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About the exercise that appears in TAOCP volume one, "Exercise Notes,"

In the TAOCP vol 1 section in the "Exercise Notes" section, there is a question that looks something like this:

"Prove that 13 ^ 3 = 2197. Summarize your answer. (This is a terrible problem that the author tried to avoid).

Questions:

  • How would you really prove it? (Direct multiplication is one way, another way is to use the formula (a + b) ^ 3). Is a solution required to use some method that allows us to make some kind of generalization?

  • What is the generalization here?

  • Why is this a terrible problem?

  • What are some other similar terrible issues you know about?

Please rate any answers.

PS We apologize if the above statement of the problem makes it look like a domestic problem, but it does not exist. Ask people not to flag this as a home problem so more people can give answers.

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3 answers

I would suggest that he is hinting that he might prove it, starting with Peano's axiom . Then, constructing integers and continuing to formally show that 13 ^ 3 = 2197 is a natural logical conclusion that follows from the definition of exponentiation,

We could generalize to show that for given a and b there exists some integer c, that is, a ^ b.

This is a terrible problem because most people find it uninteresting.

Similar problems can be found in the analysis course (along with some more interesting ones).

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I initially thought this as follows:
n 3 = n * n * n
log n (n 3 ) = log n (n * n * n)
log n (n 3 ) = log n (n) + log n (n) + log psub> (n)
3 = 1 + 1 + 1
3 = 3

This seems pretty circular in using logarithmic identities, but considering where I am in my research on algorithms, it was strangely comforting.

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Stuck in one exercise and "decided" it like this: a ^ b = mult (i = 1 - b) a

After a little thought, I came to the conclusion that this is a simple factorization (both 13 and 3 are prime numbers). Look at the little farm theorem.

(I know this is an old thread, but maybe it will help someone who is also looking for an answer to this option.)

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