Interesting Algebra Problem

In summary, the conversation discusses a problem involving three variables and their sum, square sum, and cube sum. The question asks for the value of higher powers such as x^5 + y^5 + z^5. One solution is provided using Girard-Newton Identities, while another solution uses elementary methods but involves complex numbers. The conversation also mentions the use of Wolfram Alpha in finding a complex solution.
  • #1
PeroK
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I found this interesting video from Presh Talwalkar:

Problem Statement. If:
$$x + y + z = 1$$$$x^2 + y^2 + z^2 = 2$$$$x^3 + y^3 + z^3 = 3$$ Then, find the value of the higher powers such as $$x^5 + y^5 + z^5$$
The solution posted there uses the full Girard-Newton Identities. Here is an elementary solution using the same ideas:
First, note that $$x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + xz + yz)$$ and, for ##n \ge 3## (which is easy to verify): $$x^n + y^n + z^n = (x + y +z)(x^{n-1} + y^{n-1} + z^{n-1}) - (xy + xz + yz)(x^{n-2} + y^{n-2} + z^{n-2}) + xyz(x^{n-3} + y^{n-3} + z^{n-3})$$ Plugging the values we have in the first equation gives:
$$2 = 1 - 2(xy + xz + yz)$$ Hence $$xy + xz + yz = -\frac 1 2$$ Then, for ##n = 3## we have:
$$3 = (1)(2) - (-\frac 1 2)(1) + xyz(3)$$ Giving $$xyz = \frac 1 6$$ Then, for ##n = 4## we have:
$$x^4 + y^4 + z^4 = 3 - (-\frac 1 2)(2) + \frac 1 6 = \frac{25}{6}$$ And, for ##n = 5## we have:
$$x^5 + y^5 + z^5 = \frac{25}{6} - (-\frac 1 2)(3) + (\frac 1 6)(2) = 6$$
In general we have $$x^n + y^n + z^n = (x^{n-1} + y^{n-1} + z^{n-1}) + \frac 1 2(x^{n-2} + y^{n-2} + z^{n-2}) + \frac 1 6 (x^{n-3} + y^{n-3} + z^{n-3})$$ And that, in fact, ##x^n + y^n + z^n## must be rational for all ##n##.

Finally, note that we can generalise this procedure for any initial values of the first three expressions.
 
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  • #2
WolframAlpha only finds a complex solution. All the steps are possible with complex numbers, of course.
x = 1.431, y = -0.215-0.265i, z = -0.215+0.265i
 
  • #3
mfb said:
WolframAlpha only finds a complex solution. All the steps are possible with complex numbers, of course.
x = 1.431, y = -0.215-0.265i, z = -0.215+0.265i
Yes, I forgot to add a note to say that the underlying solutions for ##x, y, z## are complex. Early in the video he uses Wolfram Alpha as well!
 

Related to Interesting Algebra Problem

1. What is an interesting algebra problem?

An interesting algebra problem is a mathematical equation or puzzle that requires the use of algebraic principles and techniques to solve. These problems often involve variables, equations, and unknowns, and can be challenging and thought-provoking.

2. Why are algebra problems important?

Algebra problems are important because they help develop critical thinking and problem-solving skills. Additionally, algebra is the foundation for many other branches of mathematics and is used in various fields such as science, engineering, and economics.

3. How can I improve my skills in solving algebra problems?

One way to improve your skills in solving algebra problems is to practice regularly. You can also seek help from a tutor or join a study group to get a better understanding of the concepts and techniques. Additionally, using online resources and solving a variety of problems can also help improve your skills.

4. What are some common strategies for solving algebra problems?

Some common strategies for solving algebra problems include identifying and simplifying expressions, using the order of operations, solving equations using inverse operations, and using substitution or elimination to solve systems of equations. It is also helpful to check your solutions and use algebraic properties to manipulate equations.

5. Can algebra problems have real-life applications?

Yes, algebra problems can have real-life applications. For example, they can be used to solve problems related to finances, such as calculating interest rates or creating a budget. Algebra can also be used in science to model and analyze real-world situations, such as the growth of populations or the spread of diseases.

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