Expand x cos 2x in a power series in x

Expand $x\cos 2x$ in a power series in $x$.

Explanation of Method

Substituting $t=2x$ into the $\cos t$ series and multiplying by $x$ shifts every exponent up by one, so no new convergence issues arise — the result is entire, just like $\cos 2x$ itself.

Substitute $t=2x$ into the series for $\cos t$.
Multiply by $x$.
State the region of convergence.

Step 1 · Substitute

$$ \cos 2x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(2x)^{2n} = \sum_{n=0}^{\infty}\frac{(-1)^n4^n}{(2n)!}x^{2n}. $$

Step 2 · Multiply by $x$

$$ x\cos 2x = \sum_{n=0}^{\infty}\frac{(-1)^n4^n}{(2n)!}x^{2n+1} = x-2x^3+\frac{2x^5}3-\frac{4x^7}{45}+\frac{2x^9}{315}-\cdots $$

Step 3 · Region of convergence

Since $\cos 2x$ is entire and multiplying by $x$ does not affect convergence: $$ \text{valid for all } x\in\mathbb R. $$

Summary Table

Property	Value
Series	$\sum(-1)^n4^nx^{2n+1}/(2n)!$
Symmetry	Odd function
Region of validity	$\mathbb R$

Additional Notes

Since $\cos 2x$ is even and multiplying by $x$ flips parity, $x\cos2x$ is an odd function — consistent with only odd powers appearing in the series.
This substitution-then-multiply technique is the most basic building block of the whole "expand using known expansions" family; more elaborate versions appear in Problems 273 ($(1+x^2)\arctan x$) and 274 ($(1+e^x)^2$).
The zeros of $x\cos2x$ occur at $x=0$ and $x=\pm(2k+1)\pi/4$; the coefficient pattern's alternating sign directly encodes the oscillation of $\cos2x$.

Result

$$ x\cos 2x = \sum_{n=0}^{\infty}\frac{(-1)^n4^n}{(2n)!}x^{2n+1}, \quad x\in\mathbb R. $$