**Using Wikis in Mathematics Classes**

**Professor David Richeson **

**Department of Mathematics and Computer Science**

** Dickinson College, Carlisle, PA 17013 **

• Students collaborate to create a homework “solutions manual” on the wiki.

• You can teach students how to write mathematical arguments (proofs).

• Can choose problems that do not have answers in the back of the book.

• All of the problems are ones that are assigned for homework: **after** the students turn in their homework, list the problems on the wiki.

• Each student contributes **one major edit** and **two minor edits** to the wiki. Examples of major edits include:

• **Major Edit: **Typing in the statement of a problem and a complete solution or making substantial changes to an existing solution. __Must be modified in a fundamental way.__

• Examples of **minor edits **include:

• Typing in the statement of a problem and a complete solution to a non-assigned problem.

• Correcting all mathematical errors in a solution.

• Correcting all typographical errors in a solution.

• Fixing badly-coded text.

• Modifying text to conform to our style guidelines.

• Starting a new solution, but not making much progress.

• Adding a picture that they created.

**Periodically teacher reviews wiki, reads solutions, comments on the “talk page” and added banners**

• **Error: **There are one or more errors in this solution. There is probably a discussion of the error on the talk page. When you have corrected the error, you should remove this banner.

• **Messy**: This text is messy. It might need better math formatting or structural repair. There may be typographical errors. Look on the talk page for more information. Once you have cleaned things up, you should remove this banner.

• **Suggestion**: I have suggested improvements for this text. The suggestions are described on the talk page. Please read the discussion there for ideas on improving the writing. Once you make the changes, you should remove this banner.

• **Not started: **This homework problem (or a part of this homework problem) has not been started. After you have written the solution, you should remove this banner.

• **Checkmark: **(If the solution to a problem is correct, I put a red checkmark underneath it. Once I signed off on it, they do not have to edit it further.)

• Students graded on **participation**, not on the quality of their edits.

• Only keep track of whether they did the **required major and minor edits**

• **Wikis have excellent version histories**

• Students quickly realize incentive to make edits early. If they wait until the night before edits are due, only the more challenging (more time-consuming) problems are available.

• Closer to exam times can assign extra minor edits so that solutions can be used as a reliable study guide.

• Creates a nice resource for the students to study from.

**Sample Wiki: Created by Caleb Kelley, Class of 2011, New Smyrna Beach High School**

**1/10/11 Discussion Problem: the tangent of a triangle is 12/5, find: the exact value of sec2x (where x is the central angle)**

** **

**Caleb: **Since the tangent of a circle is sin/cos , using the Pythagorean Theorem (A^{2}+B^{2}= C^{2}) where C^{2} is the hypotenuse. We find that the hypotenuse is 13, we could just look at it from the perspective of a Pythagorean triple, 5:12:13, is a common triple. Once we have that, we must see that the sec2x is the same as 1/cos2x. Cos2x =Cos^{2}x-Sin^{2}x, which simplifies down into (cosx)(cosx) – (Sin^{2}x), and further into (cosx)(cosx) – (1- Cos^{2}x ), and further into (cosx)(cosx) – 1+ (cosx)(cosx). Once we have this, we already know that the cos= 5/13, so substitution yields:

(5/13)(5/13)-1+(5/13)(5/13)= 25/169-169/169+25/169= -119/169

**Mr. Jones: correct, so far, but don forget this is sec2x, not simply cos2x. Sec2x=1/cos2x **

**Jenifer: **How did you find the equivalent of cos2x?

**Caleb:** Using cos2x=cos^{2} x-sin^{2}x

**Ben:** Since the tangent of a circle is sin/cos , using the Pythagorean Theorem (A^{2}+B^{2}= C^{2}) where C^{2} is the hypotenuse. We find that the hypotenuse is 13, we could just look at it from the perspective of a Pythagorean triple, 5:12:13, is a common triple. Once we have that, we must see that the sec2x is the same as 1/cos2x. Cos2x =1-2Sin^{2}x, which simplifies down into (1-2(1-cos^{2}x), and further into (1-(2-2cos^{2}x) = -1+(2cos^{2}x), which also equals -1 + (2Cosx) (Cosx) = -1 + 2(5/13)(5/13) = -169/169 + 50/169 = -119/169, since sec2x= 1/cos2x: sec2x= 169/-119

Mr. Jones: correct, nice alternative to Caleb’s post!