Approximate Complicated by Simple

Wishful Thinking

One seemingly trivial yet useful strategy for doing math or problem solving in general is wishful thinking.

Wishful thinking is when you simplify/change the problem to another problem that you wish it looked like so you could solve it easily. After getting your "wish", you try to build on your evolved form to answer the original problem.

The MAA describes wishful thinking by this question in this article: "If you could change the question, what change would make it ridiculously easy?"

Example of Wishful Thinking

A simple example in which you could use wishful thinking would be to factor $x^4+x^2+1$. It's a fourth degree polynomial, which is hard to solve, so you wished it was a quadratic because quadratics are easier to solve. But wait! It's an even polynomial, so the polynomial could be rewritten as a quadratic: $(x^2)^2+x^2+1$.

If you define $y = x^2$, you can rewrite the polynomial as $y^2+y+1$. Guess factoring leads nowhere, and you could only wish that it was a perfect square. Aha! We could force the polynomial to contain a perfect square by rewriting the expression as $(y^2+2y+1)-y = (y+1)^2 - y$. Since $y=x^2$, you can use difference of squares to factor it.

Here is the complete solution:

Let $y = x^2$.

$\begin{array}{lr} \:\:\:\:\:\: x^4+x^2+1 \\ = (x^2)^2+x^2+1 \\ = y^2+y+1 \\ = (y^2+2y+1)-y \\ = (y+1)^2 - y \\ = (y+1)^2 - x^2 \\ = (y+1+x)(y+1-x) \\ = (x^2+x+1)(x^2-x+1) \end{array}$

More Resources

For more resources on wishful thinking, you can watch this video by the Mathematical Association of America here, or read The Art and Craft of Problem Solving (Chapter 2) by Paul Zeitz.

Problem Solving: Substitution

Many times when factoring polynomials, or solving complicated algebra equations, it is useful to make a clever substitution to make life easier for us. Example

Solve $x^4+5x^2+6=0$.

This is a quartic. Normally, we try to find two binomials that multiply to this. With some guessing and checking, we find that $(x^2+3)(x^2+2)=0$ so our roots are $\pm 3i$ and $\pm 2i$. However, there is a simpler way to do this. Let $a=x^2$. We can express the equation in terms of $a$. $$x^4+5x^2+6=\left(x^2\right)^2+5(x^2)+6=a^2+5a+6.$$ Now this is a quadratic we know how to handle. We can factor this as $(a+2)(a+3)=0$ so $a=-2,-3$. Solving for $x$ yields $$\begin{align} x^2=-2\implies x&=\pm 2i \\ \\ x^2=-3\implies x&= \pm 3i \end{align}$$ This is a much easier way to do this. Of course this is just the beginning of how substitution can greatly simplify problems.

The solution to the previous problem is as follows. Let $q=\sqrt{4x-3}$. Now all we have to solve is $$q+\frac{10}{q}=7$$ which is simple. Multiplying by $q$ for each side yields $$\begin{align}q^2+10=7q \\ &\implies q^2-7q+10=0 \\ &\implies (q-5)(q-2)=0 \\ &\implies q=2,5.\end{align}$$

Now we equate each value of $q$ to $\sqrt{4x-3}$. So $$\begin{align}&q=2\implies 2=\sqrt{4x-3}\implies 4=4x-3\implies x=\frac{7}{4} \\ \\ &q=5\implies 5=\sqrt{4x-3}\implies 25=4x-3\implies x=7.\end{align}$$ Therefore, $\boxed{x=\frac{7}{4},~7}$.

Solution to previous problem

Let $u=\log x$. We have $$\begin{align} 12u^2+11u+2=0 &\implies 12u^2+3u+8u+2=0 \\ &\implies 3u(4u+1)+2(4u+1)=0 \\ &\implies (4u+1)(3u+2)=0 \\ &\implies u=-\frac{1}{4},~-\frac{2}{3} \\ \\ &\implies \begin{cases} \log x=-\frac{1}{4} \\ \log x=-\frac{2}{3} \end{cases} \\ \\ &\implies \begin{cases} x=10^{-1/4} \\ x=10^{-2/3} \end{cases} \\ \\ &\implies \boxed{ \begin{cases} x\approx 0.56 \\ x\approx 0.22 \end{cases}} \end{align}$$

Awww! MASSAGE is very fun! AWWW!!

In this article we are going to talk about a simple yet powerful tactic, massaging. That is a pretty bad name, doesn't it? (Maybe you think it is a good name, I don't care) But actually, if you massage an expression or an equation or even a term, you are "loosen them up" so that you can solve it easier (that's why I said that this is a great tactic). But you must have no idea on how to massage an expression by only reading the long and trashy text above, right? So, let's look at an example, shall we?

Example. Let $A=\displaystyle\sum^{10,000}_{n=1}\frac{1}{\sqrt n}$. Find $\lfloor A \rfloor$ without using a calculator. (Even you use a calculator, you also have no enough time to press up to 10,000, except you use Wolfram Alpha :P)

Solution. We can use massaging here since we don't need an exact value, luckily. Through massaging, we can make the sum telescope. We have $\frac{1}{\sqrt n}=\frac{2}{2\sqrt n} < \frac{2}{\sqrt n +\sqrt {n-1}} = 2(\sqrt n -\sqrt {n-1})$

(In this step, we rationalised the denominator)

On the other hand,

$\frac{1}{\sqrt n}>\frac{2}{\sqrt {n+1}+\sqrt n}=2(\sqrt{n+1}-\sqrt n)$

Implying

$2(\sqrt{n+1}-\sqrt n)<\frac{1}{\sqrt n}<2(\sqrt n-\sqrt{n-1})$

That looks great, right? We had just bound the deadly term $\frac{1}{\sqrt n}$ above and below by the terms that will telescope greatly when summed. Therefore,

$\displaystyle\sum^{10,000}_{n=1}2(\sqrt{n+1}-\sqrt n)=2\sqrt{10,001}-2$ and $\displaystyle\sum^{10,000}_{n=1}2(\sqrt n-\sqrt{n-1})=2\sqrt{10,000}-\sqrt 0$

Implying

$2\sqrt{10,001}-2<\frac{1}{\sqrt n} < 2\sqrt{10,000}-\sqrt 0 \Longrightarrow \lfloor A \rfloor= 198 \text{ or } 199$

Let me ask you something, which is the correct answer, 198 or 199, in your opinion? You can't decide it right? So, let's refine this estimate! When n=1, we get $2(\sqrt 2 -1)<1<2\sqrt 1$. Lower limit is good, but that's too terrible for upper limit! How about we start the summation at n=2? We have

$A=1+\displaystyle\sum^{10,000}_{n=2}\frac{1}{\sqrt n}$

And we estimate

$2\sqrt{10001}-2\sqrt 2<\displaystyle\sum^{10,000}_{n=2}\frac{1}{\sqrt n}<2\sqrt{10,000}-2\sqrt 1$

The lower limit is about 197, while the upper limit is 198. So, A is somewhere at 198 and 199, implying $\lfloor A \rfloor = 198$.