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Alternating Series estimation theorem vs taylor remainder

1. The problem statement, all variables and given/known data

Let Tn(x) be the degree n polynomial of the function sin x at a=0. Suppose you approx f(x) by Tn(x) if abs(x)<=1, how many terms are need (what is n) to obtain an error less than 1/120

2. Relevant equations
Rn(x)=M(x-a)^(n+1)/(n+1)!

sin(x)=sum from 0 to ∞ of (-1)^n*x^(2n+1)/(2n+1)!

3. The attempt at a solution
For the alternating series test, I plugged in 1 for x in the sin x macluarin series and got 1/(2(n+1)+1)! <= 1/120 and got n=1
Then I used the Taylor's remainder theorem, got 1/(n+1)! <= 1/120 so n=4. why am I getting a different answer from Alternating series estimation theorem?
 
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LCKurtz

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1. The problem statement, all variables and given/known data

Let Tn(x) be the degree n polynomial of the function sin x at a=0. Suppose you approx f(x) by Tn(x) if abs(x)<=1, how many terms are need (what is n) to obtain an error less than 1/120

2. Relevant equations
Rn(x)=M(x-a)^(n+1)/(n+1)!

sin(x)=sum from 0 to ∞ of (-1)^n*x^(2n+1)/(2n+1)!

3. The attempt at a solution
For the alternating series test, I plugged in 1 for x in the sin x macluarin series and got 1/(2(n+1)+1)! <= 1/120 and got n=1
Then I used the Taylor's remainder theorem, got 1/(n+1)! <= 1/120 so n=4. why am I getting a different answer from Alternating series estimation theorem?
Let's write out a few terms of the two series. The alternating series is$$
x -\frac{x^3} 3 +\frac {x^5} 5~...$$
The Taylor series is$$
0+x+0-\frac {x^3} 3 + 0 +\frac{x^5} 5~...$$Does that give you an idea?
 

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