: [a,b] Y Values: / [c,d] Result: [e,f] [1,3] / [2,4] [. In this module you will be asked to calculate the sample size for 6 situations. Therefore, let's consider the function over the closed interval If the maximum value occurs at an interior point, then we have found the value in the open interval that maximizes the area of the garden. Local Extreme Values of a Function Let c be an interior point of the domain of the. (a) Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places. 20 at x = –0. By using this website, you agree to our Cookie Policy. Find the values of f f f at the critical numbers of f f f in (a, b). To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. Explain your reasoning. Significance Levels The significance level for a given hypothesis test is a value for which a P-value less than or equal to is considered statistically significant. The triangular distribution, along with the PERT distribution, is also widely used in project management (as an input into PERT and hence critical path method (CPM)) to model events which take place within an interval defined by a minimum and maximum value. The concern is that a dose given too soon after the previous dose may reduce the response. Pick the largest and smallest. Find the local or absolute minimum or maximum of an equation using a graphing calculator; Determine the intervals on which a function is increasing, decreasing, or constant using a graphing calculator (for precalculus) Determine an appropriate viewing rectangle for the graph of an equation; Match an equation to its graph; Graph an equation on. Finding the absolute max and min is a snap. (a) When is the object at rest? (b) Evaluate 6 1 ∫ vt dt(). An absolute minimum occurs at c if for all x in the domain of f. This is achieved with a minimum of manual intervention. Sample size calculation for trials for superiority, non-inferiority, and equivalence. Please answer the following questions about the function Instructions: If you are asked to. The range spread then uses the range to find a percentage that the maximum is greater than the minimum, using the minimum as a base. Final the absolute maximum and minimum values on the given interval. We usually distinguish between local and global (or absolute) extreme values. Find the extreme values of f on the boundary of D. 23] is -80 at x = 20 and maximum value of f(x) in interval [0,23] is at zero which is 0. Sometimes it's important to consider points which are only largest or smallest in small parts of a graph. Fold Unfold. Find the maximum / minimum absolute values with Formulas. f (1) = −5 f ( 1) = - 5. Here is the code to do that. Find the absolute maximum and absolute minimum values of f on the given interval. org are unblocked. Enter DNE if the absolute maximum or minimum does not exist. 388360 # Get 3 integers from 0 to 100 # Use max=101 because it will. Finding Extrema on a closed interval: 1. To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. You have 3 solutions: x=0, x=1, and this one. )Given the function 𝑓(𝑥= 𝑥2+ 𝑥+ , chose values for a, b, and c in that could work for the graph shown. If we break down the formula we can see why it gets its strange name. Before Using this Calculator. These points are sometimes referred to as max, min, extreme values, or extrema. In a blank cell, enter this formula =Max(ABS(A1:D10)), see screenshot: 2. Therefore, we are trying to determine the maximum value of A(x) for x over the open interval $$(0,50)$$. Answer: First, ﬁnd the critical points by ﬁnding where the derivative equals zero: f0(x) = (x2 +4)(2x)−(x2 −4. ) Find the absolute max/min values of f(x) = x2 4 x2+4 on the interval [ 4;4]. The generic word for minimum or maximum is extremum. An absolute minimum is the lowest y value or output value a. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. F INDING a maximum or a minimum has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. 80 at x = –1. ) Find the absolute max/min values of f(x) = x2 4 x2+4 on the interval [ 4;4]. Find the maximum / minimum absolute values with Formulas. a local (relative) maximum 6. Which method do you prefer? f (x) = 1 + 3x^2 - 2x^3. f(x) = x^2 + 250/x on the open interval (0,infinity ) I know that the absolute max is the answer NONE but I can not figure out the absolute min can someone help please thanks. The “V” = volume per time. # The text file for eval. Where does it flatten out? Where the slope is zero. There is 95% confidence that the constructed interval includes the population mean. In worst case, if all intervals are from 'min' to 'max', then time complexity becomes O((max-min+1)*n) where n is number of intervals. It then evaluates the function at these extrema and at the endpoints of the interval, and determines the minimum over all these values. So, f(b) is a relative maximum of f. Recommended and Minimum Ages and Intervals Between Doses of Routinely Recommended Vaccines1,2,3,4 Vaccine and dose number minimum interval between doses is equal to the greatest interval of any of the individual components. For a strictly unimodal function with an extremum inside the interval, it will find that extremum, while for an interval containing multiple extrema (possibly including the interval boundaries), it will converge to one of them. So the absolute max value is 19 and the absolute min value is 1. The range spread then uses the range to find a percentage that the maximum is greater than the minimum, using the minimum as a base. Absolute Maximum and Absolute Minimum This page is intended to be a part of the Real Analysis section of Math Online. We do not know that a function necessarily has a maximum value over an open interval. In this module you will be asked to calculate the sample size for 6 situations. These values correspond to the probability of observing such an extreme value by chance. Solution for Find the absolute maximum and minimum of the function f (x) = x³ - x+2 on the interval [0, 3] %D. Find the absolute maximum and absolute minimum values of f on the given interval. If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0. For instance, in the example at. f (c) is called the global (absolute) maximum value. Thread starter denbal87; Start date Nov 4, 2014; D. If you finish a job in less than 25% of the time allotted, you will be paid a Time Bonus, so try to finish as quickly as possible! The maximum Time Bonus is a 25% boost to your Base Reward. Calculate Field honors the Transfer Field Domain Descriptions environment. Explain your reasoning. Enter the equation in the Y= section for Y1. In the single-variable case, it is known, by the Extreme Value Theorem, that if f is continuous on a closed interval [a;b], then it has has an absolute maximum and an absolute minimum on [a;b]. ) On the other hand, it is possible to see directly that. How to calculate a confidence interval? First, you need to calculate the mean of your sample set. An absolute minimum occurs at c if for all x in the domain of f. An absolute maximum is the highest y value or output value a graph has over a specific interval. 140 of 155. The golden-section search is a technique for finding an extremum (minimum or maximum) of a function inside a specified interval. Let this index be 'max_index', return max_index + min. Infinite Discontinuities: both one-sided limits are infinite. This lesson will focus on the maximum and minimum points. Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value. (1 point) Let g(s) = i on the interval [0, 1. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval. (𝑓𝑥)=5 Two solutions. Pick the largest and smallest. This corresponds to zero Kelvin, or minus 273. We know that the absolute max/min values of f(x) will occur either at an endpoint or a critical number. summary_interval # How often to write checkpoints (rounds up to the nearest statistics # interval). Calculate the range for your confidence statistics. Audio dithering. This is defined everywhere and is zero at $\ds x=\pm \sqrt{3}/3$. Looking first at $\ds x=\sqrt{3}/3$, we see that $\ds f(\sqrt{3}/3)=-2\sqrt{3}/9$. The maximum will occur at the highest f (x) f ( x) value and the minimum will occur at the lowest f (x) f ( x) value. The sample maximum and minimum provide a non-parametric prediction interval: in a sample from a population, or more generally an exchangeable sequence of random variables, each observation is equally likely to be the maximum or minimum. Sometimes it's important to consider points which are only largest or smallest in small parts of a graph. f x x3 2 on 3,1 > @ 15. find the absolute maximum and absolute minimum values of the fuction f(x)=2x-13ln(3x) on interval [1,8] 2. Time needed: 10 minutes. The calculators will allow you to convert any heart rate between 63% and 102% of your maximum heart rate to a percentage of your VO2max , or any percentage of VO2max. X Values: [a,b] Y Values: / [c,d] Result: [e,f] [1,3] / [2,4] [. On a closed interval these points are referred to as absolute or global minimum/maximum points. f(5) = (5)^4 + 8*(5)^3 -32*(5)^2 = 825 Thus function has absolute minimum value at x = 2 and absolute maximum value at x = 5 in the interval [-1, 5]. By practicing these kinds of problems you can understand this topic clearly. Thus if one has a sample {, …,}, and one picks another observation +, then this has / (+) probability of being the largest value seen so far. Calculus Refresher by Paul Garrett. Relative Minimum - The lowest point on an interval of a curve. It is a measure of your capacity for aerobic work and can be a predictor of your potential as an endurance athlete. The maximum acceleration attained on the interval 03ddt by the particle whose velocity is given by v t t t t( ) 3 12 4 M 32 is A)9 B)12 C)14 D)21 E)40 3. f(x) = x^2 + 250/x on the open interval (0,infinity ) I know that the absolute max is the answer NONE but I can not figure out the absolute min can someone help please thanks. Local Extreme Values of a Function Let c be an interior point of the domain of the. An extreme value, or extremum (plural extrema), is the smallest (minimum) or largest (maximum) value of a function, either in an arbitrarily small neighborhood of a point in the function's domain — in which case it is called a relative or local extremum — or on a given set contained in the domain (perhaps all of it) — in which case it is called an absolute or global extremum (the latter. To find the extreme values of a function (the highest or lowest points on the interval where the function is defined), first calculate the derivative of the function and make a study of sign. It can be used as a worksheet function (WS) in Excel. A continuous function f(x) on a closed and bounded interval [a,b] has both an absolute min-imum and an absolute maximum on the interval. Locate the maximum or minimum points by using the TI-83 calculator under and the “3. (c) Find all intervals on which the graph of f is concave up and also has positive slope. We know that the absolute max/min values of f(x) will occur either at an endpoint or a critical number. Q: Determine the absolute maximum and minimum values of the function on the given interval. Find the absolute maximum and minimum values on theinterval: f(x) = x - 2cosx [-pi, pi] f ' (x) = 1+2sinx f (-pi)= -pi - 2cos(-pi) = -pi - 2. 5] The calculator performs interval arithmetic operations and computers interval version of mathematical functions. Finding Minimums and Maximums. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval. By using this website, you agree to our Cookie Policy. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. a local (relative) minimum 5. To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. absolute minimum value at x = 2 is -48. We know that the absolute max/min values of f(x) will occur either at an endpoint or a critical number. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. Example 2: Locate the value(s) where the function attains an absolute maximum and the value(s) where the function attains an absolute minimum, if they exist. We usually distinguish between local and global (or absolute) extreme values. An extreme value, or extremum (plural extrema), is the smallest (minimum) or largest (maximum) value of a function, either in an arbitrarily small neighborhood of a point in the function's domain — in which case it is called a relative or local extremum — or on a given set contained in the domain (perhaps all of it) — in which case it is called an absolute or global extremum (the latter. f(x) = 4x^3 – 6x^2 – 144x + 9, [-4, 5] 13. The student familiar with the sum formula can easily prove that. (c) For any. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval. These are the examples in the topic increasing and decreasing intervals. Maximum intervals — they don't exist. Let f be a function defined and. It is a greatest value in a set of points but not highest when compared to all values in a set. Sample size calculation for trials for superiority, non-inferiority, and equivalence. Local Extreme Values of a Function Let c be an interior point of the domain of the. Thus, to find the absolute maximum (absolute minimum) value of the function, we choose the largest and smallest amongst the numbers f(a), f(c 1 ), f(c 2. Absolute minimum/maximum _____ d. Before going to class, some students have found it helpful to print out Purplemath's math lesson for that day's topic. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval. For each x value: Determine the value of f '(x) for values a little smaller and a little larger than the x value. f(x) = x4 – A: Plot the graph for f(x) in the interval for [-2, 0]. If you finish a job in less than 25% of the time allotted, you will be paid a Time Bonus, so try to finish as quickly as possible! The maximum Time Bonus is a 25% boost to your Base Reward. Generating random numbers Problem. Use the Stefan Boltzmann relationship (I = σT 4 where σ = 5. minimum" or "4. Find the maximum / minimum absolute values with Formulas. in some open interval containing c. The local maximum and minimum are the lowest values of a function given a certain range. A closed interval like [2, 5] includes the endpoints 2 and 5. B) No absolute extrema. (To make the distinction clear, sometimes the ‘plain’ maximum and minimum are called absolute maximum and minimum. Step 6: As mentioned earlier, A (x) is a continuous function over the closed, bounded. Finding the absolute max and min is a snap. Let this index be ‘max_index’, return max_index + min. checkpoint_interval # Where to write out summaries. It is important to understand the difference between the two types of minimum/maximum (collectively called extrema) values for many of the applications in this chapter and so we use a variety of examples to help with this. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Maxima and minima are points where a function reaches a highest or lowest value, respectively. Let Purplemath help you always be prepared! Go to the lessons!. sin(x * y) I have an interval for x [-1, 1] and y [-1, 1]. 487] Calculating confidence intervals: Calculating a confidence interval involves determining the sample mean, X̄, and the population standard deviation, σ, if possible. summary_interval = FLAGS. The maximum will occur at the highest value and the minimum will occur at the lowest value. A point at which a function attains its maximum value among all points where it is defined is called a global (or absolute) maximum. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. f x x x 32 3 on 3,1> @ 12. Similar topics can also be found in the Calculus section of the site. between -30 to 20 function is decreasing because there are no local minima and maxima in between them. Let this index be ‘max_index’, return max_index + min. Therefore, f achieves its absolute minimum of −14 at x = −1 and its absolute maximum of 6 at both x = 1 and x = 4. 80 at x = –1. The absolute maximum value of f x x x( ) 3 12 M 32 on the closed interval > @M2, 4 occurs at x = A) 4 B) 2 C) 1 D) 0 E) M2 2. Mean number of days ≥ 30, 35 or 40 °C The average number of days in the period when the daily maximum air temperature was equal to, or exceeded 30, 35 or 40 °C. A point at which a function attains its minimum value among all points where it is defined is a global (or absolute) minimum. 14 If the first dose of recombinant zoster vaccine (Shingrix) is administered to someone 18-49 years of age, the dose does not need to be repeated. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. 1 absolute maximum or minimum. f (1) = −5 f ( 1) = - 5. Question 203087: Find the absolute maximum and absolute minimum values of the function below. In this module you will be asked to calculate the sample size for 6 situations. summary_interval # How often to write checkpoints (rounds up to the nearest statistics # interval). 2 Maximum and Minimum on an Interval. Theorem (Extreme Value Theorem) If f is continuous on a closed interval [a,b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some. The absolute minimum of the function f(x) = x2-9 on the interval - 4 Sxs 3 has a value of (Simplify your answer. In the single-variable case, it is known, by the Extreme Value Theorem, that if f is continuous on a closed interval [a;b], then it has has an absolute maximum and an absolute minimum on [a;b]. Any global maximum or minimum must of course be a local maximum or minimum. (1 point) Let g(s) = i on the interval [0, 1. Absolute & Local Minimum and Maximum Values - Relative Extrema, Critical Numbers / Points Calculus - Duration: 1:10:05. The student familiar with the sum formula can easily prove that. so minimum value of f(x) in interval [0. pow(y, 2)) * -1 return math. Extreme values ©2010 Iulia & Teodoru Gugoiu - Page 1 of 2 3. For instance, in the example at. How to use absolute minimum in a sentence. the absolute minimum age of 50 years when evaluating records retrospectively. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. (a) Interval : [—8, 0]. The smallest y -value is the absolute minimum and the largest y -value is the. 1 absolute maximum or minimum. Fermat's Theorem. Generating random numbers Problem. Help with finding absolute max/min values for a function. in some open interval containing c. Local minima and maxima (First Derivative Test) by Paul Garrett is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4. Let f be a function defined and. $f(x) = x^5 - x^3 + 2$, $-1 \leqslant x \leqslant 1$. 1 absolute maximum or minimum. The absolute max occurs at S = The absolute min occurs at S =. (a) fxc 0 at x 3, 1, 4 f c changes from positive to negative at 3 and 4. For each x value: Determine the value of f '(x) for values a little smaller and a little larger than the x value. between -30 to 20 function is decreasing because there are no local minima and maxima in between them. have both an absolute maximum and an absolute minimum. Find the extreme values of f on the boundary of D. An absolute extremum is an absolute maximum or an absoute minimum, and absolute extrema are absolute maximum and absolute minimum. Zoom in on the interval [-2,2] using the x-axis. In worst case, if all intervals are from 'min' to 'max', then time complexity becomes O((max-min+1)*n) where n is number of intervals. The MAX function is a built-in function in Excel that is categorized as a Statistical Function. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. minimum” or “4. Find the absolute maximum and absolute minimum values of f on the given interval. Find the absolute maximum and minimum values on theinterval: f(x) = x - 2cosx [-pi, pi] f ' (x) = 1+2sinx f (-pi)= -pi - 2cos(-pi) = -pi - 2. The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. Find the absolute maximum and. By using this website, you agree to our Cookie Policy. denbal87 New member.


# How to round 0.4999… ? Is it 0 or 1?

If you want to round the repeating decimal 0.4999... to a whole number what is the right answer? Is it 0 or 1?

On the one hand you only look at the first digit when you round numbers which in this case is 4 so the answer should be 0.

On the other hand 0.4999... is only another representation for 0.5 which makes the result 1.

My question
Which of these rules wins? (My gut feeling is that the most consistent result should be 1)

Edit
Just for clarification: What is meant by rounding here is replacing a fractional decimal number by one with fewer digits with the "Round half up"-Tie-breaking method.

-
It appears that the rule is inconsistent. –  Isaac Solomon Oct 21 '12 at 9:04
Doesn't that depend on what type of rounding you are using? If you're using Bankers Rounding, you would round towards 0. –  Fake Name Oct 21 '12 at 12:22
@FakeName: Fair enough. What is obviously meant here is replacing a fractional decimal number by one with fewer digits with the "Round half up"-Tie-breaking method: en.wikipedia.org/wiki/Rounding –  vonjd Oct 21 '12 at 13:50
This question should NOT be closed; it has a definitive answer - with references "out there" to support it. At the point when this was closed, the issue on the table (discussion!) was if 0.499... was 0.5 or not: reference is en.wikipedia.org/wiki/0.999... –  Richard Sitze Oct 22 '12 at 0:08
I agree with Richard. Given that we have a method for rounding midpoint values, and the author of the question has supplied a method, it follows that this question has an answer. –  Doug Spoonwood Oct 22 '12 at 2:52

It's $1$, because $0.49\ldots$ is the same as $0.5$. If rounding is to be well-defined, it can't map one real number to two integers, so whatever it maps $0.49\ldots$ to, it better maps it to the same integer as $0.5$. You could round both to $0$, of course, but that wouldn't then be the way we usually round.

What this shows you is that rounding doesn't commute with limits, i.e. there's a difference between find the limit of a sequence and then rounding, and rounding first and then finding the limit. As you correctly observed, all values $a_n = 0.4\underbrace{9\ldots 9}_{n\text{ times}}$ are rounded down to zero. Thus, $$\lim_{n\to\infty} \text{round }(a_n) = 0$$ On the other hand, $\lim_{n\to\infty} a_n = 0.5$, and thus $$\text{round }\left(\lim_{n\to\infty} a_n\right) = 1$$

There's another word for functions which don't commute with limit - they're called non-continuous. So what you have discovered is simply that rounding is not a continuous function.

-
Perhaps I'm nitpicking, but is the result of rounding the limit perhaps $1$ instead? –  pimvdb Oct 21 '12 at 10:05
@pimvdb: I went ahead and edited the answer, since it's obviously what was intended. (And so many people seem to agree.) –  ShreevatsaR Oct 21 '12 at 14:30
+1 especially for pointing out the connection with continuity. –  Neal Oct 21 '12 at 16:00
@pimvdb Indeed the limit should be $1$, and I wouldn't call pointing out blatant errors nitpicking ;-) Thanks for noticing, and thanks to ShreevatsaR for fixing this! –  fgp Oct 21 '12 at 19:42
@jmoreno - but it is, in fact the same as 0.5: en.wikipedia.org/wiki/0.999... –  Richard Sitze Oct 22 '12 at 0:01

For $0\le x\le 1$, we round $x$ to $1$ if $x\ge \frac12$ and to $0$ if $x<\frac12$ (though there are many conventions, see e.g. Wikipedia on rounding; the section "Table-maker's dilemma" a bit further down may also be interesting). Since $0.4\bar9=\frac12$, we should round to $1$. Another way of looking at this is that we always consider only the standard decimal expansion (i.e. we prefer $\bar0$ over $\bar 9$), and we are allowed to treat the first decimal as $4$ only if we know that it cannot turn out as $5$ "later". Thus if an inexact measurement gives us that $0.495\le x\le0.5$, we cannot say definitely, what $\operatorname{round}(x)$ should be (we could if the measurement resulted in $0.495<x<0.5$). This is not different from the fact that we cannot say definitely what $\operatorname{round}(x)$ should be if our measuremen merely says that $0.4997<x<0.5003$.

-

You're right. Since $.4\bar{9}=.5,$ if you want your rounding function to be well-defined you'll have to require an exception: round based on the first digit after the one you're rounding to, unless it's a $4$ followed by infinitely many $9$s.

-

It is not obvious that 0.5 should be rounded to 1. Obviously 0.49999... should be treated the same because it is the same number. Wikipedia rounding article

-

$1 - 0.5 =0.5 \quad$ and $\quad 1 - 0.4999\ldots = 0.500\ldots = 0.5 \quad$ so $\quad 0.5 = 0.4\overline9$

Well, the above line says the proof / intuition. From our information... $0.5$ well rounds to $1$.

-
Are you sure that $1-0.4999\ldots\ne 0.500\ldots1$? –  MJD Nov 18 '12 at 16:46
@MJD: Not sure if serious... –  Clive Newstead Nov 18 '12 at 16:47
@MJD: In 0.500...1, the zeroes never end, so the $1$ never comes ;) –  Parth Kohli Nov 18 '12 at 16:52

I find very convenient to define rounding as follows: $$round(x) = \left \lfloor x + \frac{1}{2}\right \rfloor$$

Considering this rule, what follows is understanding the context. If you are working with a computer, you have only a fixed number of decimal positions to work with, in which case, if x=0.4999999999999999 (16 decimal positions), then

round(x)=floor(0.4999999999999999 + 0.5)=
=floor(0.9999999999999999)=
=0

Of course, if you are talking about a real number, then 0.4999... equals 0.5, in which case $round(x)=1$.

So, the real question here is: are you working with a fixed number of decimal positions or not? If the answer is 'yes', then definitely round(x)=0. If the answer is 'no', then you are talking about a number which is equal to 0.5, and $round(x)=1$

-

@fgp answered this well but there is another aspect that I don't think has been touched on in any of the answers. What is the type of thing that we should consider as an input to the process of rounding? If it is a real number, we are unable to apply the rule without a specific digit sequence representing that number. If instead we interpret rounding as an operation on a possibly infinite digit sequence, we can now apply the rule, but its result has no meaning for general real numbers. One way out which has been discussed is to make a bijection between digit sequences and reals by disallowing digit sequences ending in $999...$, and this view effectively unasks the question by taking the position that $0.4999...$ is not a valid representation of a real number, but I think this is an overly burdensome technicality.

I think an easier way to look at it is just to say that rounding is an operation that applies to finite digit sequences and not to real numbers in general nor to infinite digit sequences. When we say something like "$\pi$ rounds down to $3$" it is a lazy way of saying that all sufficiently precise approximations do. But when we say "$0.5$ rounds up to $1$", exactly that is meant, because we don't mean to suggest that $0.4 + \sum_{k=2}^{n}{9 \cdot 10^{-k}}$ will round up to $1$ for sufficiently large $n$. And without context, if it is said about a real number $x$ that "$x$ rounds up to $1$" then that can be read as "$x$ has a finite decimal representation which rounds up to $1$, or $x$ has no finite decimal representation and all sufficiently precise finite decimal approximations to $x$ round up to $1$". In this way we can meaningfully discuss rounding real numbers in terms of rounding finite digit sequences, and without explicitly addressing the problem of sequences ending in $999...$.

-

If you consistently use the round-up half method you don't always look at the first digit of the numeral when rounding numbers. You go by the method. The method allows that if we have anything else than a half-way value, then you can round by the rule "if (working in base 10) the first digit of the numeral after the decimal point belongs to {0, 1, 2, 3, 4}, then round down, if the first digit of the decimal point belongs to {5, 6, 7, 8, 9}, then round up." But, that rule does not say anything about half-way values. Going by q=FLOOR(y + .5) we can reason as follows: q=FLOOR(.499... + 5)=FLOOR(.999...)=FLOOR(1)=1, since .999...=1. So, that method yields that .499... rounds to 1.

-

It is the same as 0.5 by definition of decimal numbering.

-
And rounding takes by definition the first digit which is 4 in this case. So this is exactly where the two definitions collide! –  vonjd Oct 21 '12 at 18:36
@vonjd I disagree that rounding by definition takes the first digit and then rounds. I ask you to try and round 1/3 without converting it into a decimal. Do you not round 1/3 down to 0, since it comes as clear that 1/3 lies closer to 0 than to 1? Or I suggest you round 12/11 to the nearest whole number. Or round 155/66 to the nearest whole number. Finding the multiples of a denominator d will help here, and what half of d equals. If the denominator equals 7, then if the numerator of "the remainder" equals 1, 2, or 3 we round down, if the numerator equals 4, 5, or 6 we round up. –  Doug Spoonwood Oct 21 '12 at 19:17
E. G. Say we have 13/7. 13/7=7/7+6/7, correct? So, 13/7 has "remainder" of 6/7, or in other words it has a fractional part (search that term on Wikipedia) of 6/7. Thus, 13/7 rounds up to 14/7 (6/7 lies closer to 7/7 than to 0/7). You could also try and rounding numbers in binary, trinary, or any n-ary base system. E. G. round 1.34523 in base 7. In base 7, 1.4142 rounds up to 2. So, I simply do not see how the digit makes the difference here. More goes on than that! –  Doug Spoonwood Oct 21 '12 at 19:23
@vonjd this is an exception. rounding takes first digit EXCEPT for values like this, where fraction is written in limitless form. –  Suzan Cioc Oct 21 '12 at 19:25

This is just an obersvation that I was missing in the many answers already given.

While there is absolutely no doubt that the mathematical function $\Bbb R\to\Bbb Z$ defined by $x\mapsto\lfloor x+\frac12\rfloor$ when applied to the real (in fact rational) number represented by the repetitive infinite decimal sequence $0.4\overline9$ takes as value the integer$~1$, saying that this is what happens when rounding the repeating decimal $0.4999\ldots$ is misleading. That is because "rounding" suggests an operation done during actual computation. This ignores the often overlooked fact that it is impossible to do arithmetic computations with infinite decimals. This does not mean of course that there isn't a class of infinite decimals with which computations are possible; for instance if all decimals are repeating, then one can replace them by rational numbers and compute with those, which is definitely possible. However the misconception is that, because we can do arithmetic on numbers with any fixed number of decimal places and the procedures for doing so are always basically the same, therefore one can apply these same procedures to infinite decimals. However just looking at these procedures will reveal that this is plainly false: all of these procedures need at some point (often right at the beginning) to locate the final decimal position, and for an infinite decimal such a position simply does not exist. This means that in practice whenever one needs rounding decimal numbers as a computational operation, then those decimals are necessarily finite, and the number $0.4999\ldots$ simply cannot occur.

In fact there is no effective system for computing with (general) true real numbers at all, for the simple reason that the majority of the real numbers cannot be represented in any way (a concrete representation having to be, by the nature of things in the real world, finite). One can however restrict to the (countable) subset of real numbers that do allow some kind of representation (which is less brutal that restricting to some arbitrarily chosen finite subset of numbers declared representable, as is done in numeric computation), and then arithmetic and other kinds of computation do become possible, and indeed can be done without having to introduce approximations. This has been studied in constructive mathematics, but it turns out that for this purpose representing real numbers by infinite decimals, which might be produced on demand by a finite algorithm, is not a viable option. Even just finding the first decimal of the sum of two numbers so given by infinite decimals cannot be assured by an algorithm; this is in fact closely related to issues invoked in this question and the related $0.9999\ldots$ problematic. This in contrast to computations with formal power series with integer (or rational) coefficients: these suffer from the same "most elements cannot be represented" problem as the real numbers, but if one sticks to series whose coefficients can be computed by an algorithm, then all arithmetic (and many other) operations on them can be performed algorithmically without problem.

A more workable way to represent real numbers is by a sequence of definite approximations (rational numbers with a given maximal error), with some kind of guarantee that the error tends to $0$. But even then there are sacrifices one has to make: it is impossible to computations with discontinuous functions (like the floor or round-to-integer operations): in the constructive theory of real numbers there only exist continuous functions. So all of this is rather far removed from what one does in numeric computation, where real numbers are necessarily approximated, the number of decimals finite, and the question of how to round $0.4\overline9$ moot; the rule for rounding of looking at the first digit to be discarded in the rounding is perfectly valid in practice.

-

If we round c to d, then c approximately equals d. If d also consists of an integer, then c lies closer to d than any other integer i. In other words for all integers i such that i does not equal d, ABS(ci-)>ABS(cd-), where ABS(ci-) indicates the number by which c and i differ by (the absolute value of the difference of c an i). This implies that we can use the distance metric between real numbers to see how rounding works as follows.

Consider the set S=={[l, lu+2/), (lu+2/, u]} where ul-=1. In other words, l and u differ by 1, and we consider the set of all numbers between x and y except for the midpoint m==lu+2/ between them. We can use the metric function D(x,y)=|xy-|=ABS(xy-), where x belongs to S and y belongs to {l, u}, to determine how to round all members of S by following rule:

If D(x, l)>D(x, u), then round to u. If D(x, u)>D(x, l), then round to l.

Since .5=.49999..., it follows that D(.49999..., 1)=D(.49999...., 0)=.4999...=.5. Thus, the distance function D(x, y)=ABS(xy-) does not give us any guide as to how to round here.

Consequently, we have several different choices that we can make which will not go against our intuition of rounding as follows:

1. Choose to round .49999... to 1 (maybe you value larger numbers in general),
2. Choose to round .49999... to 0 (maybe you value smaller numbers in general),
3. Choose to allow .49999... to get rounded to both as one pleases or sees fit at particular points in time, or
4. Choose to round .49999... to neither 0 nor 1 (maybe you don't want to make a choice here).

None of those possibilities win or lose until we have a standard to measure a winner or loser for the rules. There also doesn't exist a "right answer" here unless we have a means to determine what properties a "right answer" has in this context. I suspect that many people would resist 3. with the possibility that one could round .4999... to 1 at one time, and then round .4999... to 0 at another as one pleases, thus indicating arbitrariness. But, it does not follow that such comes as an improper procedure unless one first rates such "arbitrariness" as bad, and something so utterly awful and terrible that we or others must never, ever engage in such "arbitrariness". Perhaps many people would see 4. as "evasive", but unless there exists a logical basis to picking either 0 or 1, anyone accused of "evasiveness" by picking 4. could very well respond to such a charge, that anyone who sees choosing 4. as indicating "evasiveness", insists dogmatically that all such questions absolutely must have answers, and/or that the people claiming "evasiveness" have copped out of making choices on the basis of logic as much as possible.

So, what sort of choice do you want to make?

-

As a programmer, 0.4999... is closer to 0 than it is to 1, so I would always round down to 0.

Keep it simple! Rounding is always to the closest whole number. You only need to apply special cases when there are two closest whole numbers.

But there are situations where this would be the wrong choice, so you need to do some research every time you apply rounding function to a value.

For example, the tax department in the country I live in does not always acknowledge cents. Under legislation, there are times when the only valid rounding function is floor.

-
I think we all agree that $0.4999$ should round down to $0$. The question is whether $.499999999......$ should (with infinitely many $9$s). –  Jason DeVito Oct 21 '12 at 18:16
How close $0.4999...$ is to $0$ or $1$ is independent on whether or not you are a programmer. In fact, it is the same distance from $0$ and $1$. If the sequence of digits were to terminate after finitely many (e.g. in computing) then fine, but here we have infinite precision and the number is exactly equal to $0.5$. –  Clive Newstead Oct 21 '12 at 18:20
@AbhiBeckert: The case in point is that it is indeed equal to 0.5, as is 0.999... equal to 1: en.wikipedia.org/wiki/0.999... –  vonjd Oct 21 '12 at 18:32
Dear Abhi, I have downvoted your answer, since it it seems to be premised on a confusion with regard to the fact that $0.4999\cdots$ (infinitely many $9$'s) is equal to $0.5$. Regards, –  Matt E Oct 21 '12 at 19:56
@AbhiBeckert: If you accept that $0.4999...$ is a real number, and you also claim it's less than $0.5$, then (1) you should be able to say what is the distance between them (for any two real numbers $x<y$, their distance is another real number $y-x$: and no, "infinitesimal" is not a number), and (2) you should be able to exhibit a number that lies between them (for any $x<y$, there exist infinitely many real numbers between, e.g. $x<(x+y)/2<y$). The fact that there is no number between $0.4999...$ and $0.5$ should convince you that they are two representations of exactly the same number. –  ShreevatsaR Oct 23 '12 at 5:16


# How to translate mathematical intuition into a rigorous proof?

I've seen a lot of questions about how to develop mathematical intuition, but often I have the opposite problem.

Several times I have run into a situation where I want to solve a math problem, and I play around with it until my intuition reaches the point where I have a good idea of how the complete proof might be structured. But then when it comes to actually write out the proof, I have trouble translating this intuition into a rigorous proof. I want to give the following example, from this PDF of Putnam training problems.

1.13. Prove that for every $n\ge 2$, the expansion of $(1+x+x^2)^n$ contains at least one even coefficient.

When first thinking about this problem, I wanted to look at the polynomials $\pmod{2}$ so that "even coefficient" is simplified to "zero coefficient."

Then from doing a few computations, I notice a pattern.

\begin{align*} (1+x+x^2)^2&=(1+x+x^2)+x(1+x+x^2)+x^2(1+x+x^2)\\ &=1+x+x^2\\ &+x+x^2+x^3\\ &+x^2+x^3+x^4\\ &=1+x^2+x^4 \end{align*}

I notice that multiplying a polynomial by $(1+x+x^2)$ is like adding that polynomial with itself three times, each shifted over. The $x$ shifts it over by one place and the $x^2$ shifts it over by two places.

Continuing this pattern, we can get the coefficients of $(1+x+x^2)^3$ as follows:

\begin{array}{ccccccc} 1&0&1&0&1&&\\ &1&0&1&0&1&\\ &&1&0&1&0&1\\ \hline 1&1&0&1&0&1&1 \end{array}

So $(1+x+x^2)^3\equiv 1+x+x^3+x^5+x^6 \pmod{2}$

Let's make a triangle of a few more results:

\begin{array}{cccccccccccc} 0:&&&&&&1&&&&&\\ 1:&&&&&1&1&1&&&&\\ 2:&&&&1&0&1&0&1&&&\\ 3:&&&1&1&0&1&0&1&1&&\\ 4:&&1&0&0&0&1&0&0&0&1&\\ 5:&1&1&1&0&1&1&1&0&1&1&1\\ \end{array}

We see that the structure of our problem is essentially the same as an elementary cellular automaton (specifically, rule 150). I will come back to this later.

I notice another peculiar pattern with the $1$st, $2$nd, and $4$th rows: There are only $1$s on the ends and in the center. Perhaps this pattern continues for all powers of $2$. And it does, which is not hard to prove with induction.

Claim: $\forall n\ge 0,\ (1+x+x^2)^{2^n}\equiv 1+x^{2^n}+x^{2^{n+1}}\pmod{2}$

Base case: $(1+x+x^2)^{2^0}\equiv 1+x+x^2\equiv 1+x^{2^0}+x^{2^1}\pmod{2}$

Inductive Step:

\begin{align*} (1+x+x^2)^{2^n}&\equiv ((1+x+x^2)^{2^{n-1}})^2\\ &\equiv (1+x^{2^{n-1}}+x^{2^n})^2\\ &\equiv 1+x^{2^n}+x^{2^{n+1}}\pmod{2} \end{align*}

This is where the intuition comes in that is hard for me to express rigorously. On one of the $2^n$th rows, look at the $0$ equidistant from the leftmost $1$ and the $1$ in the center. If the rightmost $1$ wasn't there, then that $0$ would stay a $0$ forever because of symmetry: any effect from the left is cancelled by an effect from the right.

However, the rightmost $1$ does exist, so it will eventually change that $0$. But since effects are local in this automaton, it will require over $2^n$ more rows until the rightmost $1$ affects the $0$ we are looking at (there are over $2^n$ spaces between them). In other words, we have shown that the $0$ will stay a $0$ for all rows up to the $2^{n+1}$th row. Applying this logic on each power of two row, we show that there is always a zero coefficient from row $2$ onwards.

Is there a more rigorous way I can express those last two paragraphs? And in general, what advice do you have for translating intuition to a rigorous proof?

• In your example you changed the question. Originally it asked for a $0$ in every row. Your version only talks about rows of the form $2^n$, but you show a very special form for those rows. Your induction is fine and proves what you wanted to prove, but it is not the original question. It might be a step along the way I can imagine (but I am speculating) that you can show a partial row starting with $1$, a bunch of $0$s and a $1$ will always have a $0$ somewhere. If you can do that you are done in conjunction with what you have done. – Ross Millikan Jun 25 '18 at 4:15
• @RossMillikan I did not change the question in any way. The observation I made about the $2^n$ rows was a stepping stone in my proof. Look at the last three paragraphs. – Riley Jun 25 '18 at 4:18
• You said you would come back to rule 150 but I don't think you did – Mark Jun 25 '18 at 4:49
• @Mark I meant that I will come back to framing the problem in terms of cellular automata, which I did when discussing the "effects" of the $1$s, and the locality of the automaton. – Riley Jun 25 '18 at 4:51
• [Just a comment as the question is about the automaton-intuition being turned into a proof.] Another way of thinking about this expression is, the coefficient of $x^m$ in the expansion shows the number of ways to achieve a total of $m$ by rolling a die with sides $0, 1, 2$ a total of $n$ times. To see why, try making a three by three times table model to see the result of squaring the expression; then, erase the base of $x$. – Benjamin Dickman Jun 25 '18 at 5:17

You have proved that these binary strings are symmetric so it suffices to prove the result for only the "half-strings." Imagine cutting your triangle down the middle. Let $x_n$ be the $n^{th}$ bit string, which is length $n+1$.

We have $x_0 = 1$, $x_1 = 11$, $x_2 = 101$, etc.

Now you can address the $j^{th}$ bit inside $x_n$ as $x_{n,j}$.

This addressing system will make it easier to talk more concretely about the action of the automaton.

You can pin down a definition of "locality of action" by saying that there is some function $f(a, b, c)$ such that $\forall n, j: x_{n,j} = f(x_{n-1, j-1}, x_{n-1,j}, x_{n-1,j+1})$ (with some additional nuisance conditions about the boundary). This means that the $j^{th}$ bit in the current string can be computed by neary-by bits in the previous string. Here $f$ represents the action of this rule 150 you mention.

Now use $f$ again to show $x_{n,j} = f(f(x_{n-2, j-2}, x_{n-2,j-1}, x_{n-2, j}), f(x_{n-2, j-1}, x_{n-2,j}, x_{n-2, j+1}), f(x_{n-2, j}, x_{n-2,j+1}, x_{n-2, j+2}))$

Now you can see when we repeat this procedure $k$ times we will end up with a trinary abstract syntax tree depth $k$. To avoid having to talk about the exact structure of the expression at the $k^{th}$ level, we can reason about this substitution process purely formally as a process operating on the (countably infinite) set of symbols $\{ \underline{f}, \underline{(}, \underline{)} \} \cup \{ \underline{x_{i,j}} \}_{(i,j) \in \mathbb{N}^2}$ (here the commas are not meant to be part of the symbol set, they are just there to separate symbols from each other). You can prove by induction, that for any $k$, that $x_{n,j}$ is computed by an expression using the symbols $\{ \underline{f}, \underline{(}, \underline{)} \} \cup \{\underline{ x_{n-k, j-k}},\underline{ x_{n-k, j-k+1}}, …,\underline{ x_{n-k, j+k-1}},\underline{ x_{n-k, j+k} }\}$ (the $2k+1$ bits in the $(n-k)^{th}$ string centered on the $j^{th}$ bit) intermixed with function applications of $f$.

Fix $n$ and let $k_n$ be the distance to the previous power of $2$. (So if $n=13$, $k_n=5$ since the previous power of $2$ is $8$). And define $j_n \equiv (n-k_n)/2$. This is the index of the central bit of the $(n-k_n)^{th}$ bit string.

You have demonstrated that those bits at indices $\{ (n-k_n, m) \}_{1 \le m \lt n }$ are all zero since $n-k_n$ is a power of $2$. So any well-formed expression using only the symbols $\{ \underline {f}, \underline{(}, \underline{)}\} \cup \{ \underline{x_{n-k_n, j_n-k_n}}, \underline{x_{n-k_n, j_n-k_n+1}}, …,\underline{x_{n-k_n, j_n+k_n-1}}, \underline{x_{n-k_n, j_n+k_n}}\}$ will evaluate to zero by $f(0,0,0) = 0$, and in particular the expression representing $x_{n,j_n}$ (created from the $k_n$ iterations of the formal symbol substitution procedure) evaluates to $0$.

So as $n$ increments, $x_{n,j_n}$ traces out the path of the central bits you were referring to, all of which are zero.

General thoughts about formalization: There are certain technical devices which are work horses of the formalization process. Two of them I have used in this proof:

• indexing time and space, with a numeric coordinate system
• encoding a time evolving system as function application

A third device which I used is not as common, and comes from computer science and logic:

Something that I have been realizing is that these devices are not obvious (at least to me). Humanity was around for a long time before the Cartesian coordinate system was discovered and used to formalize geometrical intuition. It took even longer to develop the idea that almost every mathematical object could be encoded as an intricate tower of sets (ZFC).

I guess for me it's a matter of building up a library of these devices and binding them to the correct intuitions by repeated use.

In certain areas like computer programming, there are very few tools for converting intuition into proof. There were a couple of very good devices created like

• loop invariants
• Hoare logic

but multithreaded and heap-allocating programs are still an active area of research.

• By the way, I have not verified the claim about the sequence actually evolving as the cellular automata $f$ but I'm trusting it's true. – Mark Jun 25 '18 at 7:01
• I think your definition of $j$ doesn't quite match the way you use it. You're using it as if $j$ represents the $x$ coordinate in the triangle rather than the $j$th bit. Also I think you need to consider whole strings instead of half strings for the automation to work. And I think you need to define $x_{n,j}=0$ when $j$ is out of bounds. Otherwise a great answer. – Riley Jun 25 '18 at 12:50
• Yeah, for it to make sense, we need to start counting things at $0$ as we do in computer programs. $x_{n,j} \cong x[n][j]$. As for the whole strings vs half strings, maybe it will be convenient to use negative indices to talk about the left half of the triangle. – Mark Jun 25 '18 at 14:48

Here is one answer to the question. If $$\, n\ge 2\,$$ then the first $$4$$ entries in each row, namely, $$\, 1 0 1 0, 1 1 0 1, 1 0 0 0, 1 1 1 0, \dots, \,$$ repeat in a period of $$4$$. There is a $$0$$ in each of these first $$4$$ entries. This was not the proof you wanted because it is too simple.

Your proof is essentially the following. You noticed something about a $$0$$ in row $$\, 2^n \,$$ and that this $$0$$ continues up to the row $$\, 2^{n+1}. \,$$ We formalize that observation by working with polynomials over the field with two elements. Define the Laurent polynomials $$\, y(n) := x^n + x^{-n} \,$$ and since the characteristic of the field is $$2$$, $$\, y(0) = 1 + 1 = 0. \,$$ Using simple algebra of exponents we get that $$\, y(n)y(m) = y(n+m) + y(n-m), \,$$ and the special cases $$\, y(n)^2 = y(2n) \,$$ and $$\, y(2^n) = y(1)^{2^n}.$$

Define the row polynomials $$\, r(m) := (1 + y(1))^m. \,$$ The general row is $$\, r(m) = r(2^n+k) \,$$ where $$\, 0\le k < 2^n. \,$$ Notice that $$\, r(2^n) = 1 + y(1)^{2^n} = 1 + y(2^n). \,$$ Also $$\, r(m) = 1 + \sum_{i=1}^m c_i y(i), \,$$ where $$\, c_i \,$$ is the coefficient of $$\, y(i). \,$$ You noticed that $$\, c_{2^{n-1}} = 0 \,$$ but how to prove it? We calculate that $$\, r(m) = (1 \!+\! y(2^n)) r(k) = (1 \!+\! y(2^n))(1 \!+\! cy(2^{n-1}) \!+\! t) = (1 \!+\! y(2^n))(1 \!+\! t) \!+\! c y(3\,2^{n-1}) \,$$ where $$\, t \,$$ is all the terms with $$\, y(i), \,$$ and $$\, i\ne 2^{n-1}, \,i < 2^n. \,$$ But $$\, y(2^n)y(i) = y(2^n-i) \!+\! y(2^n+i) \,$$ and since $$\, i \ne 2^{n-1} \,$$ we get no $$\, y(2^{n-1}) \,$$ terms in the product $$\, (1 \!+\! y(2^n))(1 \!+\! t). \,$$ That proves it.

This answers your particular question, but what about intuition versus rigor? I think there are no good answer to that question. You have to always look for any patterns or regularities and simple or edge cases. You have to be lucky and it helps to have a lot of background experience to draw upon.

• I recognize that there is a simpler answer to the question. It is given in the PDF I liked to. I am not asking for the solution, I am asking about how to translate my ideas into a rigorous proof. – Riley Jun 25 '18 at 4:06
• I thought I was translating your ideas combined with ideas of my own into a proof. But intuitions are tricky to express clearly. Maybe you wanted an "automaton theoretic" proof? – Somos Dec 3 '18 at 19:55


# Equivalence of inner products?

So, let $\langle \cdot, \cdot \rangle_{1}$ and $\langle \cdot, \cdot \rangle_{2}$ be inner products on a real vector space $V$. Assume that \begin{align} \langle v, v \rangle_{1} = \langle v, v \rangle_{2} \end{align} for any $v \in V$.

Is it true that $\langle v, w \rangle_{1} = \langle v, w \rangle_{2}$ for all $v,w \in V$?

I can only think of two inner products on a real vector space $V$ are the standard dot product and the scaled dot product. This holds for these two since the scaled dot product would require all scalars to be 1. Can anyone think of another inner product on $V$ to test this with, or should I start trying to prove it?

• Take a basis $v_1,\dots ,v_n$ for $V$. Then $\langle v_i,v_i\rangle_1=\langle v_i,v_i\rangle_2$ for all $1\leq i\leq n$ implies that $\langle v,w\rangle_1=\langle v,w\rangle_2$.
– DKal
Mar 17 '14 at 21:09

For every $v,w \in V$ and $c$ real $$\langle v+cw,v+cw\rangle_1=\langle v+cw,v+cw\rangle_2$$ or $$\langle v,v\rangle_1+2c\langle v,w\rangle_1+c^2\langle w,w\rangle_1= \langle v,v\rangle_2+2c\langle v,w\rangle_2+c^2\langle w,w\rangle_2$$ and hence, as $\langle v,v\rangle_1=\langle v,v\rangle_2$ and $\langle w,w\rangle_1=\langle w,w\rangle_2$, then $$\langle v,w\rangle_1=\langle v,w\rangle_2.$$

Even though Yiorgos S. Smyrlis' answer is simpler, I think it is useful having a little different proof, just to add another point of view of the same argument, which is always good :)

Consider the norm $\| \cdot \|_1$ induced by $\langle \cdot, \cdot \rangle_1$ and the norm $\| \cdot \|_2$ induced by $\langle \cdot , \cdot \rangle_2$. By hypothesis and definition of induced norm $$\| v \|_1^2 = \langle v, v \rangle_1= \langle v, v \rangle_2= \| v \|_2^2$$ $\forall v \in V$. So the two norms are the same (the power doesn't make any problem thanks to the positivity of the norm.)

Then we want to do the reverse reasoning, if two product-induced-norms coincide on every element of the space, what can we say about inner products?

It's immediate (and left as little exercise for the reader - just write down the definitions and use bilinearity-) to prove that given a real inner product and its norm the following equivalence is true $$\langle x,y \rangle = \dfrac{1}{4} \left( \|x+y \|^2 - \|x-y\|^2 \right)$$ This formula is called the polarization identity (for more see here)

So by the above identity and the hypothesis we have, $$\langle x,y \rangle_1 = \dfrac{1}{4} \left( \|x+y \|_1^2 - \|x-y\|_1^2 \right) = \dfrac{1}{4} \left( \|x+y \|_2^2 - \|x-y\|_2^2 \right) = \langle x,y \rangle_2$$ $\forall x,y \in V$ and so we are done.

I wanted to stress the depth correlation between an inner product and its induced norm, and show some useful tools. Hope it helps :)