Proving sin^0 (x): Is it 0, x, or Something Else?

  • Thread starter Thread starter PhysForumID
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around the interpretation of the expression sin^0(x) and its implications in the context of proving a mathematical statement involving integrals of sin^n(x). The participants explore the meaning of applying the sine function zero times and how this relates to the broader problem of induction starting from n=0.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question the meaning of sin^0(x) and whether it should equal 0, x, or 1. There is also a discussion about the definition of natural numbers and the implications for the variable n in the context of the problem. Some participants express confusion over the notation used for sine functions raised to powers and the distinction between functional composition and multiplication.

Discussion Status

The conversation is ongoing, with participants actively questioning assumptions and clarifying definitions. Some guidance has been offered regarding the properties of sine functions and the notation used, but there is no explicit consensus on the interpretation of sin^0(x) or the classification of natural numbers.

Contextual Notes

There is a noted confusion regarding the definition of natural numbers, particularly whether 0 is included. The integral I_n is given, and participants are tasked with showing a relationship between I_0, I_1, I_2, etc., but the specifics of the integral's properties are still under discussion.

PhysForumID
Messages
7
Reaction score
0
What would sin^0 (x) mean? sin^n (x) means using the sine function 'n' times on x, so what does it mean to use it zero times? does sin^0 (x) then equal 'x' or '0' or... ?

The context of this question is that I have to prove that:
I_n = integral from zero to pi/2 of sin^n (x) with respect to 'x'

I am proving this by induction starting with n=0, assuming true for n=n and showing it is true for n=n+1
 
Physics news on Phys.org
sin^{0}(x) = (sin0)^{0} = 1
Is n a natural number? Then start with n = 1.
 
sorry maybe I've got more than that confused in my head... I always understand sinn (x) to mean you use the function 'sine' 'n' times on x, rather than take sin(x) and multiply it by 'n'... am I wrong there? Surely not because sin(sin(pi/2)) >< {sin(pi/2)}^2

and yes n is a natural number but starting from 0
 
PhysForumID said:
sorry maybe I've got more than that confused in my head... I always understand sinn (x) to mean you use the function 'sine' 'n' times on x, rather than take sin(x) and multiply it by 'n'... am I wrong there? Surely not because sin(sin(pi/2)) >< {sin(pi/2)}^2

sin^{2}(x) is shorthand for sin(x)sin(x), and so on for arbitrary n. A value raised to the power of 0 equals 1 due to the fact that x^{n} = x*x^{n-1}, so...
x^{0} = x*x^{-1} = x*\frac{1}{x} = 1

and yes n is a natural number but starting from 0

Zero isn't a natural number. This is me being pedantic, of course, and you can still begin with n=0 if you like. Can you elaborate on what you're trying to prove? What in the integral supposed to equal?
 
Last edited:
wow I have no idea how I got this far in uni making that mistake about what sin^2(x) was... thanks number nine :)

and sorry, that was my mistake for saying it was a natural number. n = {0,1,2...}

the integral is given and we have to show that I_0 > I_1 > I_2 > ... etc
 
Can you show that \sin^{n}(x) &gt; \sin^{n+1}(x) for all n? Once you do, can you see how to use this to solve the problem?

Also, I'd like to say that I think the notation \sin^2(x) to mean (\sin(x))^2 is very unfortunate. It is often the case that f^2(x) is taken to mean f(f(x)) as you had thought, PhysForumID. This is almost always the case with the exponent -1, since f^{-1} usually denotes the inverse of f with respect to functional composition, not multiplication. One great confusion people often have while learning trigonometry is that \sin^2(x) = (\sin(x))^2, but \sin^{-1}(x) \neq (\sin(x))^{-1}. Rather \sin(\sin^{-1}(x))=x, since here the exponent refers to functional composition and not multiplication.

Zero isn't a natural number. This is me being pedantic, of course, and you can still begin with n=0 if you like.

There is no consensus on whether or not 0 is a natural number. From http://en.wikipedia.org/wiki/Natural_number" :
Including 0 is now the common convention among set theorists, logicians, and computer scientists. Many other mathematicians also include 0, although some have kept the older tradition and take 1 to be the first natural number.

You can use either convention as long as you're consistent. If you really want to be unambiguous, you can say "non-negative integers" and "positive integers."
 
Last edited by a moderator:
Moderator's note: thread moved from "General Math" to "Homework & Coursework Questions". Rules for homework help are in effect.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
7K
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K