October 14 Information

I am looking for a website that does the following. (1) I enter a linear equation (e.g., y = 2x + 5) and the website graphs the correct line. Or (2) I enter the data points (ordered pairs such as (1,1), (2,2), (3,3), etc.) and the website graphs the correct line. Also, I'd like to be able to print the results. Finally, is there a way in Excel (or Word) to do something similar?

The gist of my problem is this. I have to graph several equations. And I'd want to submit the work so that it is neat (i.e., computer-generated or word-processed) and not hand-written. Any suggestions? I want it to look like the graphs of linear equations that you might see in a textbook, and not one that is written by hand. Like, for example, the two graphs shown at the very bottom of this page: [1]. Or this page: [2]. There must be something out there that does this. That makes this task easier. Any suggestions? Thank you. — Preceding unsigned comment added by 2602:252:D13:6D70:19F4:A168:19D0:22A1 ( talk) 04:26, 14 October 2015 (UTC) reply

Excel can do this. Create a column of x coordinates. In the cells beside them either enter the y values or enter the formula to generate the values ( your example "= 2 * A1 + 5" in B1 etc). Then highlight the entire table, select graph and choose scatter plot. There are loads of options, you'll have to read up on Excel to find out more. -- SGBailey ( talk) 05:45, 14 October 2015 (UTC) reply

Wolfram Alpha can do this sort of thing: e.g. plotting y = 2x + 5; and pairs of points. For printing you can either print the page as it is, or take a screenshot or snip and paste the result into your application of choice. AndrewWTaylor ( talk) 14:54, 14 October 2015 (UTC) reply

• I use Apple Grapher; haven't tried printing from it. — Tamfang ( talk) 07:01, 21 October 2015 (UTC) reply

Related question

OK. Thanks. I tried it in Excel. It seems to work OK. So, another question about the Excel process (Excel 2013, if it matters). How can I "format" the Cartesian Plane so that it always extends the "x" variable (horizontal line) from -10 to +10, and so that it always extends the "y" variable (vertical line) from -10 to +10? I explored the options, and I can see where I can control one number line, but not the other. The actual Cartesian Plane "boundaries" seem to change, depending on what the actual line looks like. I'd like to get a uniform plane of 10 by 10 for both axes. Is that possible? Thanks. — Preceding unsigned comment added by 2602:252:D13:6D70:9433:3444:F704:E383 ( talk) 17:36, 14 October 2015 (UTC) reply

In other words, I want the Cartesian Plane to look like this: [3], a "standardized" plane that always extends from -10 to +10 on both axes. And I want to draw the line on that standard template, wherever the line may happen to fall, within that grid. Thanks. 2602:252:D13:6D70:9433:3444:F704:E383 ( talk) 17:41, 14 October 2015 (UTC) reply

The following works for at least some versions of Excel (though I haven't tried it on Excel 2013): Double-click the x-axis of the graph. It will bring up a dialog box with multiple tabs. Click the tab labeled "Scale" and then set the options accordingly. When you're done with the x-axis, double-click the y-axis of the graph and do the same thing. — SeekingAnswers ( reply) 18:32, 14 October 2015 (UTC) reply

Will try that. Thanks. — Preceding unsigned comment added by 2602:252:D13:6D70:9433:3444:F704:E383 ( talk) 18:59, 14 October 2015 (UTC) reply

Think I got it to work! Thanks! (Excel 2013 is a bit different than what you described. There is no "scale" wording anywhere. But there are similar concepts, and I was able to make it work.) Thanks. — Preceding unsigned comment added by 2602:252:D13:6D70:9433:3444:F704:E383 ( talk) 19:26, 14 October 2015 (UTC) reply

Thanks, all. — Preceding unsigned comment added by 2602:252:D13:6D70:4C55:4EEE:A64B:6776 ( talk) 16:38, 19 October 2015 (UTC) reply

Resolved

Assuming that we have a set, A, what is the minimum of the intersection of its subsets? Formally: Let ${\displaystyle n\in \mathbb {N} ,m\in \mathbb {N} ,k\in \mathbb {N} }$ be three numbers. Let A be a set of size n (|A|=n). Let P(A) denote the power-set of A. Let ${\displaystyle B=\{A'|A'\subseteq P(A),|A'|=m\},C=\{B'|B'\subseteq P(B),|B'|=k\}}$. What is the value of ${\displaystyle \min _{B\in C}\max _{b1,b2\in B}|b1\cap b2|}$ ?

--Edit-- I am not intersted in the the exact value, but only in the answer for the following question: Assuming also that n is determined first, and then m, k are determined as functions of n (so, I shall denote their corresfonding functions as ${\displaystyle f_{m}(n),f_{k}(n)}$ respectievly, and I shall denote the desired value above as ${\displaystyle f(n)=\min _{B\in C}\max _{b1,b2\in B}|b1\cap b2|}$). My question is: Are there some constant c < 0.5, and functions ${\displaystyle f_{m}(n),f_{k}(n)}$ and polynomial ${\displaystyle p(n)}$ such that ${\displaystyle \forall nf_{m}(n)\leq p(n),f_{k}(n)\leq p(n)}$ and ${\displaystyle f^{n}(n)<cn}$ (peforming f for n times keeps the result < cn)

Thank you! 31.154.92.193 ( talk) 07:58, 14 October 2015 (UTC) reply

Maybe this will clarify a bit. Given a set of n points, find a collection B of subsets of order m, so that B has order k and any two elements of B have at most λ points in common, having λ is a small as possible. An example might be a projective plane of order 2, where n=k=7, m=3 and λ=1; here B is the collection of lines. You can have λ=0 if n≥mk. I'm guessing you meant to include b1≠b2 above since otherwise the answer is trivially k. The second part isn't clear to me since it seems to me you have a function of three variables f(m, k, n)=λ that you're trying to estimate. The theory of block designs uses counting arguments that might be useful for getting some bounds. I think the general term what you're talking about is Incidence structure or hypergraph depending on the context. -- RDBury ( talk) 17:47, 14 October 2015 (UTC) reply

Thanks for the extra and helpful comments, guys. Gurumaister ( talk) 15:28, 15 October 2015 (UTC) reply

After some efforts I succeded to rephrase my question: Given v, k. Is there ${\displaystyle \lambda \leq k^{2}/v}$, and a ${\displaystyle (v,k,\lambda )}$ abelian difference set, D ? 212.199.149.142 ( talk) 08:30, 16 October 2015 (UTC) reply