Trigonometry
How is sin(y) when y is in radians equal to y - (y^3)/3! - (y^5)/5! .......?
I know it gives the correct answer but I wanna know how it is derived...
Thanks in advance...
I know it gives the correct answer but I wanna know how it is derived...
Thanks in advance...
Original post by Chrysoberyl Rove
How is sin(y) when y is in radians equal to y - (y^3)/3! - (y^5)/5! .......?
I know it gives the correct answer but I wanna know how it is derived...
Thanks in advance...
I know it gives the correct answer but I wanna know how it is derived...
Thanks in advance...
https://en.wikipedia.org/wiki/Taylor_series
Original post by Chrysoberyl Rove
How is sin(y) when y is in radians equal to y - (y^3)/3! - (y^5)/5! .......?
I know it gives the correct answer but I wanna know how it is derived...
Thanks in advance...
I know it gives the correct answer but I wanna know how it is derived...
Thanks in advance...
and the third term is +... rather than - ...
That's called the power series (or Taylor series) for the sine function. It's not really possible to give a rigorous derivation without appealing to university level maths (analysis), which I'm not sure whether you're familiar with. But generally speaking, most smooth functions - that is, functions that can be differentiated as many times as you like - are 'analytic' in the sense that you can write
or more generally,
for any number (the above case is ). These expressions will typically hold true for all in some interval around , but in the case of sine it actually holds for all . You can check the values of the derivatives of sine at 0 to verify this.
The rigorous derivation uses something called the 'Mean Value Theorem' which is one of the most important results in calculus (and is actually quite intuitive).
or more generally,
for any number (the above case is ). These expressions will typically hold true for all in some interval around , but in the case of sine it actually holds for all . You can check the values of the derivatives of sine at 0 to verify this.
The rigorous derivation uses something called the 'Mean Value Theorem' which is one of the most important results in calculus (and is actually quite intuitive).
Original post by S2JN
That's how you define sine.
It is one way to define it, yes, but certainly not the most enlightening way. You can't read off from the series representation that sine is periodic or oscillates between -1 and 1 which are the key features. The vast majority of mathematicians define it geometrically with the unit circle and then derive the series representation from there. Defining it as the series and proving some of the standard properties does come up in exercise sometimes though.
Original post by Legomenon
It is one way to define it, yes, but certainly not the most enlightening way. You can't read off from the series representation that sine is periodic or oscillates between -1 and 1 which are the key features. The vast majority of mathematicians define it geometrically with the unit circle and then derive the series representation from there. Defining it as the series and proving some of the standard properties does come up in exercise sometimes though.
No, they don't. The power series definition is the standard definition.
Original post by S2JN
No, they don't. The power series definition is the standard definition.
I'm not sure what you're talking about. Almost no one teaches trigonometry by first introducing the power series. If you mean at research level then you can't say there is a standard definition because it is taken for granted that the definitions are equivalent.
Edit: I suppose you mean because it extends to complex number inputs, which is reasonable enough. I meant the way that it is most commonly introduced. Certainly in the context of this thread it is clear that sine is not being defined as the series.
(edited 3 years ago)
Original post by Legomenon
I'm not sure what you're talking about. Almost no one teaches trigonometry by first introducing the power series. If you mean at research level then you can't say there is a standard definition because it is taken for granted that the definitions are equivalent.
Edit: I suppose you mean because it extends to complex number inputs, which is reasonable enough. I meant the way that it is most commonly introduced. Certainly in the context of this thread it is clear that sine is not being defined as the series.
Edit: I suppose you mean because it extends to complex number inputs, which is reasonable enough. I meant the way that it is most commonly introduced. Certainly in the context of this thread it is clear that sine is not being defined as the series.
To be fair, your previous response referred to the "vast majority of mathematicians", which is a bit different from considering how trigonometry is taught!
In school it is conventional to introduce trigonometry from the point of view of the right-angled triangle, followed by an extension to the unit circle definition. But a mathematician's definition will start from the power series, because that's what allows you to derive the full range of analytic properties of the function with a rigorous foundation
Original post by davros
To be fair, your previous response referred to the "vast majority of mathematicians", which is a bit different from considering how trigonometry is taught!
In school it is conventional to introduce trigonometry from the point of view of the right-angled triangle, followed by an extension to the unit circle definition. But a mathematician's definition will start from the power series, because that's what allows you to derive the full range of analytic properties of the function with a rigorous foundation
In school it is conventional to introduce trigonometry from the point of view of the right-angled triangle, followed by an extension to the unit circle definition. But a mathematician's definition will start from the power series, because that's what allows you to derive the full range of analytic properties of the function with a rigorous foundation
Yeah that's fair
Thank you everyone for your replies. As I understand it, the power series isn't the best way to define sine, but one of the ways to estimate its value. Thanks again =)
Original post by Legomenon
Yeah that's fair
Original post by Chrysoberyl Rove
Thank you everyone for your replies. As I understand it, the power series isn't the best way to define sine, but one of the ways to estimate its value. Thanks again =)
Well, from an analyst's point of view, it precisely is the best way to define sine, because it's the theory of power series (and other results from Analysis) that allow us to make powerful statements about the function. But it's not always easy to reconcile the results from Analysis with the familiar properties learned at school
Oh alright ... Thank you for the clarification . And yeah, you are right about how conflicting it is at times to put together analytical and theoretical properties of something... =)
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