If f(x) is positive for all positive values of X and...

Key Concepts

Monotonic Decreasing Function

A monotonic decreasing function is one in which each subsequent value is less than or equal to the preceding one. This property is critical when comparing discrete sums to continuous integrals because it provides a natural ordering that allows for establishing inequalities—specifically, it helps in determining whether a sum over integer values forms an upper or lower bound of an integral over an interval.

Convex Function

A convex function, characterized by a positive second derivative, curves upward, meaning that the slope of the function is increasing. This property is important for error estimation when approximating sums with integrals, as it guarantees that the linear approximations (or tangents) under- or overestimate the function in a controlled manner, which leads to tighter integral bounds for the sum.

Integral Approximation of Sums

The technique of approximating sums by integrals involves using the area under the curve of a function to estimate the sum of its values at discrete points. For monotonic functions, particularly, one can use the integral test to bound the sum by comparing it with the corresponding definite integral over a continuous interval, thereby providing a framework to derive inequalities linking sums and integrals.

Riemann Sums and Error Analysis

Riemann sums, which represent the approximation of an integral via finite sums of function values times interval widths, play a crucial role in the analysis of discrepancies between sums and integrals. By considering left-hand or right-hand Riemann sums, especially for decreasing convex functions, one can accurately assess and bound the error between the discrete sum and its continuous integral approximation. This method is instrumental in establishing the precise inequalities presented in the problem.

Transcript

00:01 If a function is positive and its second derivative is negative for roll x on a closed interval ab, we will show that the integral from a to b of the function f is greater than or equal to t sub n and less than or equal to m sub n.

00:22 Where t sub n and m sub n are a trapezoidal rule and midpoint rule approximations to the interoperative.

00:31 Of the function over ab with any number of sub -intervals n and a natural number.

00:38 That is, tn, the tropositoral rule with n -sub -intervals, is less than or equal to the integral of the function over ab, and that integral is less than or equal to the midpoint rule approximation.

00:54 And that, when the function has a second derivative, which is negative on the whole interval ab.

01:01 So what we have here, the context will be, we have any number of sub -intervals that is n belonging to n to the natural numbers.

01:16 With that and the lower and upper limits of integration ab, we calculate the step -size delta -x is b -minus a -over -n.

01:27 And with that, we know that the nodes determined by these step -size are x -bibals.

01:35 A plus i times delta x for the index i from zero up to n these are the notes and the midpoints are x of i bar which is x of i plus x of i plus 1 over 2 and in this case the index goes from 0 up to n minus 1 that is we have n midpoints and n plus 1 nodes.

02:16 These n plus 1 nodes, equally spaced, determine n's of intervals and these midpoints are the midpoints of those of intervals determined by the nodes.

02:29 So this is a general regular partition of the interval of integration ab.

02:37 And in that context, we are going to prove this inequality here.

02:44 And so we are going to look at the basic property that will give us the result, and that is that the second derivative is a negative function in the whole interval ab.

03:00 So if the second derivative of f is negative for all x in ab, then it means that the graph of the function f is negative.

03:20 Is concave down.

03:27 That is, it has this type of curvature.

03:32 That is, is a curve like this.

03:35 This is concave down.

03:37 Remember, concave up would be this.

03:41 So, concave down is the other convexity.

03:46 And this property means, okay, it's important to notice that this property of course in the whole interval ab.

03:56 So if we take any sub interval, the property will be the same, that is a function will be concave down on any sub interval we take on ab.

04:06 And this property means two things.

04:12 The first is that if we draw any line joining two points of the graph of the function, that line is below, completely below the graph of the function.

04:30 So that's the first thing, that is any line segment joining, let me put that better here, any line segment joining any two points on the graph of f lies completely below the graph.

05:36 That's the first thing we can take from the fact that the function is concave down.

05:43 The second one would be, again, we do a concave -down graph, and now we can say that if we take any tangent line at any point on the curve, that is the tangent line at any point of the curve of the graph, that tangent line lies above the graph.

06:09 And that's in fact a way of defining the concave down property of the function, that is the tangent line at any point on the graph of f lies completely above the graph.

06:47 So first property, any segment joining two points on the graph, lies below of the graph.

06:54 And any tangent line at any point of the graph lies above of the graph.

07:00 It's easy to remember that two properties if we draw a concrete -down graph.

07:06 In that case, we draw the tangent lines or the segment joining points and we immediately see these two properties.

07:16 And we're going to use them two to prove the inequalities.

07:20 So we start with the property number one.

07:29 So using property 1, property 1, if we consider any sub -interval of the partition we talk about above, x0x0i plus 1, i is any index from 0 to n minus 1, we will have that the second, that the second, joining the points x of i f of x of i and x of i plus 1 f of x of i plus 1 f of x of i plus 1 the line segment joining these 2 points lies below the graph of f that is and now we are going to interpret this property in terms of inequalities.

09:24 It means that the function is greater than the segment, that is the equation of that segment line for any x in the sub interval.

09:38 We get to put greater than or equal because at the two points, the segment and the curve coincides, so they are equal.

09:54 So this is the inequality that holds in this sub -intelial x -i x -o -i plus 1, where l is the function, is the equation of the segment, where l of x is the equation of the line joining the points x -5, x -5 -1.

10:34 So, f -f of x -of -i, and x -o -i.

10:49 And it's image f of x i plus 1.

10:54 And we can write that line.

10:55 It's very important that line is infinite in fact, but we are going to take only the segment, that is, the x belongs to the sub -intervals.

11:05 Very important that problem.

11:08 Then the line can be written as f of x of i.

11:17 That is, i'm taking this point, plus the slope of these line is f of x x of i plus 1 minus s f of x of i divided by the difference x of i minus x of i but remember the difference between two consecutive nodes is delta x times x minus x so i have used the coordinates of this point and the slope is given as this well now so we have that the function is above that is the function is greater than this equation, f of x of i plus f of x of i plus 1 minus f of x of i over delta x times x minus x of i.

12:18 For all x in the sub interval, x of i, x of i plus 1.

12:24 It's very important that we are just looking at that inequality on this subinterval.

12:33 And that for any of these of intervals.

12:35 It doesn't matter which interval we are talking about.

12:38 And it means that if it is true for every point in this of interval, then the integral over that two interval, that is the integral from x of i to x of i plus 1 of the function is greater than or equal to the integral from x of i, x of i plus 1 of this equation, f of x of x.

13:07 Of i plus f of x of i plus 1 minus f of x of i over delta x times x minus x of i differential of x and now we got to calculate this and now we are going to say that let's call this i i is equal to now we separate the integral and we know that this is a constant so we get f of x of i times the length of the interval from x of i to x w x w.

13:59 That is xy plus 1 minus x of i is delta x plus this expression here is a constant.

14:07 So we get out of the integral f of x of i plus 1 minus f x of i over delta x times the integral from x sub i to x of i plus 1 of x minus x sub i.

14:26 So we get x of i, f of xx y delta x plus this constant here.

14:34 It's a good idea to put it like that.

14:38 F of x of i plus 1 minus f of x of i over delta x times...

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