Substituting particular values, and the coefficient identities that drop out

0:00So far there is one theorem, and it is about a bracket with two terms in it, raised to a positive whole power. What comes next adds no machinery at all. It puts particular things into that one statement and reads off what falls out. Replace the second term by its negative, and the signs alternate. Set the first term to one, and the letters fall off the coefficients. Set the remaining letter to one as well, and the whole thing becomes arithmetic about a row of numbers.

0:35Each of those is the same sentence with something specific put into it. Nothing is proved twice. And the return on that is out of all proportion to how cheap the move is. By the end you will have a way to raise a large number to a power, a way to settle a comparison without computing anything, and a proof about remainders that covers every power at once. Start with the first substitution.

1:05The theorem talks about a first quantity and a second quantity, and it never asked what they were. So put in the negative of the second quantity. Every term of the expansion is a selection count, times a power of the first quantity, times a power of the second. That last power is now a power of a negative quantity. And a power of a negative quantity is negative exactly when the power is odd.

1:36So the term at position nought keeps its sign, the term at position one reverses, position two keeps, position three reverses, and on down the row. Nothing was proved there. The statement was asked about a different pair of quantities, and it answered. Across four such substitutions, scored at a hundred and seventy-six positions against actually multiplying the brackets out, there is not one disagreement. It is worth being exact about where the alternation comes from, because this is the step people lose.

2:15It is not that subtraction alternates. It is that each position carries its own power of minus one, and an even power of a negative quantity is positive. Check that against the substitution itself rather than against a slogan. Out to the tenth power there are sixty-five positions, and thirty of them come out with their sign reversed. The rule that turns the odd positions gets all sixty-five right. Now here is a rule that is wrong: say instead that everything after the first position is negative.

2:54That gets forty of the sixty-five right and twenty-five wrong. Forty out of sixty-five is a rule that will feel correct for a long time and then quietly cost you a term in the middle of a sixth power. Here is the move on a real bracket: the first quantity a letter, the second twice another letter, with a minus between them, raised to the fifth. The coefficients are the fifth row: one, five, ten, ten, five, one.

3:27Each of those gets multiplied by a rising power of two, because the second quantity carries a two. That gives one, ten, forty, eighty, eighty, thirty-two. And the signs come out of the substitution: positions one, three and five enter negative. So the expansion is x to the fifth, minus ten x to the fourth y, plus forty x cubed y squared, minus eighty x squared y cubed, plus eighty x y to the fourth, minus thirty-two y to the fifth.

4:03Do the signs last. Get the magnitudes right first, from the row and from the powers, and only then ask which positions turn. Trying to carry a minus sign through a compound term is how a sixth power goes wrong. The second substitution is even cheaper. Set the first quantity to one. Every power of one is one, so every one of those factors disappears from every term. What is left is the coefficients standing bare against rising powers of the remaining letter.

4:40That is the form everything after this uses. Scored at forty-four positions out to the eighth power, the numbers left standing are exactly the rows they should be. It is not a new expansion. It is the same expansion with the scenery taken away. Now set the remaining letter to one as well. On the left the bracket is one plus one, so the left side is two raised to the power.

5:10On the right every power of one is one again, so the right side is the bare sum of the row. So a row of the triangle adds up to a power of two. The fourth row is one, four, six, four, one, and that adds to sixteen. The fifth row is one, five, ten, ten, five, one, and that adds to thirty-two. Out to the twelfth power there is not one failure.

5:36And notice what that is: not a separate fact about the triangle, but the expansion read at a single point. The identity and the theorem are the same statement. Do it once more, with the second quantity taken as the negative of the letter. Then set the letter to one. On the left the bracket is one minus one, which is nothing, so the left side is nothing raised to the power.

6:06On the right the row appears with alternate signs attached to it. So a row with alternate signs adds up to nothing. One minus four plus six minus four plus one is nought. One minus five plus ten minus ten plus five minus one is nought. Out to the twelfth power, again, not one failure. The whole of that alternating structure is one substitution, and it cost nothing. Those two identities look like a matched pair, and they are not.

6:41Take the very top row, the one with a single entry in it. That row adds to one, and two raised to the power nought is one, so the doubling identity survives up there. With alternate signs the same row still adds to one, and it is supposed to add to nothing. So the signed identity fails at the top row. And the reason is visible in the substitution itself.

7:08It asks for nothing raised to the power nought, and nothing raised to the power nought names nothing at all. Of the thirteen indices from nought to twelve, the doubling identity covers all thirteen and the signed one covers twelve. The one it misses is exactly the one where its own left side does not exist. Now the same move used as a tool. Raise ninety-eight to the fifth power.

7:38Do not multiply it out. Write it as a hundred less two, and expand. The six magnitudes are ten billion, one billion, forty million, eight hundred thousand, eight thousand, and thirty-two. Every one of those is a small number followed by zeros, because a hundred has powers you can write down without thinking. Positions one, three and five are subtracted, so bank the kept ones on one side and the taken ones on the other, and take the difference.

8:13The answer is nine billion, thirty-nine million, two hundred and seven thousand, nine hundred and sixty-eight. And this is why you split about a hundred rather than about ninety: the split about a hundred costs seven digits to write down, and the split about ninety costs twenty-nine. Same theorem, same six positions, four times the writing. Here is the move doing something a calculator will not. Which is bigger: one and one hundredth raised to the millionth power, or ten thousand?

8:50Never compute the power. Write the base as one plus one hundredth and expand it. The first term is one, and the second is a million times one hundredth, which is ten thousand. So two terms alone give ten thousand and one, which is already past the bound. And everything after them is a positive count times a positive power, so nothing that follows can pull the total back down.

9:20The comparison is settled and the power was never touched. But be exact about what that argument is: it says two terms are enough, and when they are not, it says nothing whatever. Ask it about the hundredth power of one and a tenth against a thousand, and the first two terms give eleven, which settles nothing at all, even though that power really is bigger. Now the substitution that proves something about every power at once.

9:53The claim: six raised to any positive whole power, less five times that power, leaves one when you divide by twenty-five. Six is one plus five. So put one and five into the form whose first term is one. The first term is one, the second is five times the power, and every term after those carries at least two factors of five. Take away five times the power and the second term is gone.

10:24What is left is one, plus a tail whose every term is divisible by twenty-five. Pull twenty-five out of the tail and what remains is a whole number, exhibited rather than assumed. Scored at twenty powers, twenty-five times that factor plus one is the quantity every time — and a factor one too big misses every time, which is how you know the scoring is doing something. At the first power the tail is empty and the factor is nought, and that is worth saying out loud, because the claim still holds there — there is simply nothing left over to multiply.

11:08One more consequence, and it is the one that makes awkward expressions tractable. Take a bracket and its twin with the sign flipped, raised to the same power. They agree at the positions that kept their sign and disagree at the ones that turned. So subtract them and the agreeing positions vanish; add them and the disagreeing ones do. Out to the eighth power there are forty-four positions altogether, and subtracting leaves twenty standing while adding leaves twenty-four.

11:43It does not double everything: it doubles half and destroys the other half, and which half depends on whether you added or subtracted. The fourth power of a sum, less the fourth power of the difference, is eight a cubed b plus eight a b cubed. The sixth power of a letter plus one, added to the sixth power of that letter minus one, is two x to the sixth, plus thirty x to the fourth, plus thirty x squared, plus two.

12:15And that is what makes square roots tractable: put the square root of three and the square root of two in, subtract the fourth powers, and every awkward term cancels, leaving forty times the square root of six. Take the sixth powers instead and it comes to three hundred and ninety-six times the square root of six. Step back and look at what has actually happened. There is one theorem.

12:45Three substitutions were made into it, and nothing else was added. Put a negative in the second slot, and the signs alternate. Put one in the first slot, and the coefficients stand bare. Put one in both, and a row adds to a power of two — or, signed, to nothing. Everything after that is one of those forms handed particular numbers. A hundred less two gives a fifth power; one plus a hundredth gives a comparison; one plus five gives a remainder of one under twenty-five; one plus eight gives a divisibility claim about sixty-four; and one plus three turns a row weighted by powers of three into a power of four.

13:34The divisor is an input, and the method is the content. So when a set of results in front of you looks like a list of separate facts, ask what was substituted — and the list usually collapses back to one line.