Re: Simplifying
- From: "James Hess" <mysidia@xxxxxxxxx>
- Date: 24 May 2005 18:10:55 -0700
John den Haan wrote:
> Hello there!
>
> It's been ages since I did algebra, and now I am presented with the
> following term as the right-hand part of an equation:
>
> (n+1)^2 + n*(n+1) * (2n+1) / 6
I assume you're working in the real number system.
One brute force method, which is important to be aware of
is straightforward application of distributivity and addition rules:
multiply out
(n+1)*(n+1)
Then add to the result of what follows:
Multiply out n*(n+1) and then multiply the result by (2n+1)/6
adding up all terms that are alike.
in an expression that is not equal to but looks structurally like
5x + 6xy + x^2 + x + 10,
5x, 6xy, x^2, and x are called the terms of the polynomial...
the leading number each term is multiplied or divided by are called
coefficients, i.e. 5, 6, 1, and 1 above.
You can simplify to '6x+6xy+x^2+10' by addition but no further,
using '5x+x=6x'.
To multiply out and thus simplify (a+b)*(c+d) add together
every combination of the terms.
a*c + b*c + a*d + b*d
for
(a+b)*(c+d)*(e+f)
you can do
(a*c + b*c + a*d + b*d)*(e+f)
m*(g+h+i)(j+k)
is m*g*j + m*g*k + m*h*j + m*h*k + m*i*j + m*i*k
And arbitrarily large expressions can be multiplied, but it can
be a pain to do so by hand: fortunately your current expression
of interest is simple.
Sometimes the factored version is the simpler for your problem,
but you may need to expand under multiplication to see things that
will cancel out.
By convention, in simplification you are preferring a sum of basic
terms over the factors of polynomials however...
You go from x^2(A+B) to Ax^2+Bx^2
and x^2(A+B) considered factored but not simplified
whereas (Ax^2 + Bx^2) is considered simplified but not factored.
Ex: suppose you had (a+b)(c+d)
with ... a = -c and b = d
You could simplify in general..
(a)*c + b*c + a*d + (b)*d
= -c*c + b*c + a*d + d*d
= -c*c + d*c + -c*d + d*d
= -c*c + d*d
And rewrite as: d^2 - c^2
Which is better than (a+b)(c+d),
(a+b)(d-a), or (b-c)(c+d), right?
(You have fewer operations on individual numbers)
Incidentally,
-(a+b)(c-d) = (a+b)(d-c) = (-a-b)(c-d) for any real a,b,c,d
--
To take advantage of situations a simple polynomial gives you and
to make it a little faster, I suggest... first decide what combinations
and powers terms your outcome will have.
Simplification of expressions this simple should be a fairly
straightforward mental process. If you find that you need pencil
and paper.. to worry about long-hand expanding the expressions,
then I think you could use some practice :)
In short: group and sum the terms that would appear after
the expansion.
This method may help reduce the likelihood of entering a term of the
wrong degree later during a drawn out solution process, so if nothing
else, consider using it as a check.
1. Mentally break up the sum into a list of terms
"(n+1)^2" before simplification is one term,
"n*(n+1) * (2n+1)/6" is another, etc
2. Find what degrees of terms and combinations
of your variables will be generated by the expansions
(either ahead of time or with a running mental count)
(If there are too many combinatinos like xy x^2y xy^2 x^7
I would write these down though, and create some sort of
table to keep track of what terms I have seen...
you just need to make sure you 'see' and count all the terms)
Save other multiples [of all the terms] for the end, and multiply
by the sum.
Ok, for the present situation:
Observe that When expanded
(n+1)^2 will generate terms that are multiples of 1, n, and n^2
You can see this using the standard and fairly fundamental rule for
expanding the square of a binomial
(a+b)^2 = a^2 + 2ab + b^2
when a=n, b=1
Similarly n*(n+1) will yield terms that are multiples of 1, n,
and n^2 and therefore (n*(n+1)) * (2n+1) will yield terms that
are multiples of 1, n, n^2, and n^3.
It can be rather important to keep the exponent rules like
x^a * x^b = x^(a+b)
in mind to do this computation (about which degrees to consider).
3. Next write the sum of the coefficients for each
of these term combinations you know the result will
have, distributing as necessary:
i.e. Take the first degree, imagine which terms
of (n+1)^2 will have degree 1, forget the others
until you are considering the second degree terms,
etc...
Order | (n+1)^2 . n*(n+1) * (2n+1)/6
| | = (n^2+n) * (2n+1)/6
n^0 | 1 | 0
n^1 | 2 | 1/6
n^2 | 1 | 3/6
n^3 | 0 | 2/6
---------------
This table gives a crude picture of one approach for the
simplification process that I find useful for gathering
the result quickly, sometimes, though I'd rarely need write
any part of this kind of table in counting for simple
polynomials;
what you need to keep track of as you go is the current sum on
each term, and examine which coefficients for each term will be
generated by each term (prior to simplification).
Add them up in this case.
I would write:
(2/6 + 0)n^3 + (3/6 + 1)n^2 + ...
-Mysid
.
- References:
- [Basic algebra] Simplifying
- From: John den Haan
- [Basic algebra] Simplifying
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