How to write taylor series in sigma notation

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The discussion focuses on how to express Taylor series in sigma notation, particularly for functions like sinh(x), sin(x), and 1/(x-1). The user successfully derived the Taylor series for sinh(x) but struggled with others, seeking guidance on general methods. They clarified their calculations for 1/(x-1) and expressed confusion about converting their series into sigma notation, specifically regarding the term 3^(n+1). The conversation also addressed the use of alternating signs in series, explaining the difference between (-1)^n and (-1)^(n+1). Overall, the user is looking for a clearer understanding of Taylor series representation and notation.
haflanagan
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Homework Statement


My Calc II final is tomorrow, and although we never learned it, it's on the review.
So I have a few examples. Some I can figure out, some I cant.

Examples: f(x)=sinh(x), f(x)=ln(x+1) with x0=0, f(x)=sin(x) with x0=0, f(x)=1/(x-1) with x0=4
The only one of those that I was able to figure out was sinh(x). The rest I don't get. I don't need someone to show me how to do all of them, but maybe just show me how to do these types of problems in general using one of those as an example?

I am really just completely lost here. I could definitely use some help.
Thank you!
And sorry if I did something wrong in terms of posting this thread, I have never used Physics Forum before.
 
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I'd be interested in knowing how you got sinh(x) and couldn't get sin(x). They are really pretty similar. Are you having trouble representing the alternating sign?
 
My bad...I actually did get sin(x)!
Okay, maybe I should be more specific.
f(x) = 1/(x-1), x0=4
f'(x) = -1/((x-1)^2)
f"(x) = 2/((x-1)^3)
f'''(x) = -6/((x-1)^4)

Plugging in 4:
f(x) = 1/3
f'(x) = -1/9
f"(x) = 2/27
f'''(x)=-6/81

So the taylor series is 1/3 + -(x-4)/9 + 2(x-4)^2/54 + -6(x-4)^3/486...

Now I just found in my notes that the format for sigma notation should be the sum from n=0 to infinity of the n-th derivative of f(a) * (x-a)^n/n!
But I just have no idea how to take that taylor series and get it into that format.
I think my brain is dead from doing too much math :(
 
Okay well the answer to that last one should be (-1)^n * (x-4)^n / 3^(n+1)
I get the (-1)^n (because the signs are alternating) and I get the (x-4)^n. But what I do not get is where the 3^(n+1) is coming from.
Also, how do you know when do use (-1)^n and when to use (-1)^(n+1)? What's the difference?
 
haflanagan said:
...

Also, how do you know when do use (-1)^n and when to use (-1)^(n+1)? What's the difference?
(-1)n is positive for even n, negative for odd n.

(-1)n+1 is positive for odd n, negative for even n.
 
Ahh okay that makes sense!
Thanks!
Well...at least if I make an educated guess I will probably be able to get a least partial credit...
 

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