# when are the successive approximations using picard's method for solving an ODE, are the terms of the taylor expansion of the solution of the ODE

The question is as stated in the title, as is "when are the successive approximations using picard's method for solving an ODE, are the terms of the taylor expansion about $x=0$ of the solution of the ODE".

In this question : Solve $y'=y$, $y(0)=1$ using method of successive approximations, obtaining the power series expansions of the solution, the approximations are indeed the terms of the taylor expansion about $x=0$ of the solution.

But, when solving $y' = y^2, y(0)=1$, the second approximation is not the same as the taylor expansion about $x=0$ of the solution of the ODE.

first approximation = 1

second approximation = $1+x$

third approximation = $1+x+x^2+x^3/3$

Any help appreciated.

• Can you explain why you think the second approximation is not the same as the Taylor expansion? It seems to me that for $y' = P(x, y)$, if $P$ is a polynomial, the $k$-th approximation should have at least all the Taylor terms up to degree $k$ (by induction). – Gribouillis Aug 12 '17 at 7:46
• @Gribouillis i will add what i found in the question. editing... – Shobhit Aug 12 '17 at 7:47
• @Gribouillis check now please – Shobhit Aug 12 '17 at 7:49

If the equation is $y'(x) = P(x, y(x))$ where $P$ is a polynomial and the approximations are defined by $y_0(x) = y(0)$ and

$$y_n(x) = y(0) + \int_0^x P(t, y_{n-1}(t)) dt$$

then, by induction $y_n(x)$ is a polynomial and its terms up to degree $n$ are equal to the Taylor coefficients of the solution at $0$ (but not necessarily the terms of degree $> n$). Indeed the approximation formula above implies that the term of degree $k$ in $y_n$ depends only or the terms of degrees $1, \cdots, k-1$ in $y_{n-1}$. It means that the term of degree $k$ is stationary for all $n \ge k$. In particular, it is the term of the Taylor series because we know that a solution exists in an interval around $0$ and that this solution is $\mathcal{C}^\infty$.

In your case, the solution is $$y(x) = \frac{1}{1-x} = 1 + x + x^2 + \cdots$$ The $y_2$ approximation has Taylor terms $1+x+x^2$ but the term in $x^3$ will only be equal to the Taylor coefficient in $y_3$.

Edit: A more detailed proof by induction (on John Ma's request)

If $q$ is any polynomial, let us denote $D_k(q)$ its coefficient of degree $k$. Let us consider the polynomial

$$r(x) = y(0)+ \int_0^x P(t, q(t)) dt$$ Then $D_k(r)$ depends only on the terms of degree $< k$ in $q$. It means that there is a function $F_k$ such that $$D_k(r) = F_k(D_0(q),\ldots,D_{k-1}(q))$$ The approximations $y_n$ above are polynomials that satisfy $$D_k(y_{n+1}) = F_k(D_0(y_n),\ldots,D_{k-1}(y_n))$$ Our induction hypothesis $\cal{H}_n$ is that $\forall i\le n$, $D_i(y_n) = D_i(y_i)$.

Obviously, $\cal{H}_0$ is true because the term of degree $0$ in all the $y_n$'s is $y(0)$. Supposing that $\cal{H}_n$ is true, then for all $i\le n+1$, one has $$D_i(y_{n+1}) = F_i(D_0(y_n),\ldots,D_{i-1}(y_n)) = F_i(D_0(y_0),\ldots,D_{i-1}(y_{i-1})) = D_i(y_i)$$ which proves that $\cal{H}_n \Rightarrow \cal{H}_{n+1}$.

Hence, $D_i(y_n)$ does not depend on $n$ when $n \ge i$. One last question remains: why is this term equal to the Taylor series term of degree $i$ in the solution? For this, let $T_k(x)$ be the Taylor polynomial up to order $k$ in the solution. One can write $y(x) = T_k(x) + x^k \epsilon(x)$ where $\epsilon(x)\to 0$ when $\epsilon \to 0$. By using $$T_k(x) + x^k \epsilon(x) = y(0) + \int_0^x P(t, T_k(t) + t^k \epsilon(t)) dt$$ it is easy to see that the coefficients of $T_k$ must satisfy $D_i(T_k) = F_i(D_0(T_k),\ldots,D_{i-1}(T_k))$ when $i\le k$. It follows easily that the coefficients of $T_k$ must be the same as the coefficients given by the approximation process.

• May you show how to use induction to show that the coefficients agrees up to degree $n$ to the Taylor coefficients? – user99914 Aug 12 '17 at 8:09
• @JohnMa see my edit above. – Gribouillis Aug 12 '17 at 9:07