Difference between revisions of "Imaginary unit/Introductory"

Revision as of 13:41, 26 October 2007

Problem

Find the sum of $i^1+i^2+\ldots+i^{2006}$

Solution

Since $i$ repeats in a n exponential series at every fifth turn, the sequence i, -1, -i, 1 repeats. Note that this sums to 0. That means that all sequences $i^1+i^2+\ldots+i^{4k}$ have a sum of zero (k is a natural number). Since $2006=4\cdot501+2$ , the original series sums to the first two terms of the powers of i, which equals $-1+i$ .

@@ Line 2: / Line 2: @@
 Find the sum of <math>i^1+i^2+\ldots+i^{2006}</math>
 == Solution ==
-Let's begin by computing powers of <math>i</math>.
+Since <math>i</math> repeats in a n exponential series at every fifth turn, the sequence i, -1, -i, 1 repeats. Note that this sums to 0. That means that all sequences <math>i^1+i^2+\ldots+i^{4k}</math> have a sum of zero (k is a natural number). Since <math>2006=4\cdot501+2</math>, the original series sums to the first two terms of the powers of i, which equals <math>-1+i</math>.
-#<math>i^1=\sqrt{-1}</math>
-#<math>i^2=\sqrt{-1}\cdot\sqrt{-1}=-1</math>
-#<math>i^3=-1\cdot i=-i</math>
-#<math>i^4=-i\cdot i=-i^2=-(-1)=1</math>
-#<math>i^5=1\cdot i=i</math>
-We can now stop because we have come back to our original term. This means that the sequence i, -1, -i, 1 repeats. Note that this sums to 0. That means that all sequences <math>i^1+i^2+\ldots+i^{4k}</math> have a sum of zero (k is a natural number). Since <math>2006=4\cdot501+2</math>, the original series sums to the first two terms of the powers of i, which equals <math>-1+i</math>.

Page

Toolbox

Search

Difference between revisions of "Imaginary unit/Introductory"

Revision as of 13:41, 26 October 2007

Problem

Solution