Solving for Maximum Value of h in \mu = \lambda h

In summary, the equation for solving for maximum value of h in μ = λ h is h = μ / λ, and it is used to find the largest possible value for h given a specific mean and standard deviation. It can be rearranged to solve for either the mean or standard deviation, making it a versatile tool for various statistical problems. Solving for maximum value of h allows us to determine the highest possible value of h consistent with the given mean and standard deviation, which is useful in statistical and scientific calculations. To use the equation, plug in the given values for μ and λ and solve for h.
  • #1
Ted123
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0

Homework Statement



I'm asked to find the maximum value of [itex]h[/itex] such that [itex]\mu = \lambda h\;\;(\lambda \in \mathbb{R})[/itex] satisfies: [tex]|1+\mu + \frac{1}{2} \mu ^2 + \frac{1}{6} \mu ^3| < 1[/tex]

The Attempt at a Solution



My hurdle is solving [itex]1+\mu + \frac{1}{2} \mu ^2 + \frac{1}{6} \mu ^3=1[/itex] to find the interval [itex]\mu\in (?,?)[/itex] which satisfies [tex]|1+\mu + \frac{1}{2} \mu ^2 + \frac{1}{6} \mu ^3| < 1[/tex]
 
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  • #2
-1 from both sides, then obviously μ=0 is a solution, and from there you should be able to find the rest of the intervals using the fact it's a positive cubic.
 

What is the equation for solving for maximum value of h in μ = λ h?

The equation for solving for maximum value of h in μ = λ h is h = μ / λ, where h represents the maximum value, μ represents the mean, and λ represents the standard deviation.

What does it mean to solve for maximum value of h in μ = λ h?

Solving for maximum value of h in μ = λ h means finding the largest possible value for h that satisfies the equation, given a specific mean and standard deviation.

How do I use the equation to find the maximum value of h in μ = λ h?

To use the equation, plug in the given values for μ and λ and solve for h. This will give you the maximum value for h that satisfies the equation.

What is the significance of solving for maximum value of h in μ = λ h?

Solving for maximum value of h in μ = λ h allows us to determine the highest possible value of h that is consistent with a given mean and standard deviation. This can be useful in various statistical and scientific calculations.

Can the equation be used to solve for other variables besides h?

Yes, the equation can be rearranged to solve for either the mean or the standard deviation, depending on the given values. This makes it a versatile tool for solving various statistical problems.

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