- #1
dimension10
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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?
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The discussion explores the relationship between the sine function and its argument as it approaches zero, specifically examining the behavior of sin(pi*10^-n). It is noted that for small values of x, sin(x) approximates x closely, leading to the limit that as x approaches zero, sin(x)/x approaches 1. This relationship is confirmed through geometric proofs. A clarification is made that while sin(0)/0 is often mentioned, it is undefined due to division by zero, emphasizing that the limit applies only when x is near zero but not equal to it. Understanding this limit is crucial for grasping the behavior of sine in calculus.
- #2
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Hi dimension10! 
Your result can be generalized. Indeed, if x is small then
\sin(x)\sim x
So for small values of x, we will have that x approximates sin(x) quite closely.
The precise result is
\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}=1
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
\sin(x)\sim x
So for small values of x, we will have that x approximates sin(x) quite closely.
The precise result is
\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}=1
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.
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- #3
dimension10
- 371
- 0
micromass said:Hi dimension10!
Your result can be generalized. Indeed, if x is small then
\sin(x)\sim x
So for small values of x, we will have that x approximates sin(x) quite closely.
The precise result is
\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}=1
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?
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- #4
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dimension10 said:
No, not at all, since you cannot divide by 0. What
\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}=1
mean is, if x is very close to 0 (but not equal to 0!), then \frac{\sin(x)}{x} 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!
Hello!
In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way:
I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true?
And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...