A long sum of inverse tangents can often be done in a line. Write each term as a difference tan−1ar+1−tan−1ar using tan−1x−tan−1y=tan−11+xyx−y. Then all the terms cancel except the first and the last. For example ∑r=1ntan−1r2+r+11=tan−1(n+1)−4π, and the infinite sum is 4π.
This is the first lesson of Inverse Trigonometric Functions: Telescoping Series and Parameter Problems, free for everyone to read.
What you will be able to do
recognise a term of the form tan−11+aba−b and split it into tan−1a−tan−1b;
add a long series of inverse tangents by telescoping, and find its limit;
use the same idea for sequences such as the Fibonacci numbers and for computing π.
One term as a difference
The difference rule from the second course says
tan−1a−tan−1b=tan−11+aba−bwhen ab>−1.
Read it from right to left. Whenever a term looks like tan−1 of a fraction whose bottom is 1 plus a product, the term splits into two inverse tangents. For positive numbers a and b the condition ab>−1 is always true.
Take the term r2+r+11. Notice that r2+r+1=1+r(r+1). So with a=r+1 and b=r:
tan−11+r(r+1)1=tan−1(r+1)−tan−1r.
This is true because
1+(r+1)r(r+1)−r=1+r(r+1)1.
A fan of five grey lines from a corner point at the bottom left to the points (1, 1), (1, 2), (1, 3), (1, 4) and (1, 5), which are marked with red dots on a vertical black line at the right. A horizontal black line runs along the bottom from the corner to the foot of the vertical line. The four thin wedges between neighbouring lines are coloured blue, orange, green and purple from the bottom, and are labelled tan inverse of one third, one seventh, one thirteenth and one twenty-first. The title says that the four small angles add up to the big one, tan inverse 5 minus tan inverse 1, and that the drawing is not to scale.
Each inverse tangent appears once with a plus and once with a minus, except the first and the last. They cancel in pairs, like the sections of a telescope that slide into each other. What is left is
As n grows, tan−1(n+1) approaches 2π. So the infinite series adds up to 2π−4π=4π.
A method for any series
Look at the term. Is it tan−1 of a fraction with 1+ a product at the bottom?
Write the top as a difference. Choose a and b with a−b equal to the top and ab equal to the rest of the bottom.
Split the term into tan−1a−tan−1b.
Cancel, and keep the first and the last pieces.
Take the limit if the series is infinite. Use tan−1x→2π as x→∞.
Three more examples follow the same steps.
tan−1r22: put a=r+1 and b=r−1. Then a−b=2 and 1+ab=r2. So the term is tan−1(r+1)−tan−1(r−1). The pairs cancel two steps apart, and r=1∑ntan−1r22=tan−1n+tan−1(n+1)−4π.
tan−12r21: write 2r21=4r22 and put a=2r+1, b=2r−1. The sum from 1 to n is tan−1(2n+1)−4π=tan−1n+1n.
An arithmetic progression. If ar rises by d each step and all the numbers are positive, then tan−11+arar+1d=tan−1ar+1−tan−1ar.
A worked example
Find r=1∑∞tan−1r22, and the value of r=1∑4tan−1r2+r+11 as one inverse tangent.
The first sum. Each term is tan−1(r+1)−tan−1(r−1). The partial sum to n terms is tan−1n+tan−1(n+1)−tan−11−tan−10. As n→∞ this tends to 2π+2π−4π−0=43π.
The second sum. It equals tan−15−tan−11=tan−11+55−1=tan−132.
Practise with new numbers
Try it in Python
The program adds the terms one at a time. It prints the running sum next to the telescoped answer tan−1(n+1)−4π, and then it adds a thousand terms.
Common mistakes
Cancelling the wrong terms. For tan−1(r+1)−tan−1(r−1) the cancelling pairs are two steps apart, so two terms survive at each end.
Forgetting the first term. The sum to n terms is tan−1(n+1)−tan−11. The tan−11=4π does not cancel.
Using the rule when ab<−1. The splitting needs ab>−1. For positive numbers it is always true.
Taking the limit too early. Find the partial sum first, and then let n grow.
Remember these
tan−11+aba−b=tan−1a−tan−1b for ab>−1.
tan−11+r(r+1)1=tan−1(r+1)−tan−1r, so r=1∑n=tan−1(n+1)−4π.
tan−1r22=tan−1(r+1)−tan−1(r−1).
In an infinite series, tan−1 of a growing number tends to 2π.
Check yourself
Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.
Write tan−1131 as a difference of two inverse tangents.
Find r=1∑3tan−1r2+r+11 as one inverse tangent.
What does the infinite series r=1∑∞tan−1r2+r+11 add up to?
Write tan−192 as a difference of two inverse tangents with whole numbers.
Why is the splitting of a term allowed when a and b are positive?
Liked this lesson? There is more
Inverse Trigonometric Functions: Telescoping Series and Parameter Problems is a Standard course. It adds 2 more lessons, practice and tests on top of the free chapter.
12Problems2Python Labs1Question Set1Mock Test
Range and graph questions: To find the range of an expression built from inverse functions, work out what it equals on each piece of its domain, because the branches often change the formula. tan−1x+tan−1x1
takes only the two values
±2π
, and
sin−11+x22x
rises to
2π
at
x=1
and then falls again. Count solutions by sliding a horizontal line across the graph. For a periodic composite, the period of a sum is the least common multiple of the periods.
Equations with a parameter: In an equation such as (sin−1x)2+(cos−1x)2=a, write t=sin−1x and cos−1x=2π−t. The equation becomes a quadratic in t, where t must stay in [−2π,2π]. Different x give different t, so the number of solutions x is the number of roots t in that interval. The vertex and the end values of the parabola give the range of the left side and the counts for each value of the parameter.