If f(x) is a polynomial of degree k and phi(x) is

If $f(x)$ is a polynomial of degree $k$ and $\varphi(x)$ is an arbitrary (sufficiently differentiable) function, can the degree-$n$ Maclaurin polynomial of $f(x)\varphi(x)$ be obtained by multiplying $f(x)$ by the degree-$(n-k)$ Maclaurin polynomial of $\varphi(x)$?

Step 1: What Problem 631 already tells us

Since $f$ is a polynomial of degree $k\le n$, its own degree-$n$ Taylor polynomial at $0$ is exact: $T_1(x)=f(x)$, with zero remainder. Problem 631 then gives, immediately,

\[ T(x) = f(x)\cdot Q_n(x) - P(x), \tag{1} \] where $Q_n(x)$ is the degree-$n$ Maclaurin polynomial of $\varphi$ (not degree $n-k$), and $P(x)$ collects the terms of $f(x)Q_n(x)$ of degree $\gt n$. This is the correct recipe: multiply by the full degree-$n$ series of $\varphi$, then discard the overflow.

Step 2: Testing the proposed shortcut

The question asks about a cheaper recipe: multiply $f$ by only the degree-$(n-k)$ polynomial $Q_{n-k}$, which already has degree exactly $n-k+k=n$ after multiplying by $f$, so no discarding would be needed. Test it on $f(x)=1+x$ ($k=1$), $\varphi(x)=e^x$, $n=3$ (so $n-k=2$):

\[ Q_2(x)=1+x+\frac{x^2}2, \qquad f(x)Q_2(x) = \frac{x^3}2+\frac{3x^2}2+2x+1. \tag{2} \]

But the true degree-3 Maclaurin polynomial of $(1+x)e^x$, computed directly, is

\[ \frac{2x^3}3+\frac{3x^2}2+2x+1. \tag{3} \]

The cubic coefficients disagree ($\tfrac12$ versus $\tfrac23$): the shortcut fails.

Step 3: Why it fails

The missing contribution is exactly the term $P(x)$ from (1): multiplying $f$'s lower-degree coefficients (here, the constant term $a_0=1$) by $\varphi$'s coefficients of degree strictly between $n-k$ and $n$ (here, $\varphi$'s own $x^3$ coefficient, $\tfrac16$) still lands at degree $\le n$ and must be kept. Those coefficients of $\varphi$ live beyond degree $n-k$, so $Q_{n-k}$ never sees them. Only the single term multiplying $f$'s leading coefficient $a_k x^k$ is correctly captured by $Q_{n-k}$; every lower-degree term of $f$ needs more of $\varphi$'s series than $Q_{n-k}$ supplies. Precisely: $\varphi(x)-Q_{n-k}(x)=o(x^{n-k})$, so $f(x)\bigl[\varphi(x)-Q_{n-k}(x)\bigr]=o(x^{n-k})$ — an error of order $n-k$, not the required order $n$, whenever $k>0$.

Step 4: Verification

\[f=1+x,\ \varphi=e^x,\ n=3\text{: naive }fQ_2\text{ at }x=0.1\text{ gives }1.2155,\ \text{true degree-3 poly gives }1.215\overline6,\ \text{true }f\varphi=1.21569\ldots\text{: the naive value is the worse approximation}\ \checkmark\]

\[\text{Monomial check: }f=x\ (k=1,a_0=0),\ \varphi=e^x,\ n=3\text{: }fQ_2=x(1+x+\tfrac{x^2}2)=x+x^2+\tfrac{x^3}2\text{ exactly matches the true degree-3 poly of }xe^x\ \checkmark\]