SFNQuestions 17 Posted February 14, 2017 Share Posted February 14, 2017 It doesn't seem right, but I could have sworn I remember seeing instances in statistics where a summation was literally just swapped with an integral sign without anything else changing, except adding a dx. Can you really just swap a sum and an integral? I mean, the sum of a linear variable k is a second degree polynomail, and the integral of a continuous variable k would be a 2nd degree polynomial, so maybe there's something to it. Link to post Share on other sites

Xerxes 45 Posted February 14, 2017 Share Posted February 14, 2017 The answer is yes, under certain circumstances. The conventional way to define the Riemann definite intgral of a function[math]f(x)[/math] over a close interval [math][a,b][/math] is to divide this interval into a number of non-overlapping interval [math][x_0,x_1),[x_1,x_2),....,[x_k,x_{k+1}),....,[x_{n-1},x_n][/math] where [math]a \equiv x_0 <x_1 <.....<x_n \equiv b[/math]. You form the so-called Riemann sum [math]\sum\nolimits_{k=0}^{n-1} f(\xi_k)(x_{k+1}-x_k)[/math] where [math]\xi_k[/math] denotes a point in the interval [math][x_k,x_{k+1})[/math]. Now you let the number of intervals increase without bound, so that [math]x_{k+1}-x_k \to 0[/math], then provided the limit of the Riemann sum exists, then this goes over to the integral [math]\int\nolimits_a^b f(x)\,dx[/math] Link to post Share on other sites

Country Boy 70 Posted February 17, 2017 Share Posted February 17, 2017 You can approximate an integral by a Riemann sum. Is that what you mean by "replace"? Link to post Share on other sites

SFNQuestions 17 Posted February 25, 2017 Author Share Posted February 25, 2017 (edited) The answer is yes, under certain circumstances. The conventional way to define the Riemann definite intgral of a function[math]f(x)[/math] over a close interval [math][a,b][/math] is to divide this interval into a number of non-overlapping interval [math][x_0,x_1),[x_1,x_2),....,[x_k,x_{k+1}),....,[x_{n-1},x_n][/math] where [math]a \equiv x_0 <x_1 <.....<x_n \equiv b[/math]. You form the so-called Riemann sum [math]\sum\nolimits_{k=0}^{n-1} f(\xi_k)(x_{k+1}-x_k)[/math] where [math]\xi_k[/math] denotes a point in the interval [math][x_k,x_{k+1})[/math]. Now you let the number of intervals increase without bound, so that [math]x_{k+1}-x_k \to 0[/math], then provided the limit of the Riemann sum exists, then this goes over to the integral [math]\int\nolimits_a^b f(x)\,dx[/math] That's just the details of the formal definition of an integral, that doesn't really answer anything. You can approximate an integral by a Riemann sum. Is that what you mean by "replace"? No I mean flat out swap, as in the only thing that changes is the summation sign turning into an integral sign and vice versa. The only time I've seen it is when you have the integral of a sum, and then you can switch them, but that's not what I am referring to. Edited February 25, 2017 by SFNQuestions Link to post Share on other sites

zztop 11 Posted February 28, 2017 Share Posted February 28, 2017 It doesn't seem right, but I could have sworn I remember seeing instances in statistics where a summation was literally just swapped with an integral sign without anything else changing, except adding a dx. Can you really just swap a sum and an integral? I mean, the sum of a linear variable k is a second degree polynomail, and the integral of a continuous variable k would be a 2nd degree polynomial, so maybe there's something to it. Yes, you can, this is used routinely for calculating series limits. Link to post Share on other sites

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now