Sums of weird series

**DecayProduct** · Jun16-09, 01:25 PM

I have a rudimentary understanding of integration as it applies to finding the area under a curve. I get the idea of adding up the areas of progressively smaller rectangles to approach the area, and that at an infinite number of rectangles the areas would be exactly the same. Right now I'm just playing around with the idea and I'm curious about how to sum up n number of things if the ratio between each one changes.

For example, I've drawn a graph of f(x) = $LaTeX Code: \\sqrt{x}$ between 0 and 1. Now this isn't like a geometric series where I can find the sum using $LaTeX Code: S_{n}=a_{1}(1-r^{n})/1-r$ , because r changes. I have discovered that $LaTeX Code: a_{n} = a_{1}\\sqrt{n}$ . I have played around with the ratios and discovered some interesting patterns that emerge, and I have found a complicated way to sum up two objects, but it is really more work than just doing the sum directly, and there'd be no way to do it when $LaTeX Code: n=\\infty$ .

Sorry for such a basic question, but how are things like this summed?

**HallsofIvy** · Jun16-09, 02:32 PM

The idea of integration as the limit of Riemann sums can be used to determine what functions are integrable and can be used as a guide to setting up integrals in applications. But in fact, for all except the simplest examples, we use the "Fundamental Theorem of Calculus"-
$LaTeX Code: \\int_a^b f(x) dx= F(b)- F(a)$
where F(x) is any function having f(x) as derivative- F is an "anti-derivative" of f.

In this particular case, to find
$LaTeX Code: \\int_0^1\\sqrt{x}dx= \\int x^{\\frac{1}{2}}dx$
I would note that the derivative of $LaTeX Code: (2/3)x^{3/2}$ is $LaTeX Code: (2/3)(3/2)x^{3/2- 1}= x^{1/2}$ so I can take $LaTeX Code: f(x)= x^{1/2}$ and $LaTeX Code: F(x)= (2/3)x^{3/2}$ .
$LaTeX Code: \\int_0^1\\sqrt{x}dx= (2/3)(1^{3/2}- 0^{3/2})= 2/3$

**DecayProduct** · Jun16-09, 03:09 PM

Originally Posted by HallsofIvy

In this particular case, to find
$LaTeX Code: \\int_0^1\\sqrt{x}dx= \\int x^{\\frac{1}{2}}dx$
I would note that the derivative of $LaTeX Code: (2/3)x^{3/2}$ is $LaTeX Code: (2/3)(3/2)x^{3/2- 1}= x^{1/2}$

Thanks for the response. I sort of understand it, except for where the derivative of $LaTeX Code: (2/3)x^{3/2}$ comes from. I mean, is it just something that one has to work out until you find an antiderivative that equals $LaTeX Code: \\sqrt{x}$ , or is there some more basic and natural way that this comes about?

**Cyosis** · Jun16-09, 03:53 PM

It is exactly for this reason that finding a primitive is a lot harder in general than finding a derivative. However in this case you can just use the power rule for integrating polynomials.

$LaTeX Code: <BR>\\int x^n dx=\\frac{1}{1+n} x^{n+1},\\;\\;n \\neq -1<BR>$

In word, raise the power of your integrand by 1 then divide through the new power.

**chiro** · Jun17-09, 12:34 AM

Originally Posted by DecayProduct

Thanks for the response. I sort of understand it, except for where the derivative of $LaTeX Code: (2/3)x^{3/2}$ comes from. I mean, is it just something that one has to work out until you find an antiderivative that equals $LaTeX Code: \\sqrt{x}$ , or is there some more basic and natural way that this comes about?

I guess one way you could think of it is that if you had a derivative function which measures all the changes between two points, adding up all the changes between these points gives the normal function value. In this case the derivative function is just that and the anti derivative is the function value. Its kind of intuitive if you think of it in that way and then calculus just has to make sense.