YT2095 Posted July 22, 2004 Share Posted July 22, 2004 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` Link to comment Share on other sites More sharing options...
bloodhound Posted July 22, 2004 Share Posted July 22, 2004 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. Link to comment Share on other sites More sharing options...
jordan Posted July 22, 2004 Share Posted July 22, 2004 No you don't, it's inclusive. What do you mean? Link to comment Share on other sites More sharing options...
Freeman Posted July 22, 2004 Author Share Posted July 22, 2004 Thanks bloodhound! How do I calculate partial sums with mathematica??? Link to comment Share on other sites More sharing options...
bloodhound Posted July 22, 2004 Share Posted July 22, 2004 dont know mate, never used it in my life Link to comment Share on other sites More sharing options...
jordan Posted July 22, 2004 Share Posted July 22, 2004 [math]\sum_{q=0}^n (q+1)^2[/math'] if n was 2 This is really bothering me now. What is the answer here? Link to comment Share on other sites More sharing options...
Freeman Posted July 22, 2004 Author Share Posted July 22, 2004 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 Link to comment Share on other sites More sharing options...
jordan Posted July 22, 2004 Share Posted July 22, 2004 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] Link to comment Share on other sites More sharing options...
bloodhound Posted July 22, 2004 Share Posted July 22, 2004 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] Link to comment Share on other sites More sharing options...
timo Posted July 22, 2004 Share Posted July 22, 2004 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. Link to comment Share on other sites More sharing options...
jordan Posted July 22, 2004 Share Posted July 22, 2004 So the answer is 14, correct? Link to comment Share on other sites More sharing options...
timo Posted July 22, 2004 Share Posted July 22, 2004 no, it´s 42 ... sry, I couldn´t resist yes: 1+4+9 = 14 Link to comment Share on other sites More sharing options...
Dave Posted July 23, 2004 Share Posted July 23, 2004 rofl Link to comment Share on other sites More sharing options...
jordan Posted July 23, 2004 Share Posted July 23, 2004 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. Link to comment Share on other sites More sharing options...
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 accountSign in
Already have an account? Sign in here.
Sign In Now