calculuspower series
I recently had a problem. I know how to evaluate power series but I cannot seem to find an expansion for $\sqrt{x+1}$.
I've tried differentiating it, in order to bring it in reciprocal form but that didn't help. Due to the presence of square root, I cannot change it in the form of $1/(x+1)$.
Kindly help.
The first thing I should do is to make $x^2=t$ and to consider $$f(t)=\tan ^{-1}\left(\frac{1-t}{2+t}\right)$$ So, as you thought about it, differentiate $$f'(t)=-\frac{3}{2 t^2+2t+5}$$ Computing the complex roots of the denominator and using partial fraction decomposition $$f'(t)=\frac{i}{2 \left(t+\left(\frac{1}{2}-\frac{3 i}{2}\right)\right)}-\frac{i}{2 \left(t+\left(\frac{1}{2}+\frac{3 i}{2}\right)\right)}$$ $$f'(t)=\sum_{n=0}^\infty -\frac{i}{2} \left(-\frac{1}{5}-\frac{3 i}{5}\right)^{n+1} t^n-\sum_{n=0}^\infty -\frac{i}{2} \left(-\frac{1}{5}+\frac{3 i}{5}\right)^{n+1} t^n$$ $$f'(t)=\sum_{n=0}^\infty \frac{i}{2}\left(\left(-\frac{1}{5}+\frac{3 i}{5}\right)^{n+1} -\left(-\frac{1}{5}-\frac{3 i}{5}\right)^{n+1}\right)t^n$$ Using de Moivre, this write $$f'(t)=\sum_{n=0}^\infty -\left(\frac{5}{2}\right)^{-\frac{n+1}{2}} \sin \left((n+1) \left(\pi -\tan ^{-1}(3)\right)\right) t^n$$ Expanding the sine $$f'(t)=\sum_{n=0}^\infty \left(-\sqrt{\frac{2}{5}}\right)^{n+1} \sin \left((n+1) \tan ^{-1}(3)\right)\, t^n=\sum_{n=0}^\infty a_n t^n$$ Do not worry about the $a_n$'s; they are rational numbers generating the sequence $$\left\{-\frac{3}{5},\frac{6}{25},\frac{18}{125},-\frac{96}{625},\frac{12}{3125}, \frac{936}{15625},-\frac{1992}{78125},-\frac{5376}{390625},\frac{30672}{1953125},- \frac{7584}{9765625},\cdots\right\}$$ So, integrating termwise $$f(t)=\tan ^{-1}\left(\frac{1}{2}\right)+\sum_{n=0}^\infty \frac {a_n}{n+1} t^{n+1}$$ Back to $x$ $$x \tan ^{-1}\left(\frac{1-x^2}{x^2+2}\right)=\tan ^{-1}\left(\frac{1}{2}\right)x+\sum_{n=0}^\infty \frac {a_n}{n+1} x^{2n+3}=\tan ^{-1}\left(\frac{1}{2}\right)x+\sum_{n=0}^\infty b_n x^{2n+3}$$ The $b_n$'s generate the sequence $$\left\{-\frac{3}{5},\frac{3}{25},\frac{6}{125},-\frac{24}{625},\frac{12}{15625}, \frac{156}{15625},-\frac{1992}{546875},-\frac{672}{390625},\frac{3408}{1953125} \right\}$$
$$e^z=\sum_{n=0}^\infty \frac{z^n}{n!}\qquad\forall z\in\mathbb{C}$$
and so:
$$e^{-x^2}=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{n!}$$
the problem is that the terms alternate between positive and negative, so if you cut the series off at any arbitrary point (which is going to be small compared to the infinite number of terms). To demonstrate this point look at the graph below:
The red line represents the actual $\exp(-x^2)$ function whilst blue is $n\in[0,5]$ and green is $n\in[0,10]$ and for your integral you want to integrate this series and then take the limit for $n\to\infty$ whilst will just accentuate any "small" error you see here
Best Answer
Here's one way. Start with the expansion you want, using $a_0$, $a_1$, $a_2$, $a_3$, etc. for the unknown coefficients:
$$\sqrt{x+1}\;=\;a_{0}\;+\;a_{1}x\;+\;a_{2}x^2\;+\;a_{3}x^3\;+\;a_{4}x^4\;+\;a_{5}x^5\;+ ...$$
Finding $a_0$: Plugging in $x=0$ on both sides leads to $a_{0}=1$.
Finding $a_1$: Differentiate both sides of the expansion. This gives
$$\frac{1}{2}\left(x+1\right)^{-\frac{1}{2}}\;\;=\;\;a_{1}\;+\;2a_{2}x\;+\;3a_{3}x^2\;+\;4a_{4}x^3\;+\;5a_{5}x^4\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{1}=\frac{1}{2}$.
Finding $a_2$: Differentiate 2-times both sides of the expansion. This gives
$$\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{3}{2}}\;\;=\;\;2a_{2}\;+\;(2)(3)a_{3}x\;+\;(3)(4)a_{4}x^2\;+\;(4)(5)a_{5}x^3\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{2}=-\frac{1}{8}$.
Finding $a_3$: Differentiate 3-times both sides of the expansion. This gives
$$\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{5}{2}}\;\;=\;\;(2)(3)a_{3}\;+\;(2)(3)(4)a_{4}x\;+\;(3)(4)(5)a_{5}x^2\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{3}=\frac{1}{16}$.
Finding $a_4$: Differentiate 4-times both sides of the expansion. This gives
$$\left(-\frac{5}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{7}{2}}\;\;=\;\;(2)(3)(4)a_{4}\;+\;(2)(3)(4)(5)a_{5}x\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{4}=-\frac{5}{128}$.
Finding $a_5$: Differentiate 5-times both sides of the expansion. This gives
$$\left(-\frac{7}{2}\right)\left(-\frac{5}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{9}{2}}\;\;=\;\;(2)(3)(4)(5)a_{5}\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{5}=\frac{7}{256}$.
Keep going to get as many coefficients as you want. If you keep careful track of the numbers without reducing the fractional expressions for the coefficients, you can easily determine a pattern (a pattern that can be proved by mathematical induction if you're so inclined).