JC1 H2 Mathematics: Maclaurin Series

Find the first two non-zero terms

Let $y=\tan^{-1}(2x)$. Show that $(1+4x^2)y'=2$, and use this to find the first two non-zero terms of its Maclaurin series.

At $x=0$, $y(0)=0$ and $y'(0)=2$. Differentiate the relation rather than differentiating the inverse tangent repeatedly:

Insert the derivatives with their factorials:

These are the first two non-zero terms, not the first two powers of $x$. The usual series condition is $|2x|<1$. The derivative of the polynomial at zero is 2, checking the linear coefficient.

Keep the product rule when differentiating

Given $xy+\ln y=\ln2$, find the Maclaurin series for $y$ through $x^2$.

Since $\ln y$ is present, $y>0$. At $x=0$, $\ln y=\ln2$, so $y(0)=2$.

At zero:

Differentiate the original first-derivative relation:

Substitute the known values only after differentiating:

Check by substitution: using $y=2(1-2x+6x^2+\cdots)$, the logarithm is $\ln2-2x+4x^2+\cdots$. Adding $xy=2x-4x^2+\cdots$ cancels the first two powers, as required.

Compose sine and logarithm, then deduce cosine

Find $\sin[\ln(1+x)]$ through $x^3$. Hence deduce $\cos[\ln(1+x)]$ through $x^2$.

Put $u=\ln(1+x)$ and $S(x)=\sin u$.

Differentiate. The chain rule produces a factor $1/(1+x)$, so multiply by $1+x$ to isolate the requested cosine.

Write $C(x)=\cos[\ln(1+x)]$. Then

Multiply the series by $1+x$:

To deduce a degree-2 series by differentiation, the previous series must extend to degree 3. These are expansions about zero; $|x|<1$ is a sufficient interval for this use of the logarithmic series.

Factor out the constant before using the logarithmic series

Use standard series to find $\ln(1+e^x)$ through $x^2$.

Put $u=x/2+x^2/4+\cdots$. Thus

Now the logarithm's replacement tends to zero. Keep both degree-2 contributions:

Directly writing $e^x-(e^x)^2/2+\cdots$ and retaining a few terms is unsuitable: every power of $e^x$ has a non-zero constant and can affect the requested low-degree coefficients. The result checks at zero: $\ln(1+e^0)=\ln2$.

Approximate an integral and measure the error

Expand $\int_0^x t\cos(t^2)\,dt$ through $x^{10}$, and use it to approximate the integral when $x=1/2$.

The lower limit zero makes the integration constant zero. Equivalently, for an indefinite antiderivative, use $I(0)=0$.

Substitution $u=t^2$ gives the exact check:

The percentage error, calculated using unrounded values, is

Let $A$ be the approximation and $E$ the exact value.

This approximation is accurate here because the substituted angle is only $1/4$ radian and the first omitted term is very small. Adding another non-zero term improves it. A Maclaurin polynomial used over an interval far from zero needs a fresh accuracy check even when the function itself is defined there.

Use the sine rule before expanding

In triangle $ABC$, $AC=2$, $\angle A=\pi/3$ and $\angle B=\pi/6+\theta$. Find an approximation for $BC$ through $\theta^2$, for small $\theta$ in radians.

Use $\cos\theta\approx1-\theta^2/2$ and $\sin\theta\approx\theta$. Put $v=\sqrt3\theta-\theta^2/2$.

At $\theta=0$, this gives $BC=2\sqrt3$, agreeing with the exact sine-rule formula.

As $\theta$ grows, compare the two lengths instead of assuming the neglected powers are harmless. An angle in degrees must be converted to radians before substitution.