# Gamma plus or minus gamma-beta, two different outcomes?

• ANvH
In summary: I am looking forward to your further posts, on the definitions of hyperbolic angles, and much else. Thanks for the help, In summary, the conversation discusses the correct equation for γ(1+β) and how to derive it using trigonometry. One person made a mistake due to using Microsoft Equation Editor, but was corrected by others through the use of pen and paper. They also mention a possible discussion on the definitions of hyperbolic angles.
ANvH
In Wikipedia it reads that γ(1+β) = $\sqrt{\frac{1+β}{1-β}}$, however, if I did my homework correctly I get γ(1+β) = $\sqrt{\frac{1+β^{2}}{1-β^{2}}}$. Digging more deeply into why Wikipedia is listing it as such I found that it is based on the hyperbolic angles:γ=coshΘ. But it leads to definitions, not an explanation. More interestingly, WikiPedia is equating γ(1-β) with $\sqrt{\frac{1-β}{1+β}}$. However, my calculations lead to

γ(1-β) =γ-γβ=$\sqrt{\frac{1}{1-β^{2}}}-\sqrt{\frac{β^{2}}{1-β^{2}}}=\sqrt{\frac{1-β^{2}}{1-β^{2}}}$,

which is equal to unity. I guess I am doing something utterly wrong and it feels like as if I have to start over learning calculus. I truly hope someone is setting me straight,

thanks,
Alfred

$$\gamma = \frac{1}{\sqrt{1 - \beta^2}} = \frac{1}{\sqrt{\left( 1 - \beta \right) \left( 1 + \beta \right)}}$$

$$\gamma \left( 1 + \beta \right) = \gamma \sqrt{\left( 1 + \beta \right) \left( 1 + \beta \right)} = \sqrt{\frac{\left( 1 + \beta \right) \left( 1 + \beta \right)}{\left( 1 - \beta \right) \left( 1 + \beta \right)}} = \sqrt{\frac{1 + \beta}{1 - \beta}}$$

ANvH said:
I get γ(1+β) = $\sqrt{\frac{1+β^{2}}{1-β^{2}}}$.

This is where a mistake is; look at what Peter wrote. I am not sure, but did you make the mistake

$$\left( 1 + \beta \right)^2 = 1^2 + \beta^2?$$

In general, ##\left( x + y\right)^2 = x^2 + 2xy + y^2 \ne x^2 + y^2##.

Also, below, you seem to be saying ##\sqrt{x} - \sqrt{y} = \sqrt{x -y}##, which is not correct.

ANvH said:
$\sqrt{\frac{1}{1-β^{2}}}-\sqrt{\frac{β^{2}}{1-β^{2}}}=\sqrt{\frac{1-β^{2}}{1-β^{2}}}$

PeterDonis said:
$$\gamma = \frac{1}{\sqrt{1 - \beta^2}} = \frac{1}{\sqrt{\left( 1 - \beta \right) \left( 1 + \beta \right)}}$$

$$\gamma \left( 1 + \beta \right) = \gamma \sqrt{\left( 1 + \beta \right) \left( 1 + \beta \right)} = \sqrt{\frac{\left( 1 + \beta \right) \left( 1 + \beta \right)}{\left( 1 - \beta \right) \left( 1 + \beta \right)}} = \sqrt{\frac{1 + \beta}{1 - \beta}}$$

Transcribing into trigonometry...

$$\cosh\theta =\frac{1}{\sqrt{1-\tanh^2\theta}} =\frac{1}{\sqrt{(1-\tanh\theta)(1+\tanh\theta)}}$$

$$\cosh\theta(1+\tanh\theta) =\cosh\theta\sqrt{(1+\tanh\theta)(1+\tanh\theta)} =\sqrt{\frac{ {(1+\tanh\theta)(1+\tanh\theta)}}{{(1-\tanh\theta)(1+\tanh\theta)}}} =\sqrt{\frac{ {1+\tanh\theta}}{{1-\tanh\theta}}}$$However, thinking trigonometrically...

Strating from the LHS:
$$\cosh\theta(1+\tanh\theta) =\cosh\theta(1+\frac{\sinh\theta}{\cosh\theta}) =\cosh\theta+\sinh\theta =e^{\theta}$$Starting from the RHS:
$$\sqrt{\frac{ {1+\tanh\theta}}{{1-\tanh\theta}}} =\sqrt{\frac{ 1+\frac{\sinh\theta}{\cosh\theta}}{{1-\frac{\sinh\theta}{\cosh\theta}}}} =\sqrt{\frac{ \cosh\theta+\sinh\theta}{\cosh\theta-\sinh\theta}} =\left(\frac{e^\theta}{e^{-\theta}}\right)^{1/2}=e^{\theta}$$

Note
$$\cosh\theta(1-\tanh\theta) =\cosh\theta-\sinh\theta =\cosh(-\theta)+\sinh(-\theta) =e^{-\theta} =1/e^{\theta}=\sqrt{\frac{1-\tanh\theta}{1+\tanh\theta}}$$

ANvH said:

γ(1-β) =γ-γβ=$\sqrt{\frac{1}{1-β^{2}}}-\sqrt{\frac{β^{2}}{1-β^{2}}}=\sqrt{\frac{1-β^{2}}{1-β^{2}}}$,

which is equal to unity. I guess I am doing something utterly wrong and it feels like as if I have to start over learning calculus. I truly hope someone is setting me straight,

Try this:

γ(1-β) =γ-γβ=$\sqrt{\frac{1}{1-β^{2}}}-\sqrt{\frac{β^{2}}{1-β^{2}}}=\left(\frac{1}{\sqrt{1-β^2}}-\frac{β}{\sqrt{1-β^2}}\right)=\frac{(1-β)}{\sqrt{1-β^2}}=\frac{\sqrt{(1-β)(1-β)}}{\sqrt{(1-β)(1+β)}}=\frac{\sqrt{1-β}}{\sqrt{1+β}}$

and

γ(1+β) =γ+γβ=$\sqrt{\frac{1}{1-β^{2}}}+\sqrt{\frac{β^{2}}{1-β^{2}}}=\left(\frac{1}{\sqrt{1-β^2}}+\frac{β}{\sqrt{1-β^2}}\right)=\frac{(1+β)}{\sqrt{1-β^2}}=\frac{\sqrt{(1+β)(1+β)}}{\sqrt{(1-β)(1+β)}}=\frac{\sqrt{1+β}}{\sqrt{1-β}}$

ANvH said:
In Wikipedia it reads that γ(1+β) = $\sqrt{\frac{1+β}{1-β}}$, however, if I did my homework correctly I get γ(1+β) = $\sqrt{\frac{1+β^{2}}{1-β^{2}}}$.

Sorry, you didn't do you homework correctly :-(

##\gamma(1+\beta) = \sqrt{\frac{1}{1-\beta^2}}(1+\beta) = \sqrt{\frac{(1+\beta)^2}{1-\beta^2}} = \sqrt{\frac{(1+\beta)(1+\beta)}{(1+\beta)(1-\beta)}} = \sqrt{\frac{1+\beta}{1-\beta}}##

I see a bunch of posts snuck in before mine

I am grateful to all of you. Yes, I made a big mistake, thanks to using Microsoft Equation Editor, where I fooled myself. Always use pen and paper. This is not meant as an excuse, I stand corrected.

## 1. What is the significance of gamma plus or minus gamma-beta in scientific research?

Gamma plus or minus gamma-beta refers to two different outcomes that can occur in a scientific experiment. These outcomes are often associated with the presence or absence of a specific factor or treatment, and can provide valuable information about the relationship between variables.

## 2. How do scientists determine which outcome is considered "positive" or "negative" in gamma plus or minus gamma-beta?

The determination of a positive or negative outcome in gamma plus or minus gamma-beta depends on the specific research question being addressed. In some cases, a positive outcome may be one that supports a hypothesis or demonstrates a desired effect, while a negative outcome may indicate a lack of effect or contradict a hypothesis. However, this can vary depending on the context and goals of the study.

## 3. Can a study have both a gamma plus and a gamma minus outcome?

Yes, it is possible for a study to have both a gamma plus and a gamma minus outcome. This may occur when there are multiple factors or treatments being investigated, and each one has its own set of outcomes. In this case, the presence or absence of a specific factor may result in a gamma plus or minus outcome, respectively.

## 4. How do researchers interpret the results of a gamma plus or minus gamma-beta analysis?

The interpretation of gamma plus or minus gamma-beta results can vary depending on the study design and research question. In general, a gamma plus outcome may be seen as evidence for a causal relationship between a factor and an outcome, while a gamma minus outcome may suggest that the factor does not have a significant effect. However, it is important for researchers to consider other factors and potential confounding variables when interpreting these results.

## 5. Are there any limitations to using gamma plus or minus gamma-beta as a statistical analysis method?

Like any statistical analysis method, there are limitations to using gamma plus or minus gamma-beta. This method may not be appropriate for all types of data or research questions, and may not provide a complete understanding of the relationship between variables. Additionally, the interpretation of gamma plus or minus gamma-beta results may be influenced by factors such as sample size and study design, so caution should be taken when applying this method to new research.

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