Finding the Taylor Series Expansion of sin x about the point x = pi/4

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SUMMARY

The discussion focuses on expanding the sine function, specifically sin x, around the point x = π/4 using Taylor Series Expansion. Participants emphasize the importance of recognizing the derivatives of sin(x) and cos(x) at π/4 to identify a pattern in the series. The hint provided in the problem suggests using cos(y + π/4), where y is a small variable, to facilitate the expansion. Ultimately, contributors agree that separating the summation into even and odd powers is a viable approach, while also exploring the relationship between the derivative patterns and the sine function's periodicity.

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Homework Statement



Expand sin x about the point x = pi/4. Hint: Represent the function as
sin x = cos (y + pi/4) and assume y to be small

Homework Equations



Taylor Series Expansion

f(x)= f(a) + f'(a) (x-a)+ (1/2!) f''(a) (x-a)^2+ (1/3!) f"'(a) (x-a)^3+...+ 1/n! f(n)(a) (x-a)^n

The Attempt at a Solution



I know how to do the Taylor expansion... but I believe he wants us to use the hint in order to write it as an infinite summation with a sigma notation.. like how we can write e^x equal to an infinite series in sigma notation. Does anybody have any ideas?
 
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Well, first figure out what the derivatives of sin(x) and cos(x) are at pi/4. That should give you a pattern that you can then use.
 
Yes I have done that and I am able to create a taylor expansion at pi/4. However since the expansion is not at zero, you will get a pattern like this: two positive terms, two negative terms, two positive terms, two negative terms... and so on.
I know about using an alternator such as (-1)^n to create an alternating pattern in the summation... however I could not find anything that would create the pattern that I have just described.
The only thing I can think of that has that pattern is the powers of i. However if I use the powers of i.. then i have extra i's for every other term that i can not get rid of.
And I do not understand the hint. Where did y come from? is it now a function of y? what happened to x?... and why cosine?
I hope I am being clear.
Thanks for the help
 
I wouldn't bother using the hint to be honest... and I think the best you can do here is write two separate summations: one for the even powers and one for the odd powers.
 
AlexChandler said:
Yes I have done that and I am able to create a taylor expansion at pi/4. However since the expansion is not at zero, you will get a pattern like this: two positive terms, two negative terms, two positive terms, two negative terms... and so on.
I know about using an alternator such as (-1)^n to create an alternating pattern in the summation... however I could not find anything that would create the pattern that I have just described.
The only thing I can think of that has that pattern is the powers of i. However if I use the powers of i.. then i have extra i's for every other term that i can not get rid of.
And I do not understand the hint. Where did y come from? is it now a function of y? what happened to x?... and why cosine?
I hope I am being clear.
Thanks for the help

Char. Limit said:
I wouldn't bother using the hint to be honest... and I think the best you can do here is write two separate summations: one for the even powers and one for the odd powers.

No, you can do better than separate summations. When you look at the derivative patterns, which are more complicated than (-1)n = cos(n pi), think about comparing your pattern with

\sin(\frac \pi 4 + \frac{n\pi}{2})

Btw, I don't understand his hint either.
 
LCKurtz said:
No, you can do better than separate summations. When you look at the derivative patterns, which are more complicated than (-1)n = cos(n pi), think about comparing your pattern with

\sin(\frac \pi 4 + \frac{n\pi}{2})

Ahh! Thank you! You know.. I think I could have figured that out.
Haha I feel a little stupid now.
 
LCKurtz said:
No, you can do better than separate summations. When you look at the derivative patterns, which are more complicated than (-1)n = cos(n pi), think about comparing your pattern with

\sin(\frac \pi 4 + \frac{n\pi}{2})

AlexChandler said:
Ahh! Thank you! You know.. I think I could have figured that out.
Haha I feel a little stupid now.

No need to feel embarrassed. Tricks like that always seem obvious once you've seen them. Next problem like that you will figure it right out.
 
Aha I thought of another one that would work.

Sgn(in)
 
haha, you are in my class. I was having trouble with that problem too.
 

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