TEAM 1111 If a New Zealand Student Uploads a Video Clip That Goes Viral, How Long Will It Take Before 1% of the World’S Population Has Seen It?
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TEAM 1111 If a New Zealand student uploads a video clip that goes viral, how long will it take before 1% of the world’s population has seen it? Introduction/abstract If a New Zealand student uploads a video clip that goes viral, how long will it take before 1% of the world’s population has seen it? Let us define viral as having an exponential growth phase. The meaning of the word viral stems from the nature of virus propagation[6]; therefore, truly viral phenomena must exhibit exponential growth to qualify as such. Assuming that this (ordinary, non-celebrity, high school) student uploaded his video to YouTube (it is, after all, certainly the most popular video sharing site in the world)[2], one hundredth of the world’s population, at the time of writing, constitutes approximately 70 million people[3]. Supposing—optimistically— that every view represents one unique viewer, this video will need to surpass 70 million views on YouTube. This is no easy feat. This is almost twice as many views as Keyboard Cat and Double Rainbow. Without introducing any mathematical model at all, it is common knowledge that YouTube view rates decline over time after having passed a certain point in their lifetimes. It is also common knowledge that New Zealand content finds it very difficult to achieve world-recognition status. As such, the answer to the question when could easily be ‘never’, in the reasonable future. In the following section, we will introduce the model to demonstrate this fact. Process & method Explain the approach used to arrive at the final conclusion, in a formal and organized fashion. Mention any assumptions made during the process and explain how they might affect the conclusion. Answer the original question. Analysis We may begin our inspection with an examination of the cumulative-view graphs of known viral videos. Gangnam Style [1] This is one of the most classic viral videos in YouTube history. By observing its graph, we may draw some conclusions about the nature of the propagation of viral videos. Section A represents the exponential curve section of the graph, the viral section; point P is a point of inflection, and thereafter the graph resembles a logarithmic curve (section B) for a time as its growth slows, before it takes on a roughly linear view aggregation pattern (section C). We notice, first of all, that the exponential phase is brief in comparison to the stable phases, and that the stable phases account for the bulk of the views that the video eventually achieves. The exponential phase, however, is unique to viral videos: videos like Gentleman, Psy’s second popular YouTube song, has only the stable curve phases. Gentleman[4] The logarithmic-then-linear stable curve seems to be common to many ‘non-viral popular’ YouTube videos, which we may define as videos that achieve high view counts largely due to an existing fanbase or initial hype. Viral propagation—propagation that produces an exponential growth curve in terms of views—only occurs when the fanbase of a particular artist or video is massively and rapidly expanded through channels such as sharing and media attention. Here is an additional view graph for a ‘truly viral’ video: Call Me Maybe[5] Both Gangnam Style and Call Me Maybe are characterized by a sharp initial climb in daily views to a single peak, as shown. This generates the exponential propagation that can be seen in the cumulative view graphs. Gangnam Style (daily views) [2] Call Me Maybe (daily views) [5] On the contrary, ‘non-viral popular’ videos tend to exhibit either a peak almost immediately after posting, as the daily view graphs for Gentleman and Taylor Swift’s Bad Blood demonstrate: Gentleman [4] Bad Blood [7] Modelling Having made these observations, we may model the cumulative view graphs of truly viral videos as follows: 0 ≤ 푡 ≤ 푃: 푒푥푝표푛푒푛푡푎푙 푔푟표푤푡ℎ 푃 ≤ 푡 < 푇: 푙표푔푎푟푡ℎ푚푐 푔푟표푤푡ℎ 푡 ≥ 푇: 푙푛푒푎푟 푔푟표푤푡ℎ where t stands for time after posting. Mathematically, this translates to a piecewise function. To generate a smooth curve with manipulatable variables, we produced the following function*: 푟(푡−푃) −푟푃 푓(푡) = 푘푒 − 푘푒 , 0 ≤ 푡 ≤ 푃 푟푘 푣(푡) = 푔(푡) = ln(푞(푡 − 푃) + 1) + 푘 − 푘푒−푟푃, 푃 ≤ 푡 < 푇 푞 ′ ′ { 푙(푡) = 푔 (푇)푡 + [푔(푇) − 푇푔 (푇)], 푡 ≥ 푇 *Refer to sub-section Derivation for derivation process. This function has 4 variables, other than t, which need to be calibrated. P, the point of inflection. r, k and q, variables which control the horizontal and vertical stretch factors of both graphs. T, the point at which linear growth begins. In real life, these variables represent the share rates of the videos, as well as how long viral fame generally endures before the video settles into a generally slower and more stable mode of view accumulation. These variables can be manipulated so as to represent any truly viral video’s growth curve. In order to calibrate them so that they accurately represent market conditions in New Zealand, we can match them to the view graph taken a video uploaded by a New Zealand student that has recently gone ‘viral’ [9]: “The importance of correctly pronouncing Maori words”: cumulative view graph [8] “The importance of correctly pronouncing Maori words”: selected points “The importance of correctly pronouncing Maori words”: MODEL* P = 2 k = 307558.14 q = 8.55 r = 0.47 *Refer to sub-section Derivation for derivation process. With this model, it takes approximately 81 years to achieve 70 million views. YouTube has only existed for 10 years; we cannot guarantee the accuracy of our model, or really any model, beyond that timeframe. Therefore, for all practical intents and purposes, this video will never achieve viewing by 1% of the world’s population. Derivation 푣(푡) 푟푘 푟(푡−푃) −푟푃 ∴ 푠 = 푙푛(10) 푓(푡) = 푘푒 − 푘푒 , 0 ≤ 푡 ≤ 푃 푞 푟푘 푟푘 = 푔(푡) = ln(푞(푡 − 푃) + 1) + 푘 − 푘푒−푟푃, 푃 ≤ 푡 < 푇 ∴ 푔(푡) = ln(10) × log(푞(푡 − 푃) + 1) + 푐2 푞 푞 푙(푡) = 푔′(푇)푡 + [푔(푇) − 푇푔′(푇)], 푡 ≥ 푇 푟푘 log(10) log(푞(푡 − 푃) + 1) { 푔(푡) = × + 푐2 We shall now attempt to explain the processes that 푞 log(푒) log(10) we took to derive these equations. log(10) 푐푎푛푐푒푙푠 표푢푡 푟푘 Firstly, we assumed that section a of the trend ∴ 푔(푡) = 푙푛(푞(푡 − 푃) + 1) + 푐 푟(푡−푃) 2 followed the equation, 푓(푡) = 푘푒 + 푐1, 0 ≤ 푞 Finally, substituting our two constant values, our 푡 ≤ 푃 where 푐1, is a constant, and the section b of the trend to be equations are: 푟(푡−푃) −푟푃 푔(푡) = 푠 log(푞(푡 − 푃) + 1) + 푐2, 푃 ≤ 푡 < 푇 where 푓(푡) = 푘푒 − 푘푒 , 0 ≤ 푡 ≤ 푃 푐 is another constant, contributing to g's horizontal 푟푘 2 푔(푡) = ln(푞(푡 − 푃) + 1) + 푘 − 푘푒−푟푃, 푃 ≤ 푡 < 푇 shift, and 푠 is the vertical stretch of the graph g. Since 푞 the graph is continuous, 푓(푃) = 푔(푃), and thus Now we shall derive the equation for the linear 푓′(푃) = 푔′(푃). We can use these two simple rules to function that we assume the views to grow at a determine the two constants. constant rate or at the same rate of change / (You may notice for this 푔(푡), the equation involves tangenet of the logarithimic graph at a chosen point 푙표푔, but our final equation is in natural logarithm, 푡 = 푇 where we believe that the graph acheives a and you shall see that the base a or 10 log turns into constant rate of growth. a natural log with appropriate simplification.) Equation of a linear equation: 푡 = 푃 푇ℎ푒 푝표푛푡 푤푒 푎푠푠푢푚푒 푡표 ℎ푎푣푒 푎 푐표푛푠푡푎푛푡 푔푟표푤푡ℎ 푠 푇, 푔(푇) 푟(푃−푃) 푓(푃) = 푘푒 + 푐1 푡ℎ푒 푔푟푎푑푒푛푡 푠 푚, 푡ℎ푒 푐표푛푠푡푎푛푡 (ℎ표푟푧표푛푡푎푙 푠ℎ푓푡) 푠 푐 푓(푃) = 푘 + 푐1 푙(푡) = 푚푡 + 푐 ′ 푔(푃) = 푠 log(푞(푃 − 푃) + 1) + 푐2 푚 = 푔 (푇) ′ 푔(푃) = 푠 log(1) + 푐2 퐴푡 푙(푇) = 푔 (푇) × 푇 + 푐 = 푔(푇) ′ 푔(푃) = 푐2 푐 = 푔(푇) − 푔 (푇) × 푇 푓(푃) = 푔(푃) 푙(푡) = 푔′(푇)푡 + [푔(푇) − 푇푔′(푇)], 푡 ≥ 푇 푘 + 푐1 = 푐2 Now we shall apply these equations to our example We shall now sensibly assume that when 푡 = 0, of the Maori video data. 푓(푡) = 0, as it makes sense that there'll be no views at the instant the video is uploaded. 푟(−푃) 푓(0) = 푘푒 + 푐1 −푟푃 푓(0) = 푘푒 + 푐1 = 0 −푟푃 푐1 = −푘푒 −푟푃 We shall let 푡, be time in days, and so we know the ∴ 푐2 = 푘 − 푘푒 points Now we shall now use 푓′(푃) = 푔′(푃) to work our 푠, (1,7200), (2,187500), (3,225000), (5,263000), 푎푛푑 (7, 281750) 푐 , and 푐 : We shall assume 푃 = 2, as it looks reasonable for the 1 2 concavity to change at that point. ′ 푟(푡−푃) 푑 푓 (푡) = 푘푒 × (푟(푡 − 푝)) (푃 = 2, 푦푃 = 187000) 푑푡 푟(2−2) −푟×2 푟(푥−푃) 187000 = 푘푒 − 푘푒 = 푘푒 × 푟 −2푟 푟(푥−푃) 푘 − 푘푒 = 187000 = 푟푘 푒 −2푟 ∴ 푓′(푃) = 푟푘 푘(1 − 푒 ) = 187000 [1] 푑 (1,7200) ′ 푟(1−2) −푟×2 푔 (푡) = 푠 × [log(푞(푡 − 푃) + 1) + 푐2] 72000 = 푘푒 − 푘푒 푑푡 −푟 −2푟 ′ 푞 72000 = 푘푒 − 푘푒 푔 (푡) = 푠 × −푟 −푟 (푞(푡 − 푃) + 1)푙푛(10) 푘푒 (1 − 푒 ) = 72000 [2] 푠푞 [2] 푔′(푡) = (푞(푡 − 푃) + 1)푙푛(10) [1] 푠푞 푘푒−푟(1 − 푒−푟) 72000 푔′(푃) = = 푙푛(10) 푘(1 − 푒−2푟) 187000 ′ ′ 퐴푝푝푙푦푛푔 푓 (푃) = 푔 (푃): 푒−푟(1 − 푒−푟) 72 푠푞 = 푟푘 = (1 − 푒−푟)(1 + 푒−푟) 187 ln(10) 푒−푟 72 = ( −푟) 푟푘 1 + 푒 187 푔′(푡) = 72 72 (푞(푡 − 푃) + 1) 푒−푟 = + 푒−푟 187 187 ′ 푟푘 115 ∴ 푔 (푡) = 푒푟 = (푞(푡 − 2) + 1) 72 (퐴)(퐵) 115 ∴ 푔′(9) = 푟 = ln ( ) = 0.468266 → (퐴) (8.55(9 − 2) + 1) 72 푔′(9) = 2366.788 187000 푊푒 푘푛표푤 푡ℎ푎푡 푙(푡) = 푔′(푇)푡 + [푔(푇) − 푇푔′(푇)], 푡 푘 = −2푟 (1 − 푒 ) ≥ 푇 푘 = 307558.1396 → (퐵) 푙(푡) = 2366.788 푡 + [256203.44 − 9 × (2366.788)] 푙(푡) = 2366.788푡 + 234902.352 We shall now work out the value of q of the log Now we need to work out the number of days graph.