Mathematica code for sigma?

[YT2095]

I`m nowhere near that advanced yet, I`m up the part of a square X and Y, and then enlarging it proportionaly so that dx and dy (in the top right) is made up of y+dy and x+dx and stuff like that. and the one for calculating the angle of a ladder and how pulling the ladder away from 0 along the x will affect the height on the Y and stuff like that, the odd part is, there`s no real numbers involved? it`s certainly a new concept to me outside of programing

 

oh yeah, I covered 1`st and 2`nd orders of `d`

[bloodhound]

I think ur talking about a sequence of partial sums. i.e if u have a [math]\sum_{k=p}^{n}a_k[/math] then the nth partial sum would be [math]S_n=\sum_{k=p}^{n}a_k[/math] for all [math]n \ge p[/math]

So now you define a sequence of nth partial sums

that is {S_n}, and that will give u the set your looking for. maths programs should have the ability to calculate partial sums.

[jordan]

No you don't, it's inclusive.

What do you mean?

[Freeman]

Thanks bloodhound! How do I calculate partial sums with mathematica???

[bloodhound]

dont know mate, never used it in my life

[jordan]

[math]\sum_{q=0}^n (q+1)^2[/math'] if n was 2

This is really bothering me now. What is the answer here?

[Freeman]

Dude, its the equation for a square with 2 units by 2 units. The equation would solve this problem, five squares! The answer is 5! The four one by one units and the one two by two square. (0+1)^2=1 and (1+1)^2=4, and 1+4=5

[jordan]

So what does dave's post mean?

 

And why does my calc book say this:

 

[math]\sum_{i=1}^6_i=1+2+3+4+5+6[/math]

[bloodhound]

oh i get u now didnt realise.

 

the series u have given [math]\sum_{q=0}^{n}(q+1)^2[/math]

is equivalent to [math]\sum_{q=1}^{n}q^2[/math]

 

which is a standard maths series whose nth partial sum can be written down explicitely as a fucntion of n

 

and its given as [math]\sum_{q=1}^{n}q^{2}=\frac{n(n+1)(2n+1)}{6}[/math]

 

THIS IS A MISTAKE, ITS FALSE

 

correct version

 

the series u have given [math]\sum_{q=0}^{n}(q+1)^2[/math]

is equivalent to [math]\sum_{q=1}^{n+1}q^2[/math]

 

which is a standard maths series whose nth partial sum can be written down explicitely as a fucntion of n

 

and its given as [math]\sum_{q=1}^{n+1}q^{2}=\frac{(n+1)(n+2)(2n+3)}{6}[/math]

[timo]

So what does dave's post mean?

It means that [math] \sum _{q=0} ^2 (q+1)^2 = (0+1)^2 + (1+1)^2 + (2+1)^2 [/math]

 

And why does my calc book say this: [...]

because what dave said is correct.

[jordan]

So the answer is 14, correct?

[timo]

no, it´s 42 ...

 

sry, I couldn´t resist

 

yes: 1+4+9 = 14

[Dave]

rofl

[jordan]

And we all make fun of me. Oh well.

 

There was much controversy over this earlier. I got myself all confused. Glad it's resolved.