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If x+(1/x) = 7. Compute x²+(1/x²) and x³+(1/x³)

  1. Jan 6, 2017 #1

    TheBlackAdder

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    1. The problem statement, all variables and given/known data
    I.M. Gelfand - Algebra: Problem 133. You know that $${x+\frac{1}{x}=7}$$ Compute (a) $$x^2+\frac{1}{x^2}$$ and (b) $$x^3+\frac{1}{x^3}$$

    2. Relevant equations
    None

    3. The attempt at a solution
    I tried solving for x, and then substituting x. But that doesn't produce the correct answer. I won't copy all my incorrect attempts to LaTeX because I haven't found a fast way to convert it yet. https://www.wolframalpha.com/input/?i=(7-1/x)²+1/(7-1/x)²

    $${x+\frac{1}{x}=7}$$
    $${x=7-\frac{1}{x}}$$
    $$(7-\frac{1}{x})^2+\frac{1}{(7-\frac{1}{x})^2}$$
    $$\frac{2402x^{4}-1372x^{3}+294x^{2}-28x+1}{x^{2}(7x-1)^{2}}$$

    I'm guessing it's incorrect because for instance
    $$
    x+\frac{1}{x}=7\rightarrow\frac{1}{x}=7-x
    \rightarrow
    1=x(7-x)
    \rightarrow
    x=\frac{1}{(7-x)}
    $$
    does not equal
    $$
    x+\frac{1}{x}=7\rightarrow x=7-\frac{1}{x}
    $$

    However, when giving WA the two rational expressions, it apparently can solve the problem by substitution. https://www.wolframalpha.com/input/?i=if+x+1/x=7+then+what+is+x²+1/x²
    $$\left\{x+\frac{1}{x}=7,x^2+\frac{1}{x^2}\right\}$$
    Substitution:
    $$x^2+\frac{1}{x^2}=47$$

    How should you tackle this problem other than haphazardly squaring both sides and finding the answer? Or is seeing this relation some mathematical intuition that develops over time? I think I'm missing something super obvious. (Solution for (b) is found by the same method)
    $$(x+\frac{1}{x})^{2}=49\rightarrow x^{2}+2+\frac{1}{x^{2}}=49\rightarrow x^{2}+\frac{1}{x^{2}}=47$$

    Self studying math continuously makes me feel retarded; trying to solve something in a way too complicated approach and finding out the answer is super simple.
     
  2. jcsd
  3. Jan 6, 2017 #2

    andrewkirk

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    Squaring both sides is not haphazard. We want to find the value of ##x^2+1/x^2## and are given an equation involving ##x## and ##1/x##. We know that squaring that will give us an equation that involves ##x^2## and ##1/x^2## as well as some other terms. If things go well, those other terms will cancel out, be constants, or allow reduction of the problem in some other way. We try it, and we find that they do.

    While it's worth having a brief think about whether the first equation can be solved for ##x##, the nature of the problem strongly hints that that solution is not available in a nice neat form, because if it were, there would be no point in asking the value of ##x^2+1/x^2##. They would just ask the value of ##x##.

    Have you managed to solve (b)? If not, try doing the analogous thing to what you did to solve (a), but with cubes instead of squares. Again, this approach is suggested by the nature of the problem.
     
  4. Jan 6, 2017 #3

    TheBlackAdder

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    Okay, that's what I meant by haphazard. Not a wild guess, but trying something that looks like it might work. I didn't feel comfortable doing that since I did not know if it was allowed. I was under the impression math prefers raw calculation instead of trying an educated 'guess'.
     
  5. Jan 6, 2017 #4

    hilbert2

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    Something that would be haphazard would be dividing an equation with something without making sure that the divisor is not zero, or squaring both sides of an inequality like ##x+4>2##.
     
  6. Jan 6, 2017 #5

    epenguin

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    Just square (x + 1/x) - what do you get?.
    After that cube (x + 1/x) and what do you get?

    It is all moderately important because it gives you a way of reducing and then possibly solving "reciprocal polynomials" - ones in which the coefficient of the i-th term and the (n - i)-th are equal. Which are more important than you might think - for example all fourth degree polynomials can be thrown into the symmetric or reciprocal form by a suitable substitution, which gives one way of solving them algebraically.
     
    Last edited: Jan 6, 2017
  7. Jan 6, 2017 #6

    haruspex

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    Not at all. Maths is about seeing patterns and connections. The simplest path to a proof generally reveals the connection most clearly.
     
  8. Jan 6, 2017 #7

    pasmith

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    How do you think research works except by "trying something that looks like it might work?" Calculation is how you determine whether it actually worked.
     
  9. Jan 6, 2017 #8

    berkeman

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    [tex]x + \frac{1}{x} = 7[/tex]

    Just curious -- why not just use the Quadratic Equation to solve for the two solutions to that equation, and substitute them into the other equations to get the answers? Or was that discussed already and I missed it? (wouldn't be the first time...) :smile:
     
  10. Jan 6, 2017 #9

    haruspex

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    No, the first would simply be an error. (I see nothing wrong with the second.) Haphazard means random.
     
  11. Jan 6, 2017 #10

    TheBlackAdder

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    Comes only later in the book :)

    Well, to be honest, sometimes I think too little. And, I'm self studying with no formal mathematical background whatsoever. The only math I've seen in the past was purely rote learning. Even though I liked math back then, only now I realize I didn't understand it at all. That's why I felt uncomfortable deviating from calculation or looking for some 'rule', that's the only thing I've encountered in the past. Anyway, I still don't understand most of it, but that's why I am here. I've got nothing to lose but my ego.
     
  12. Jan 6, 2017 #11

    berkeman

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    Ah, fair enough.

    But you can always read ahead and learn how to use it, then use it to check the answers you get here using the techniques being suggested. :smile:
     
  13. Jan 6, 2017 #12

    hilbert2

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    Sorry, I made a wrong choice of inequality. I meant something like ##x+4<-1##. If ##x=-6##, the equation is true: ##-2<-1##, but if I square the both sides without changing the comparison operator I get ##4<1## which isn't true.
     
  14. Jan 6, 2017 #13

    andrewkirk

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    Not only is it allowed, but it is the main means by which major mathematical and scientific breakthroughs are made. We see various things that we could try, which may lead to a solution or innovation, or may lead to a complicated mess that goes nowhere. So we start with the most promising lead, try that and, if it leads nowhere, move on to the next most promising. If we run out of leads, we need to do some brainstorming to try and come up with new ones. For me jogging on a beach is a good environment for that.

    It occurs to me that detective work may be similar.
     
  15. Jan 6, 2017 #14

    epenguin

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    Well i think doing as #5 will get you there quite fast.
    Edit: well it will get you some enlightenment and an answer.
    Then - bearing out Ray Vixon's and others' comments, calculate the squares and cubes of (x - 1/x) and that should give you some extra enlightenment. :oldwink:

    Alternatively you have a clue that you don't have in research or if this had come from a physical or other science problem. It has come from a chapter and section of a textbook. It is an excercise in whatever principles or techniques etc.were treated there. So that chapter or section where problem 133 appears is probably indicating how you tackle this! A similar or related problem is probably solved as an example.
     
    Last edited: Jan 7, 2017
  16. Jan 6, 2017 #15

    Ray Vickson

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    No: that is how the final solution appears. The process of arriving at the final solution is often very different: many times I have spent hours (or even many days) and numerous sheets of paper plus several sharp pencils trying to solve some problems. Tried this---it didn't work; tried that---it failed; tried a third way and got closer, but still not complete. Finally, after numerous attempts the "aha" moment came (or maybe only a "OMG, it's over at last!" moment), and the solution was then sometimes frustratingly simple looking. When writing up the solution I never showed all the failed attempts, only a logical step-by-step development of the correct solution, and that would often give a misleading impression of "ease". Fortunately, things do get a bit better with practice.

    I would bet if you asked the helpers in this forum, my experience would not be that unusual.
     
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