Newton binomial problem

So what's the smallest positive n that does that?In summary, the conversation discusses finding the minor value of a natural number n for which a complex number raised to the power of n becomes a real positive number. The conversation suggests using the binomial theorem and equating terms to cancel out imaginary components. However, this approach is incorrect as it only considers two terms and does not guarantee a positive real result. Instead, it is recommended to evaluate the complex number for successive values of n until a positive real result is obtained. Alternatively, the complex number can be rewritten in polar form and solved using the fact that n times the argument of the complex number must be a multiple of 2*pi for the result to be real and positive.
  • #1
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Homework Statement



Find the minor value of the natural number n such that [tex]\left (\frac{\sqrt{3}}{2} + \frac{1}{2}i \right )^{n}[/tex] be a real positive number.
EDIT: n must not be 0.

Homework Equations



Considering the binomial theorem as:

[tex]{\left(x+y\right)}^n=\sum_{k=0}^n{n \choose k}x^{n-k}y^k\,\![/tex]

I realized that we need to cancel the terms that we have i and -i. So we need that the terms where k = 1 and k = 3 be opposite. So we need that:

[tex]\binom{n}{1} \left( \frac{\sqrt{3}}{2} \right )^{n-1} \left ( \frac{1}{2}i \right ) = \binom{n}{3} \left( \frac{\sqrt{3}}{2} \right )^{n-3} \left ( \frac{1}{2}i \right )^{3}[/tex]

Is that right?

The Attempt at a Solution



So I tried to solve that equation as the follows:

[tex]\frac{n!}{\left( n-1 \right )!} \left( \frac{3^{\frac{1}{2}}}{2} \right )^{n-1} \left ( \frac{1}{2}i \right ) = \frac{n!}{3!\left( n-3 \right )!} \left( \frac{3^{\frac{1}{2}}}{2} \right )^{n-3} \left ( \frac{1}{2} \right )^{3}\left(-i \right )[/tex]

[tex]\frac{n!}{\left( n-1 \right )!} \ \ \frac{3^{\frac{n-1}{2}}}{2^{n-1}} \ \ \frac{i}{2} = \frac{n!}{3!\left( n-3 \right )!} \ \ \frac{3^{\frac{n-3}{2}}}{2^{n-3}} \ \ \left ( \frac{-i}{2^{3}} \right )[/tex]

[tex]n \ \ \frac{3^{\frac{n-1}{2}}}{2^{n}} \ \ \ i = \frac{n\left(n-1 \right )\left(n-2 \right )}{3!} \ \ \frac{3^{\frac{n-3}{2}}}{2^{n}} \ \ \ \left( -i \right)[/tex]

[tex]n \ \ 3^{\frac{n-1}{2}} = - \frac{n\left(n-1 \right )\left(n-2 \right )}{6} \ \ 3^{\frac{n-3}{2}}[/tex]

[tex]n = - \frac{n\left(n-1 \right )\left(n-2 \right )}{6} \ \ 3^{-1}[/tex]

[tex]18 = - \left(n-1 \right )\left(n-2 \right )[/tex]

[tex]18 = - \left( n^{2} -3n +2 \right )[/tex]

and finally

[tex]-n^{2} + 3n + 16 = 0[/tex]

But the answers to that quadratic equation aren't natural numbers. What did I do wrong?

Thank you
 
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  • #2
I don't know but try this:

[tex]\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)^{n}= p[/tex]

[tex]\frac{\sqrt{3}}{2}+\frac{1}{2}i= p^{\frac{1}{n}}[/tex]

[tex]\sqrt{3}+i= 2p^{\frac{1}{n}}[/tex]

[tex]\left(\sqrt{3}+i\right)^{n}= 2^{n}p[/tex]


[tex]2^{n}[/tex] times a positive real number is a positive real number, so find a positive real solution to the simpler equation

[tex]\left(\sqrt{3}+i\right)^{n}[/tex]
 
  • #3
follwoing on from psuedogenius, any complex number [itex] a+ bi [/itex] can be written [itex] r e^{i \theta} [/itex]

re-writing in that form should simplify the problem as when you multiply by itself you get [itex]( r e^{i \theta})^2 = r^2 e^{2 i \theta} [/itex] as [itex]r[/itex] is a real numer, so is [itex]r^2[/itex]
 
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  • #4
lanedance said:
follwoing on from psuedogenius, any complex number [itex] a+ bi [/itex] can be written [itex] r e^{i \theta} [/itex]

re-writing in that form should simplify the problem as when you multiply by itself you get [itex]( r e^{i \theta})^2 = r^2 e^{2 i \theta} [/itex] as [itex]r[/itex] is a real numer, so is [itex]r^2[/itex]

I can't solve this way because we didn't learn that on the classes. But thank you for the help.

Does anyone have another suggestions? What did I do wrong on my try?
 
  • #5
Taturana said:
I can't solve this way because we didn't learn that on the classes. But thank you for the help.

Does anyone have another suggestions? What did I do wrong on my try?

What you did wrong in your try is equating the k=1 and k=3. For one thing that doesn't make them cancel. For another thing, even it did, what about k=5, 7, etc? For another thing you also want the result to be positive, not just real. If you don't know polar form, then why not just try evaluating
[tex]
\left (\frac{\sqrt{3}}{2} + \frac{1}{2}i \right )^{n}
[/tex]
for successive values of n, until you get a value that let's you see the answer.
 
  • #6
Dick said:
What you did wrong in your try is equating the k=1 and k=3. For one thing that doesn't make them cancel. For another thing, even it did, what about k=5, 7, etc? For another thing you also want the result to be positive, not just real. If you don't know polar form, then why not just try evaluating
[tex]
\left (\frac{\sqrt{3}}{2} + \frac{1}{2}i \right )^{n}
[/tex]
for successive values of n, until you get a value that let's you see the answer.

Thank you for the reply.

I know the polar form: z = |z| * (cos a + i * sin a). But how do I solve the problem as lanedance said? I don't understand...

I consider the complex as [itex]
r e^{i \theta}
[/itex] and do what after that?
 
  • #7
Taturana said:
Thank you for the reply.

I know the polar form: z = |z| * (cos a + i * sin a). But how do I solve the problem as lanedance said? I don't understand...

I consider the complex as [itex]
r e^{i \theta}
[/itex] and do what after that?

Try to figure out what theta is. Then (r*exp(i*theta))^n=r^n*exp(i*n*theta). That's real and positive if n*theta is a multiple of 2*pi, right?
 

What is the Newton binomial problem?

The Newton binomial problem, also known as the binomial expansion or binomial theorem, is a mathematical concept that involves expanding a binomial expression (a + b)^n, where n is a positive integer, into a polynomial. It is named after Sir Isaac Newton, who first discovered the formula.

What is the formula for the Newton binomial problem?

The formula for the Newton binomial problem is (a + b)^n = ΣnCr * a^(n-r) * b^r, where nCr represents the combination of n objects taken r at a time. This formula can also be written as (a + b)^n = ∑(n, r = 0) nCr * a^(n-r) * b^r.

What is the significance of the Newton binomial problem?

The Newton binomial problem is significant in mathematics because it provides a way to expand binomial expressions to any power, making it easier to solve complex equations and problems. It also has many real-world applications, such as in probability and statistics, where it is used to calculate the probability of multiple events occurring.

What is the difference between the Newton binomial problem and Pascal's triangle?

The Newton binomial problem and Pascal's triangle are related concepts, but they are not the same. The Newton binomial problem is a formula that can be used to expand binomial expressions, while Pascal's triangle is a triangular array of numbers that can be used to find coefficients for the binomial expansion. Essentially, Pascal's triangle is a visual representation of the coefficients in the Newton binomial formula.

How is the Newton binomial problem used in real life?

The Newton binomial problem has many real-world applications, such as in finance, physics, and engineering. In finance, it is used to calculate the future value of investments and in physics, it is used to model projectile motion. In engineering, it is used to solve problems involving probability and combinations, such as in quality control processes. It is also used in computer science and data analysis to model and predict outcomes in various scenarios.

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