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I will try a standard technique for fiddling with power series.

If $g(t) =\sqrt{f(t)} $, then, differentiating, $g'(t) =\frac {f'(t)}{2\sqrt{f(t)}} =\frac {f'(t)}{2g(t)} $ so $2g'(t)g(t) = f'(t) $.

If $f(t) =\sum_{n \ge 0} a_n t^n $ and $g(t) =\sum_{n \ge 0} b_n t^n $, then $f'(t) =\sum_{n \ge 1} na_n t^{n-1} =\sum_{n \ge 0} (n+1)a_{n+1} t^{n} $ and $g'(t) =\sum_{n \ge 1} nb_n t^{n-1} =\sum_{n \ge 0} (n+1)b_{n+1} t^{n} $.

Multiplying the series for $g(t)$ and $g'(t)$, and equating this to the series for $f'(t)$ you get a recurrence which will allow the coefficients $b_n$ to be iteratively computed.

I try to solve an example for this method for $\sqrt(x)$ but i can't Please help?

behmor67
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1 Answers1

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The product of two power series can be expressed as a single power series by the convolution product. Use that to obtain a general expression for the coefficient of $t^n$ in the power series of $2g'(t)g(t)$, and use the fact that it is equal to $(n+1)a_{n+1}.$

The derivative of $g(t)^2$, expressed as a power series, is

$$2g(t)g'(t)=2\sum_{i=0}^\infty b_it^i\sum_{j=0}^\infty(j+1)b_{j+1}t^j$$

Re-arranging indices with $i+j=n$ gives

$$2g(t)g'(t)=\sum_{n=0}^\infty\left(2\sum_{j=0}^n(j+1)b_{j+1}b_{n-j}\right)t^n$$

Equating this to the expression for $f'(t)$ gives an equality of coefficients term by term

$$(n+1)a_{n+1}=2\sum_{j=0}^n(j+1)b_{j+1}b_{n-j}$$

which permits to evaluate the $b_j$ in terms of the $a_i$ recursively. The first three equations are

$$a_1=2b_0b_1$$ $$2a_2=2b_1^2+4b_2b_0$$ $$3a_3=6b_1b_2+6b_3b_0$$

(This was a quickie, I cordially invite corrections!)

Note that $b_0$ has to be determined independently as $\sqrt a_0$ because taking derivatives destroys information about constant terms. Also note that these (i.e., $a_0$ and $b_0$) cannot be zero because $\sqrt t$ is not analytic in 0.

Justpassingby
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