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Posted

Hey, is it possible to have a sigma equation to figure out all possible solutions for a problem? What would the code of that be?

Posted

Because that's the way it works... it's every premade image put on a site with a lot of bandwidth, then sent here. I think that's how it works...

Posted

Okay, allow me to restate it. The sigma code adds the solutions togather fopr the answer, e.g. [math]\sum_{q=0}^n (q+1)^2[/math] if n was 2 the answer is 5. However, is there an equation which makes the answer to this problem a set? So it would be {1,4} rather than 5.

 

If an equation exists, what would it be on mathematica? :eek::confused::):cool:

Posted

Like, if I were going to have this equation: [math]n=>2, m=>2, p=>1,

 

{p \times n}-{n \times m}[/math]

With a limit to 1000, how would I encode this??? :confused:

Posted
Okay, allow me to restate it. The sigma code adds the solutions togather fopr the answer, e.g. [math]\sum_{q=0}^n (q+1)^2[/math'] if n was 2 the answer is 5.

Sorry to digress from the topic for a minute, but wouldn't the answer be 14, or is it too late for me to be trying this stuff?

Posted

No, the limit is two. There would be two equations starting with 0, so-

(0+1)^2=1

(1+1)^2=4

1+4=5 QED

ITs a little weird...

Posted
No, you calculate it only twice... Not until q=2. I did the same thing when I learned it at first :P

 

No you don't, it's inclusive.

Posted

Dave, in my calculus book, they use a different symbol for Sum of, it looks like a long S of a fancy f without the line through the middle, is this the same as Sigma?

 

that symbol and `d` are the only I`ve read about yet, `d` being "a little bit of"

 

like dx would be a little bit of x and Sdx would be the sum of all the little bits of x

 

 

is Sigma the same as the long S?

Posted

In a sense. What you're talking about is the integral symbol - which is hopefully what I'll be covering in my calculus threads fairly soon if I can get around to writing the next one.

 

Imagine you have a curve, then draw equally spaced lines under the curve so that you can approximate the area under the curve by summing the areas of the trapezia that you've just drawn. Now if you decrease the height of the trapezia and take the limit of the series as the height tends to zero, that's your integral.

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