Substitution, or Making Things "Easy on the Eyes"

Section 2.2 Substitution, or Making Things “Easy on the Eyes”

Mathematicians do not study objects, but relations between objects. Thus, they are free to replace some objects by others so long as the relations remain unchanged.
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―Henri Poincarè (1854–1912)
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It is a common misconception that mathematicians love complicated formulas. In fact, the opposite is true. We really very much prefer to keep things as simple as possible. We only accept complexity when we are forced to. Moreover, with experience we have learned that complexity is usually borne of an inadequate understanding of the problem. Better understanding usually results in a simpler formulation.

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This works in reverse too. If we can find a simple way to express what appears to be a complex problem we can usually gain a deeper understanding of the problem and its solution. A judicious substitution is one way to re-express a complex problem.

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Example 2.2.1. A Horrible Formula.

For example, suppose you were asked to solve the following equation for \(x\text{:}\)

\begin{equation*} \frac{\frac{\pi x}{\sqrt{(28-\sqrt[3]{26})^{\frac{1}{3}}}}}{\pi^2-\sqrt{3}} - \sqrt[21]{2^{\frac{11}{2}}}= (\pi^3-\sqrt{2})x. \end{equation*}

At first glance this looks horribly complicated. But look at it again. On the right the coefficient of \(x\) is \(\pi^3-\sqrt{2}\text{.}\) That’s just a number. It is approximately \(29.6\) but who cares? For purposes of algebraically isolating \(x\) all we need to know is that it is a number. Rather than calling it \(\pi^3-\sqrt{2}\text{,}\) let’s just call it something simple, like \(A\text{.}\) Think of it as giving the expression \(\pi^3-\sqrt{2}\) a nickname; something shorter, and easier to remember.

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If we make that substitution our equation is now

\begin{equation*} \frac{\frac{\pi x}{\sqrt{(28-\sqrt[3]{26})^{\frac{1}{3}}}}}{\pi^2-\sqrt{3}} - \sqrt[21]{2^{\frac{11}{2}}}= Ax. \end{equation*}

which is a bit “easier on the eyes.”

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But wait. The expression \(\sqrt[21]{2^{\frac{11}{2}}}\) is also just a number. Let’s make it “easier on the eyes,” too. If we let \(B=\sqrt[21]{2^{\frac{11}{2}}}\) we have

\begin{equation*} \frac{\frac{\pi x}{\sqrt{(28-\sqrt[3]{26})^{\frac{1}{3}}}}}{\pi^2-\sqrt{3}} - B= Ax. \end{equation*}

which is even easier to look at. By now you see where this is going. If we let \(C=\frac{\frac{\pi}{\sqrt{(28-\sqrt[3]{26})^{\frac{1}{3}}}}}{\pi^2-\sqrt{3}}\) then our equation is \(Cx-B=Ax\) which is much “easier on the eyes” than the original equation was. In fact, we can now see that this is really a very simple equation and the solution is \(x=\frac{B}{C-A}\text{.}\) Of course we inserted \(A\text{,}\) \(B\text{,}\) and \(C\) into the problem so now we need to unwind our substitutions. When we do that we get

\begin{equation*} x=\frac{\sqrt[21]{2^{\frac{11}{2}}}}{\frac{\frac{\pi}{\sqrt{(28-\sqrt[3]{26})^{\frac{1}{3}}}}}{\pi^2-\sqrt{3}} - (\pi^3-\sqrt{2})}. \end{equation*}

If a decimal number is required we could enter this formula into whatever computational tool we’re using. Good luck with that.

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The previous example was clearly contrived. Obviously we went out of our way to make a simple problem look complicated just so we could simplify it. As a result it would be easy to dismiss this as a silly example. But it is not. There is a serious lesson to be learned here. One that we will use over and over again in the near future. By its very nature mathematical notation is not easy to read. It is so dense, with so much information packed into a few symbols, that it is easy to misinterpret. Making a formula easier to read can be very helpful. A well chosen substitution can make a complex looking formula much easier to work with.

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Example 2.2.2.

When you were learning Algebra you almost certainly encountered problems like this: Given

\begin{equation} x^{\frac{2}{3}}-4x^{\frac{1}{3}}+3=0\tag{2.2} \end{equation}

solve for \(x\text{.}\)

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At first glance, this looks like a formidable problem. However, if we look closer we see that it actually has a very familiar form. Because \(x^{\frac{2}{3}}=\left(x^{\frac{1}{3}}\right)^2\) we can rewrite equation (2.2) as

\begin{equation} \left(x^{\frac{1}{3}}\right)^2-4x^{\frac{1}{3}}+3=0\tag{2.3} \end{equation}

and if you look closely you’ll see that equation (2.3) is “quadratic in form.” That is, it is really a quadratic equation — which we know how to solve — in disguise. However it is still rather difficult to see the quadratic nature of equation (2.3). To make this very clear we let

\begin{equation*} z=x^{\frac{1}{3}}. \end{equation*}

That is, just as in Example 2.2.1 we give the expression \(x^{\frac{1}{3}}\) the nickname, \(z\text{,}\) which is shorter, easier to remember, and “easier on the eyes.” Then equation (2.3) becomes the quadratic equation:

\begin{equation*} z^2-4z+3=0. \end{equation*}

Our problem is not solved. Indeed, we haven’t even tried to solve it yet. But it is much more simply expressed. And because it is more simply expressed we can now see how to solve it. Factoring, we have

\begin{equation*} 0=z^2-4z+3=(z-1)(z-3) \end{equation*}

so we have \(z=1\) and \(z=3\text{.}\) But, of course, these are not solutions of our problem. We have found \(z\) (which we inserted into the problem) not \(x\) (which is what we need to find). To complete the solution we need to unwind the substitution we made and solve the two equations

\begin{equation*} x^{\frac{1}{3}}=1 \text{ and } x^{\frac{1}{3}}=3, \end{equation*}

yielding the solutions \(x=1\) and \(x=27\text{,}\) respectively.

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In making a substitution our goal is always to make the formulas we are dealing with simpler to understand and easier to read. In Examples 2.2.1 and 2.2.2 it was straightforward to see how to do that. There were expressions that were hard to read so we simply replaced a complicated part of the problem with something simple. It isn’t always that easy. Sometimes just finding a good substitution presents challenges of its own.

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Aside: Comment.

For example, the Quadratic Formula can be derived by making a very simple substitution that allows us to solve the general quadratic equation

\begin{equation} ax^2+bx+c=0.\tag{2.4} \end{equation}

Once the substitution is found it is very easy to work through the algebra, but finding the right substitution in the first place is not straightforward at all.

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Problem 2.2.3.

Before reading any further see if you can come up with the right substitution for equation (2.4).

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Hint.

The goal is to eliminate the linear term, \(bx\text{.}\) You can see how this would help. If \(bx\) were not present in equation (2.4) we could easily solve \(ax^2+c=0\text{.}\)

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Were you able to solve Problem 2.2.3? It’s actually very hard to do without some guidance. Here’s another hint. Try the substitution \(x=y-\frac{b}{ka}\) where \(k\) is a parameter to be determined. That is, make this substitution and find the value of \(k\) that eliminates the linear term.

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After making this substitution the problem will be given in terms of \(y\text{,}\) not \(x\text{.}\) That’s OK. Solve the problem in this form, and then unwind your substitution back to the original \(x\) variable.

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Problem 2.2.4.

Notice that after making our substitution we get:

\begin{equation*} a\left(y-\frac{b}{k a}\right)^2+b\left(y-\frac{b}{k a}\right)+c=0. \end{equation*}

which is more complicated, not less. Of course it is. Sometimes you have to complicate before you can simplify. Don’t give up. Multiply everything out to see what you’ve got.

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Once you’ve found \(k\) make the substitution and eliminate the linear term — that was the objective, remember? At that point you should see the Quadratic Formula starting to emerge. Don’t forget to unwind the substitution and solve for \(x\text{.}\)

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