MATH 1910 Workshop Solution

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1 MATH 90 Workshop Solutio Fractals Itroductio: Fractals are atural pheomea or mathematical sets which exhibit (amog other properties) self similarity: o matter how much we zoom i, the structure remais the same. The Koch sowflake, created by the ifiite step process whose first four iteratios are show below, is oe example of a fractal: The process of costructig a mathematical fractal is always ifiite. I practice, though, self similarity is limited by our perceptio ad the actual costructio or calculatio limits of the system or object. Eve the aturally occurrig fractal properties of a ocea coastlie, for istace, ca t cotiue dow to the molecular level! Despite this, self similarity ad fractals have may applicatios ragig from computer graphics (geeratig realistic ladscapes for games, etc.) to sigal compressio, to soil mechaics, to highly-efficiet atea desigs (a good atea ofte eeds to have large surface area while remaiig very compact). Goals: Practice workig with ifiite series. Uderstad the relatioship betwee volume ad surface area i 3D fractals. Problems: We ll cosider a 3 dimesioal versio of the Koch sowflake: a sphereflake! This fractal is created as follows: start with a sphere of radius. To this large sphere, attach 9 smaller spheres of radius /3. To each of these ie spheres, attach ie spheres of radius /9, ad so o. To each sphere of radius r, attach ie spheres of radius r/3, for ifiite iteratios. a) What is the total volume of the sphereflake? Hit: You may use the fact that the volume of a sphere is V = 4 3 πr3. Solutio. Sice the sphereflake is costructed i a ifiite process, we expect to represet its volume with a ifiite series. There are 9 spheres of radius i the sphereflake. 3 The volume of a sphere of radius r is 4 3 πr3, so the volume of the sphereflake is 9 4π 3 ( ) 3 = 3 4π = 4π 3 3 = 4π 3 3.

2 MATH 90 Workshop Solutio We ote that this ifiite series is a geometric series of the form Sice r <, we kow that it coverges to Fractals Cr with r = /3. Cr = C r. For r = /3 ad C =, the geometric series coverges to 3/2 ad the volume of the sphereflake is V = 4π = 2π. As a remider, here is how we fid the value of a geometric series whe r < : N Cr r N Cr = C + Cr + + Cr N (Cr + Cr Cr N+ ), Cr = lim N ( r) N N Cr = C Cr N+, Cr C( r N+ ) = lim N r = C, r <. r b) What is the total surface area of the sphereflake? Hit: You may use the fact that the surface area of a sphere is A = 4πr 2. Solutio. The surface area of a sphere of radius r is 4πr 2, so the area of the sphereflake is ( ) 2 9 4π = 4π = 4π. 3 So, a sphereflake has a ifiite surface area but a fiite volume! Remark: Although this result may seem surprisig, there is o paradox here. Remember that the sphereflake is obtaied by a ifiite process, ad its volume ad area are defied to be the limits of growig partial sums, ad those limits ca either coverge (as i the first case) or diverge (as i the secod case). At the same time, a ubouded geometric body does ot have to be fractal-like to have a fiite volume ad ifiite surface area; by ow you should have see a example of that i the homework. 2

3 MATH 90 Workshop Solutio Fractals c) To geeralize this example, suppose that the iitial sphere is of radius r 0 = R, each ext level cosists of spheres of radius r + = αr for some positive α <, ad there are m balls of radius r + attached to each ball of radius r. We will igore the possibility of spheres itersectig ( attached does ot have to mea touchig ). What relatio betwee α ad m guaratees that the sphereflake will have a fiite volume? fiite surface area? Solutio. The total umber of spheres of radius r is m. Geeralizig the above, the volume is V = m 4π 3 (α R) 3 = 4πR3 (mα 3 ) ; 3 so, m < α 3 will guaratee covergece. For the surface area, A = m 4π (α R) 2 = 4πR 2 (mα 2 ) ; so, m < α 2 will guaratee covergece. Sice α <, wheever the surface area is fiite, the volume is fiite as well. d) Without resortig to the iterwebs, braistorm with your group to come up with a list of objects or cocepts which exhibit or approximate self-similarity. Solutio. A (by o meas exhaustive) list: trees feathers blood vessels river systems ocea waves All of these examples have the same shape o may legth scales. Workshop takeaways: 3D fractals ca have ifiite surface area with fiite volume. Mathematical series ca be used to describe atural pheomea. 3

4 . (SEQUENCES) NAME: SOLUTIONS July 23, 208 REVIEW A sequece () is a list of umbers a0, a, a 2,.... It does t have to start with zero. A series (2) is the sum of the terms i a sequece: i=0 a i A sequeces is called: (a) (b) bouded mootoe (3) if there exists M such that a M for all. (4) if either a < a + or a > a + for all. If a sequece is both of the above, the it coverges. If f is cotiuous (5) ad lim a = L, the lim f(a ) = f(l). A sequece that looks like a = cr is called geometric (6). PROBLEMS () Determie the limit of the sequece or show that the sequece diverges. (a) a = e 2 SOLUTION: Note that e > 2, so e/2 >. Hece, a = e ( e ) 2 = 2 ( e ) lim a = lim =. 2 (b) b = SOLUTION: As, the top ad the bottom are both polyomial of the same degree, so oly the leadig coefficiets matter. Hece, 3 + lim = 3 2.

5 (c) c = + 4 SOLUTION: lim + 4 = lim = lim = + 0 =. (l )2 (d) c = SOLUTION: Use L Hôpital s Rule twice: (l ) 2 2(l ) lim = lim l = 2 lim = 2 lim = 0. (2) Show that the sequece give by a = 32 2 is strictly icreasig, ad fid a upper boud. + 2 SOLUTION: Cosider the fuctio f(x) = 3x2. The derivative of f is x 2 +2 f (x) = 2x (x 2 + 2) 2. For x > 0, f (x) > 0, so the fuctio is strictly icreasig. Therefore, the sequece a = f() is strictly icreasig. To fid a upper boud, observe that a = Therefore, M = 3 is a upper boud = 3(2 + 2) 2 =

6 (3) Let {a } be the sequece defied recursively by a 0 = 0, a + = 2 + a (a) Write the first four terms of the sequece. SOLUTION: a 0 = 0, a = 2, a 2 = 2 + 2, a 3 = (b) Show that the sequece {a } is icreasig. SOLUTION: We have a 0 = 0 < 2 = a. Now assume that it s true for, that is, a a +, ad show it for + that is, show that a + a +2. a + = 2 + a 2 + a + = a +2 (c) Show that the sequece is bouded above by M = 2. SOLUTION: Agai, the first case is simple sice a 0 = 0 2. Now assume that it s true for, that is, a 2, ad show it s true for +, that is, a + 2. a + = 2 + a = 4 = 2 (d) Prove that lim a exists ad compute it. SOLUTION: Sice the sequece is icreasig ad bouded above, the limit exists. To compute it, we do L = lim a = lim 2 + a = 2 + lim a = 2 + L The L = 2 + L, from which we get that either L = 2 or L =. But all terms of the sequece are positive, so it must be L = 2. 3

7 (4) Cosider the sequece {a } where a = 2 +. (a) Show that {a } is decreasig. SOLUTION: a + = 2( + ) + = < 2 + = a, so {a } is decreasig. You ca also use derivatives to show this. (b) Fid bouds M l ad M u such that M l a M u for every. SOLUTION: Note that 2 + is always positive, so we ca take M l = 0. Also, sice {a } is decreasig, we kow that the biggest term will be the first, a 0 =, so we ca take M u =. (c) Show that lim a exists without computig it. The compute it. SOLUTION: From parts (a) ad (b) we kow that {a } is a decreasig sequece that s also bouded below, so it must coverge. To compute it, lim a = lim 2 + = 0. 4

8 (5) Use the fact that si as to fid the limit of a = ( si ). SOLUTION: Multiply ad divide by + si(/): ( ) a = si + si + si ( ( = si )) + si = si + si = + 0 = 2 5

Infinite Sequences and Series

Infinite Sequences and Series Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet