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.

  • $\begingroup$ 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). $\endgroup$ – Gribouillis Aug 12 '17 at 7:46
  • $\begingroup$ @Gribouillis i will add what i found in the question. editing... $\endgroup$ – Shobhit Aug 12 '17 at 7:47
  • $\begingroup$ @Gribouillis check now please $\endgroup$ – 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.

  • $\begingroup$ May you show how to use induction to show that the coefficients agrees up to degree $n$ to the Taylor coefficients? $\endgroup$ – user99914 Aug 12 '17 at 8:09
  • 2
    $\begingroup$ @JohnMa see my edit above. $\endgroup$ – Gribouillis Aug 12 '17 at 9:07

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