272 Views
292 Views
195 Views
282 Views
241 Views
221 Views
221 Views
Let's discuss another function, which is the expansion with the help of Taylor series for \(\sin { x } \) and \(\cos { x } \). If we want to write \(\sin { x } \), then according to the expansion it is $$ \sin { x } =x-\frac { { x }^{ 3 } }{ 3! } +\frac { { x }^{ 5 } }{ 5! } -\frac { { x }^{ 7 } }{ 7! } +...=\sum _{ n }{ { \left( -1 \right) }^{ n }\frac { { x }^{ 2n+1 } }{ \left( 2n+1 \right) ! } } $$ The approximation of this series for \(x <<\) is $$ \sin { x } \sim x $$ It can be seen in the plot.
Let's start with \(\cos { x } \) now. $$ \cos { x } =1-\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 4 } }{ 4! } -\frac { { x }^{ 6 } }{ 6! } +...=\sum _{ n }{ { \left( -1 \right) }^{ n }\frac { { x }^{ 2n } }{ \left( 2n \right) ! } } $$ For \(x<<\) $$ \cos { x } \sim 1 $$ Now, we will have to follow mathematics here. For example if I say where a point \(a\) is in xy- coordinates? We can answer this question geometrically but how does mathematics do it? It says that there is a point let's say point \(a\) and suppose a point \(b\) somewhere. When the difference between these two points \(a\) and \(b\) will go to zero, then this will be actually point \(a\). $$ \left( b-a \right) \rightarrow 0 $$ But exact 0 is not possible so it will always be closer to zero. The more I get closer to zero, I will be exactly at point a but it depends on how much accuracy we meet.
That's why when we expand a function, we see that the more number of polynomials we go, we may get the function, it always depends on the accuracy. For less accuracy, we just approximate the functions.
We use these approximations in Physics a lot, for example when we calculate the simple pendulum equation, the equation for time period, you consider over there. $$ \sin { \theta } \sim \theta $$ If I give the pendulum more amplitude, \(\theta \) is no more very small. The time period that I will measure with the help of stop watch will be different from my equation, because in my equation I have used some approximation, then I've derived that equation. They will never match but if I will keep the amplitude very small, then the stopwatch measurement will match my theoretical calculations.
310 Views
330 Views
349 Views
345 Views
332 Views
Apply to join the passionate instructors who share their expertise and knowledge with the world. You'll collaborate with some of the industry's best producers, directors, and editors so that your content is presented in the best possible light..