By William Briggs
Ants, motorcycles, and Clocks is a wonderful textual content for an undergraduate problem-solving path or as a source for arithmetic educators, offering enormous quantities of mathematical difficulties that may be utilized in any direction. Mathematically the ebook is determined by semesters of calculus, even if a lot of the ebook calls for in basic terms precalculus talents.
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Extra resources for Ants, bikes, & clocks: problem solving for undergraduates
3e4x + 4e5x 0 0 1 0 ⎦ ⎣ ex − 8e3x − 5e4x ⎦ 0 e2x e−x + 12ex + 10e2x Consider the system Y = AY + G where A= 2 −17 200 and G = 1 4 160ex . 4 that P = −1 + 4i 1 −1 − 4i 1 and MZ = e(3+4i)x 0 0 e(3−4i)x and NY = P MZ . Then 0 1 1 + 4i e−(3+4i)x (1/(8i)) 0 e−(3−4i)x 1 1 − 4i (−3−4i)x (−2−4i)x + 20(4 − i)e −25e . = 25e(−3+4i)x + 20(4 + i)e(−2+4i)x NY−1 G = 200 160ex , 51 52 CHAPTER 2. SOLVING SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS Then 0 NY−1 G = (4 + 3i)e(−3−4i)x + (−4 + 18i)e(−2−4i)x (4 − 3i)e(−3+4i)x + (−4 − 18i)e(−2+4i)x and Yi = NY 0 NY−1 G −1 + 4i −1 − 4i = 1 1 −1 + 4i −1 − 4i = 1 1 −32 − 136ex .
2 into a practical one. To keep our notation simple, we will stick to 2-by-2 or 3-by-3 cases, but the principle is the same regardless of the size of the matrix. One case is relatively easy. 3. If J is a diagonal matrix, ⎤ ⎡ d1 ⎢ ⎢ J =⎢ ⎣ d2 0 ⎥ ⎥ ⎥ ⎦ 0 .. dn then eJ x is the diagonal matrix ⎡ ⎤ ⎢ ⎢ eJ x = ⎢ ⎣ ed1 x ed2 x 0 ⎥ ⎥ ⎥. ⎦ 0 .. edn x 56 CHAPTER 2. SOLVING SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS Proof. Suppose, for simplicity, that J is 2-by-2, J = . 3 0 3 = d1 , J d2 2 0 d1 2 0 Then you can easily compute that J 2 = d1 k 0 0 d2 d1 0 0 , and similarly, J k = d2 3 0 for any k.
4. THE MATRIX EXPONENTIAL so eAx = 57 1 11 −7 2 1 41 −49 3 1 0 5 −7 x + x + ... , + x+ 2 0 1 2 −4 2! 2 3! 14 −22 which looks like a hopeless mess. But, in fact, the situation is not so hard! 5. Let S and T be two matrices and suppose S = P T P −1 for some invertible matrix P . Then S k = P T k P −1 for every k and eS = P eT P −1 . Proof. We simply compute S 2 = SS = (P T P −1 )(P T P −1 ) = P T (P −1 P )T P −1 = P T I T P −1 = P T T P −1 = P T 2 P −1 , S 3 = S 2 S = (P T 2 P −1 )(P T P −1 ) = P T 2 (P −1 P )T P −1 = P T 2 I T P −1 = P T 2 T P −1 = P T 3 P −1 , S 4 = S 3 S = (P T 3 P −1 )(P T P −1 ) = P T 3 (P −1 P )T P −1 = P T 3 I T P −1 = P T 3 T P −1 = P T 4 P −1 , etc.