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Maclaurin series for tan(x), and polynomial division
By Anonymous on Friday, January 26, 2001 - 12:46 pm:

Hi there,

I was fiddling with the Maclaurin Series, or the Power Series, which ever you like to call it.

When I tried to derive the Maclaurin Series for f(x) = tan(x), it became a laborious and messy task very quickly indeed, unless you like some dirty integration!

f(x) = tan(x)
f'(x) = sec2(x)
f''(x) = 2sec(x)sec(x)tan(x) = 2sec2(x)tan(x)
.... and on and on

Is there a better (easier) way to do this?

Thanks for your help in advance.

By Kerwin Hui (Kwkh2) on Friday, January 26, 2001 - 02:04 pm:

If you only want the MacLaurin series for tan(x), then there is indeed a easier method, if you are prepared to do some algebraic manipulation.

First, note that tan(x)=sin(x)/cos(x)

Now derive/use the MacLaurin series for sin(x) and cos(x), i.e.,

sin(x)=x-x3/3!+x5/5!-...
cos(x)=1-x2/2!+x4/4!-...

Upon dividing the two series, you get

tan(x)=x-x3/3+...

Kerwin

By Anonymous on Friday, January 26, 2001 - 02:07 pm:

How do you divide two infinite series?

By Kerwin Hui (Kwkh2) on Friday, January 26, 2001 - 02:16 pm:

Just like dividing two finite degree polynomial, except that we have the series in ascending power of x rather than decending power. At each stage, we try to eliminate the lowest power of x that remains, so writing out a few terms (imagine the horizontal lines you normally write in a long division sum):
x
+  x3
3
+  2x3
15
+...
¬ ANSWER
1 -  x2
2!
 +  x4
4!
 -  x6
6!
 + ... ö
ø
x
-  x3
3!
+  x5
5!
-  x7
7!
+...
¬ sinx
x
-  x3
2!
+  x5
4!
-  x7
6!
+...
¬ x× æ
è		 1-  x2
2!
 +  x4
4!
 +... ö
ø
 x3
3
-  x5
30
+  x7
840
+...
Subtract above two lines
 x3
3
=  x5
6
+  x7
72
+...
¬  x3
3
 × æ
è		 1-  x2
2!
 +  x4
4!
 ö
ø
 2x5
15
-  x7
5040
+...
Subtract above two lines
 2x5
15
-  x7
15
+...
¬  2x5
15
 × æ
è		 1-  x2
2!
 +  x4
4!
 +... ö
ø


Kerwin
By Anonymous on Friday, January 26, 2001 - 02:19 pm:

I don't know how to divide finite polynomials!

By Kerwin Hui (Kwkh2) on Friday, January 26, 2001 - 02:34 pm:

Suppose you have two (finite) polynomials, f(x) and g(x), such that degree of f ³ degree of g. Then we write f(x) and g(x) in decending power of x. For example, let f(x)=2x2+4x+3, g(x)=x+1. Now, we write them with the usual long-hand version of division:

x + 1 ö
ø
2x2
+4x
+3
Now the purpose of division is to get a polynomial h(x) such that f(x)ºh(x)g(x)+r(x), where r(x) has degree less than that of g(x). So we first eliminate the highest power of x in f(x), i.e. the term 2x2, by an appropriate multiple of x,i.e. 2x. When we do this, we have
2x
x + 1 ö
ø
2x2
+4x
+3
2x2
+2x
2x
+3
¬ Subtract above two rows
i.e. f(x)=2x(x+1)+(2x+3)

Now repeat the procedure:
2x
+2
x + 1 ö
ø
2x2
+4x
+3
2x2
+2x
2x
+3
¬ Subtract above two rows
2x
+2
+1
¬ Subtract above two rows
to get f(x)=(2x+2)g(x)+1

You can verify that this procedure gives the normal decimal division if you substitute in x=10. In some cases these will not be equal, because we are only interested in powers of x and so the remainder will take a different form.

Kerwin
By Anonymous on Friday, January 26, 2001 - 03:20 pm:

Thank you for your help on the MacLaurin Series.