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
sorry i'm not guessing i'm just confused and need help.
stop guessing and look up the series involved to confirm your work.
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?
D. None of these
The correct series for g(x) = tan⁻¹(x) using the series for f(x) = 1 / (1 + x²) is not given in any of the options. The correct series is:
g(x) = x - x³/3 + x⁵/5 - x⁷/7 + ...
This is the Maclaurin series expansion for arctan(x).
To find the series for \(g(x) = \tan^{-1}(x)\) using the series for \(f(x) = \frac{1}{1+x^2}\), we can start by recalling the power series expansion for \(\tan^{-1}(x)\):
\[\tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \ldots\]
Now, we need to manipulate this series to match the form of \(f(x) = \frac{1}{1+x^2}\).
We know that
\[\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \ldots\]
We can rewrite this as:
\[\frac{1}{1+x^2} = 1 - (x^2 - x^4 + x^6 - \ldots)\]
Notice that the term inside the parentheses is the negative of the odd-powered terms in \(\tan^{-1}(x)\). So, if we substitute \(-x^2\) for \(x\) in the series for \(\tan^{-1}(x)\), we can get the desired series for \(g(x)\):
\[g(x) = -x^2 + \frac{x^4}{3} - \frac{x^6}{5} + \frac{x^8}{7} - \ldots\]
Comparing this series with the answer options:
A. \(C + x - \frac{x^3}{3} + \frac{x^5}{5} - \ldots\) - The terms in this series match the series for \(\tan^{-1}(x)\), not the desired series for \(g(x)\). Therefore, Option A is incorrect.
B. \(C + x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots\) - The terms in this series do not match the series for \(\tan^{-1}(x)\) or the desired series for \(g(x)\). Therefore, Option B is incorrect.
C. \(C + 2 - 2x - 3x^2\) - This series does not match either the series for \(\tan^{-1}(x)\) or the desired series for \(g(x)\). Therefore, Option C is incorrect.
D. None of these - This is the correct answer since none of the given options match the desired series for \(g(x)\).
Therefore, the correct answer is D. None of these.
Recall that ∫ 1/(1+x^2) dx = arctan(x)
so integrate the series for 1/(1+x^2) term by term