Dan's Thoughts Thinking somewhat carefully

30Mar/090

Maven2 Echoing To the Screen

<project xmlns="http://maven.apache.org/POM/4.0.0" xsi="http://www.w3.org/2001/XMLSchema-instance" schemalocation="http://maven.apache.org/POM/4.0.0     http://maven.apache.org/xsd/maven-4.0.0.xsd">
 
<modelversion>4.0.0</modelversion>
<groupid>mygroup</groupid>
<artifactid>hw</artifactid>
<packaging>pom</packaging>
<version>1.0-SNAPSHOT</version>
<name>Print Hello World</name>
 
<build>
  <defaultgoal>validate</defaultgoal>
  <plugins>
      <plugin>
          <inherited>false</inherited>
          <groupid>org.apache.maven.plugins</groupid>
          <artifactid>maven-antrun-plugin</artifactid>
          <executions>
              <execution>
                  <id>helloworld</id>
                  <phase>validate</phase>
                  <configuration>
                      <tasks>
                          <echo>Hello World!</echo>
                          <echo>Hello World!</echo>
                          <echo>JAVA_HOME=${java.home}</echo>
 
                      </tasks>
                  </configuration>
                  <goals>
                      <goal>run</goal>
                  </goals>
              </execution>
          </executions>
          <dependencies>
              <dependency>
                  <groupid>ant</groupid>
                  <artifactid>ant</artifactid>
                  <version>1.6.5</version>
              </dependency>
          </dependencies>
      </plugin>
  </plugins>
</build>
</project>

$ mvn
[INFO] Scanning for projects...
[INFO] ------------------------------------------------------------------------
[INFO] Building Print Hello World
[INFO] task-segment: [validate]
[INFO] ------------------------------------------------------------------------
[INFO] [antrun:run {execution: helloworld}]
[INFO] Executing tasks
[echo] Hello World!
[echo] Hello World!
[echo] JAVA_HOME=c:\j2sdk1.4.2_18\jre
[INFO] Executed tasks
[INFO] ------------------------------------------------------------------------
[INFO] BUILD SUCCESSFUL
[INFO] ------------------------------------------------------------------------
[INFO] Total time: 1 second
[INFO] Finished at: Tue Sep 09 10:00:44 EDT 2008
[INFO] Final Memory: 2M/5M
[INFO] ------------------------------------------------------------------------

Filed under: maven No Comments
22Mar/090

Derivation for sum of squares

My problem with mathematical induction is that a lot of the time it's simple to prove a statement is true (just do enough of the algebra calisthenics) yet just because a statement is true doesn't mean that you learned all that much about a problem. I think everyone that ever took a discrete math class had to churn out the proof for the following statement.

1 + 2 + \cdots + n = \frac{n(n + 1)}{2}

Certainly the wikipedia page on induction doesn't fail to include this as an example.
This is the canonical inductive proof. However to me it's not interesting. I think deriving the solution is easy enough. More interesting is deriving the solution to the sum of square or sum of cubes or an arbitrary higher power.

1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}
1^3 + 2^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}

I was curious as to how these formulas came to be. It's all well and good to say, "Sure, this formula does indeed work", and it's a different thing to derive it.
For convenience I will write the identities in the following way:

\displaystyle\sum_{1}^{n}{i} = \frac{n(n + 1)}{2}

The crux of the derivation for higher powers lies in the fact that:
\displaystyle\sum_{1}^{n}{(i+1)^2-i^2} = (n+1)^2-1
to see why this is so, plug in the first few terms of the expansion of (i+1)^2 as well as the last few terms:
(i+1)^2 = 2^2 + 3^2 + \cdots + (n-3)^2 + (n-2)^2 + (n-1)^2+ n^2 + (n+1)^2
i^2 = 1^2 + 2^2 + 3^2 + \cdots + (n-3)^2 + (n-2)^2 + (n-1)^2+ n^2
Subtracting all the like terms, only two terms remain: -1 and (n+1)^2
Here's the derivation:
\displaystyle\sum_{1}^{n}{(i+1)^2-i^2} = \displaystyle\sum_{1}^{n}{(i^2+2i+1)-i^2} = \displaystyle\sum_{1}^{n}{2i+1} = (n+1)^2-1
\displaystyle\sum_{1}^{n}{2i+1} = \displaystyle2\sum_{1}^{n}{i} + \displaystyle\sum_{1}^{n}{1} = \displaystyle2\sum_{1}^{n}{i} + n = (n+1)^2-1
\displaystyle\sum_{1}^{n}{i} = \frac{(n+1)^2-1 - n}{2}
\displaystyle\sum_{1}^{n}{i} = \frac{n(n+1)}{2}

By a similar argument the sum of cubes is derived as well:
\displaystyle\sum_{1}^{n}{(i+1)^3-i^3} = (n+1)^3-1
\displaystyle\sum_{1}^{n}{(i+1)^3-i^3} = \displaystyle\sum_{1}^{n}{3i^2+3i+1} = (n+1)^3-1
\displaystyle\sum_{1}^{n}{i^2} = \frac{(n+1)^3-1 - n - \frac{3n(n+1)}{2}}{3}
After plenty of simplification the simplified form does indeed come out
\displaystyle\sum_{1}^{n}{i^2} = \frac{n(n+1)(2n+1)}{6}

The same process can be used to derive sums of cubes and any other power, as long as the sums of previous powers are known.

Filed under: math No Comments
   

Pages

Categories

Blogroll

Archive

Meta