Use substitution, multiplication, differentiation or integration of known series. Keep enough terms before operating. Termwise differentiation and integration are valid inside the interval of convergence. **Operation** | **Example** Substitution | ex2=1+x2+ x42+. s Product | exsin x=x+x2+ x33+. s Differentiation | d dxln(1+x)=1-x+x2-. s= 11+x, |x|<1 Integration | _0x(1-t2+t4-. s), dt=x- x33+ x55-. s=arctan x, |x|<1 After a substitution, transform the validity condition too. For ln(1+2x), -1<2x 1 gives - 12<x <= 12. Do not carry an old interval over unchanged.

Use substitution, multiplication, differentiation or integration of known series. Keep enough terms before operating. Termwise differentiation and integration are valid inside the interval of convergence.

After a substitution, transform the validity condition too. For \(\ln(1+2x)\), \(-1<2x\le1\) gives \(-\dfrac12<x\le\dfrac12\). Do not carry an old interval over unchanged.

Maclaurin Series and Power Series

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Operation	Example
Substitution	\(e^{x^2}=1+x^2+\dfrac{x^4}{2}+\cdots\)
Product	\(e^x\sin x=x+x^2+\dfrac{x^3}{3}+\cdots\)
Differentiation	\(\dfrac{\mathrm d}{\mathrm dx}\ln(1+x)=1-x+x^2-\cdots=\dfrac1{1+x}\), \(\|x\|<1\)
Integration	\(\int_0^x(1-t^2+t^4-\cdots)\,\mathrm dt=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\cdots=\arctan x\), \(\|x\|<1\)

Obtaining Further Series

Obtaining Further Series