Sin [Pi. (x-1)] / x-1; for x =1,2,3, .N

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Discussion Overview

The discussion revolves around the expression Sin [Pi. (x-1)] / (x-1) for integer values of x ranging from 1 to N. Participants explore the behavior of this expression, particularly at the point where x equals 1, and consider the implications of limits and undefined values.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant poses the question about the value of Sin [Pi. (x-1)] / (x-1) for integer values of x.
  • Another participant notes that for x > 1, the expression can be evaluated directly, while for x = 1, it results in an undefined form due to division by zero, suggesting the use of limits.
  • A third participant reiterates the point about the undefined nature at x = 1 and suggests that for all x values, the result will be '0'.
  • A fourth participant clarifies that for integer values of x that are at least 2, the expression evaluates to 0 because sin(n pi) equals 0 for all integers n, while noting that for n = 1, the expression leads to the indeterminate form 0/0.

Areas of Agreement / Disagreement

Participants generally agree on the undefined nature of the expression at x = 1 and the evaluation of the expression for integer values greater than 1. However, there is some uncertainty regarding the interpretation of the limit and the behavior of the expression as x approaches 1.

Contextual Notes

The discussion includes assumptions about the integer nature of x and the implications of evaluating limits, particularly at the point of division by zero.

pjunky
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Sin [Pi. (x-1)] / x-1; for x =1,2,3,...N

Sin [ Pi. (x-1) ] / (x-1) = ?

for x=1,2,3,4,...N

I think for Sin (x) / x =1

but, what about that.
 
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For x > 1 it is rather obvious, you can just plug in the value.
For x = 1 you are dividing by 0, so it is undefined. You can however take the limit, as you are indicating.
 


CompuChip said:
For x > 1 it is rather obvious, you can just plug in the value.
For x = 1 you are dividing by 0, so it is undefined. You can however take the limit, as you are indicating.

for all X values will I get '0'
 


I am assuming you mean: for all X values which are integer and at least 2. Then you are right. sin(n pi) = 0 for all integers n, and since you are dividing by something non-zero for n non-zero, you have sin(n pi)/n = 0/something = 0.

For n = 1, you get sin((n - 1) pi) / (n - 1) which looks like 0/0 if you plug in the numbers.
 

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