The problem was to construct a Maclaurin series for arctan(x) out to the first two nonzero terms.
Step 1: Find the derivatives of the function.
1
f'(x) =
1 + x2
-2x
f"(x) =
(1 + x2)2
-2(1 + x2)2 + 8x2(1 + x2) -2 + 6x2 + 8x4
f(3)(x) = =
(1 + x2)4 (1 + x2)3
Without knowing the trick for simplifying these derivatives (shown below),
it is doubtful that you would be able to see the pattern emerging here.
So we will skip to step 3.
Step 3: Evaluate the function and its derivatives at zero.
f(0) = 0 f'(0) = 1 f"(0) = 0 f(3)(0) = -2
Step 4: Apply the Maclaurin formula.
-2
arctan(x) » x + x3
3!
1
arctan(x) » x - x3
3
If you use the to approximate arctan(0.3), you get 0.2910000.
The actual arctan(0.3) to seven digits is 0.2914568.
The remainder of this solution-page is optional material. But if you are familiar with complex numbers, it will probably interest you.
Do you remember learning about complex numbers when you took algebra in high school? Among real numbers, -1 has no square root. But in the complex numbers, we invent a square root for it:
__
i = Ö-1
i2 = -1
i(-i) = 1
Each complex number is can be expressed in the form of
z = a + ibwhere a and b are real numbers. Notice that if you consider only the complex numbers where
The neat thing about complex numbers is that every quadratic polynomial can be factored. This includes
x2 + 1 = (x + i)(x - i)Multiply it out for youself to see why.
This means that the derivative of arctan(x) is
1 1
f'(x) = =
1 + x2 (x + i)(x - i)
If you algebraically munge the right-hand expression enough, you can
come up with the following equivalent:
i i
f'(x) = -
2(x + i) 2(x - i)
Go ahead and put these summands over a common denominator to see why their
sum is the same as the reciprocal product of
Taking still higher derivatives of this form of the expression is much easier than taking them of the original.
-i -i
f"(x) = -
2(x + i)2 2(x - i)2
2i 2i
f(3)(x) = -
2(x + i)3 2(x - i)3
-6i -6i
f(4)(x) = -
2(x + i)4 2(x - i)4
24i 24i
f(5)(x) = -
2(x + i)5 2(x - i)5
It's not difficult to identify the general pattern as
(n-1)!i(-1)n-1 (n-1)!i(-1)n-1
f(n)(x) = -
2(x + i)n 2(x - i)n
That is, each summand has, on top, the factorial that develops from taking derivatives
of a reciprocal power, the alternating sign (denoted by the (-1)n-1)
that develops from taking derivatives of a reciprocal power, and the original
i that we started with. On the bottom you have increasing
powers of the binomials we got by breaking the reciprocal product into a sum.
Now watch what happens when you put
(n-1)!i(-1)n-1 (n-1)!i(-1)n-1
f(n)(0) = -
2in 2(-i)n
Here comes the fun part. Remember that by definition,
(n-1)!i(-1)n-1 (n-1)!i(-1)n-1
f(n)(0) = -
2(-1)n/2 2(-1)n/2
The expressions on the left and right of the minus operator are the
same. So the two terms cancel and you have to conclude that whenever
n is even,
But what about when n is odd? An odd number is always one more than an even
number. So divide an i out of in
and a -i out of (-i)n and
you can again use our observations about even powers of i and
of -i.
The result is that when n is odd you have
(n-1)!i(-1)n-1 (n-1)!i(-1)n-1
f(n)(0) = -
2i(-1)(n-1)/2 -2i(-1)(n-1)/2
The i's on the top and bottom cancel. Also observe that minus a minus
is a plus. Finally when you cancel the powers of
the (-1) in the numerator and denominator you end up with (-1)(n-1)/2 left
in the numerator.
When you do all that simplification you get
(n-1)!(-1)(n-1)/2 (n-1)!(-1)(n-1)/2
f(n)(0) = + = (n-1)!(-1)(n-1)/2
2 2
The expression on the right still contains the factorial we've been getting from
taking derivatives of a reciprocal power. And it shows a sign that alternates with
each successive odd value of n.
Here is a table of what you get for various n's:
n f(n)(0)
|
1 1
2 0
3 -2!
4 0
5 4!
6 0
7 -6!
8 0
9 8!
10 0
and so on
I think you can see the pattern now. So we put that into the
Maclaurin formula:
2! 4! 6! 8!
arctan(x) = x - x3 + x5 - x7 + x9 - ...
3! 5! 7! 9!
When you divide (n-1)! by n! you get 1/n. So the above simplifies to
1 1 1 1
arctan(x) = x - x3 + x5 - x7 + x9 - ...
3 5 7 9
In sigma notation that's
¥ (-1)k
arctan(x) = å x2k+1
k=0 2k+1
I'll state without proof (proof will be in a later section) that this series converges only for
email me at hahn@netsrq.com