- #1
dimension10
- 371
- 0
I have noticed something strange when you take the value of sin(pi*10^-n). It approaches pi*10^-n. I have attatched the file here.
What is the Relationship Between Sine and Inverse Pi as n Increases?
Click For Summary
Discussion Overview
The discussion revolves around the relationship between the sine function and the value of pi as n increases, specifically examining the behavior of sin(pi*10^-n) as n approaches infinity. The scope includes mathematical reasoning and conceptual clarification regarding limits and approximations in calculus.
Discussion Character
- Exploratory
- Mathematical reasoning
- Conceptual clarification
Main Points Raised
- One participant observes that sin(pi*10^-n) approaches pi*10^-n as n increases, suggesting a potential relationship.
- Another participant generalizes this observation, stating that for small values of x, sin(x) approximates x closely, and cites the limit lim_{x→0} (sin(x)/x) = 1 as a supporting result.
- A follow-up comment questions whether sin(0)/0 equals 1, leading to a clarification that this expression is undefined, but emphasizes that sin(x) approaches x as x approaches 0.
- There is a reiteration that the limit does not imply division by zero is valid, but rather describes behavior as x approaches zero.
Areas of Agreement / Disagreement
Participants generally agree on the behavior of sin(x) as x approaches zero, but there is a misunderstanding regarding the interpretation of sin(0)/0, which leads to clarification rather than consensus on that specific point.
Contextual Notes
The discussion includes nuances regarding the definition of limits and the behavior of functions near zero, which may not be fully resolved or understood by all participants.
- #2
- 22,170
- 3,333
Hi dimension10! 
Your result can be generalized. Indeed, if x is small then
[itex]\sin(x)\sim x[/itex]
So for small values of x, we will have that x approximates sin(x) quite closely.
The precise result is
[itex]\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}=1[/itex]
which can be proved by geometric methods. See http://www.khanacademy.org/video/proof--lim--sin-x--x?playlist=Calculus to see how to derive the result.
Your result can be generalized. Indeed, if x is small then
[itex]\sin(x)\sim x[/itex]
So for small values of x, we will have that x approximates sin(x) quite closely.
The precise result is
[itex]\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}=1[/itex]
which can be proved by geometric methods. See http://www.khanacademy.org/video/proof--lim--sin-x--x?playlist=Calculus to see how to derive the result.
Last edited by a moderator:
- #3
dimension10
- 371
- 0
micromass said:Hi dimension10!
Your result can be generalized. Indeed, if x is small then
[itex]\sin(x)\sim x[/itex]
So for small values of x, we will have that x approximates sin(x) quite closely.
The precise result is
[itex]\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}=1[/itex]
which can be proved by geometric methods. See http://www.khanacademy.org/video/proof--lim--sin-x--x?playlist=Calculus to see how to derive the result.
So that just means that sin(0)/0=1, right?
Last edited by a moderator:
- #4
- 22,170
- 3,333
dimension10 said:
No, not at all, since you cannot divide by 0. What
[tex]\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}=1[/tex]
mean is, if x is very close to 0 (but not equal to 0!), then [itex]\frac{\sin(x)}{x}[/itex] comes very close to 1.
Thus if x is very close to 0, then sin(x) comes very close to x!
The statement sin(0)/0 makes no sense, since division by 0 is not allowed!