Asked by kaa
Use the series for f of x equals 1 over the quantity 1 plus x squared to write the series for g(x) = tan–1(x).
A. C plus x minus x cubed over 3 plus x to the 5th power over 5 minus ...
B. C plus x minus x squared over 2 plus x cubed over 3 minus ...
C. C + 2 – 2x – 3x2 –
D. None of these
A. C plus x minus x cubed over 3 plus x to the 5th power over 5 minus ...
B. C plus x minus x squared over 2 plus x cubed over 3 minus ...
C. C + 2 – 2x – 3x2 –
D. None of these
Answers
Answered by
oobleck
Recall that ∫ 1/(1+x^2) dx = arctan(x)
so integrate the series for 1/(1+x^2) term by term
so integrate the series for 1/(1+x^2) term by term
Answered by
kaa
so B?
Answered by
oobleck
stop guessing and look up the series involved to confirm your work.
Answered by
kaa
sorry i'm not guessing i'm just confused and need help.
Answered by
oobleck
1/(1+x^2) = 1 - x^2 + x^4 - x^6 + ...
integrating that, you get
∫ 1/(1+x^2) dx = x - x^3/3 + x^5/5 - x^7/7 + ...
Now, if you check the series for arctan(x), you get
arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + ...
ta-daa! choice A
Now, why does it say C at the front?
integrating that, you get
∫ 1/(1+x^2) dx = x - x^3/3 + x^5/5 - x^7/7 + ...
Now, if you check the series for arctan(x), you get
arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + ...
ta-daa! choice A
Now, why does it say C at the front?
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