Daedalus

Not only do they exclude certain parts of mathematics, but also claim that their science / health books give a solid foundation in all areas of science. However, they have no problem with denying scientific evidence or bearing false witness.

I am not familiar with the particular mathematical formulas that you are working with besides that they are based on complex numbers. However, it does sound like this is homework. If that is the case, we are not allowed to give direct answers to your questions, and I ask a moderator to move this topic to the appropriate forum. Although I am not familiar with these formulas, I am sure that I or someone else will be able to help you with this problem once we have established where this thread should be located. If this is not a homework assignment, then we can answer your questions directly. However, if it is homework, then we can only provide hints and clues to help guide you to a solution. Furthermore, I noticed that you attempted to use LaTeX commands in your post. While this forum supports LaTeX, you must encapsulate the code between [math]\left [ \text{math} \right] \ \left [ \text{/math} \right][/math] tags instead of the [math]\left [ \text{tex} \right] \ \left [ \text{/tex} \right][/math] tags that you tried to use. You can refer to our Quick LaTeX Tutorial for more information. In addition, our plugin supports most LaTeX commands found here at wikipedia: http://en.wikipedia....aying_a_formula. I hope that you find this post helpful, and I look forward to helping you find the solutions to your problem.

Since no one solved this challenge (or perhaps just didn't care lol), I will go ahead and post the answer.

This is an awesome view of Mars:

You can definitely see the tail bone and spine at the top of the image.

Happy belated birthday SFN!!!

Thank you studiot.

After looking into the problem with my blog not rendering certain equations correctly, I have determined that latex can no longer be successfully embedded within a HTML table. I am currently trying to find a workaround for this problem.

You're welcome : ) I'm glad we could help.

In other words, Log[3] in Mathematica / WolframAlpha is actually [math]\text{ln} (3)[/math] where Log[10,3] is [math]\text{log}_{10} (3)[/math]: See Mathematica's help:

Here is the latest image from Curiosity - Facebook - NASA's Curiosity Mars Rover: "Me & My Shadow... & Mount Sharp. My view of the 3-mile-high mountain in the middle of Mars' Gale crater." http://www.jpl.nasa.gov/news/news.cfm?release=2012-231 Curiosity's decent to Mars: "The Curiosity Mars Descent Imager (MARDI) captured the rover's descent to the surface of the Red Planet. The instrument shot 4 fps video from heatshield separation to the ground."

This is pretty cool! NASA has provided us with an online 3D interactive view of Curiosity and the Martian terrain. http://mars.jpl.nasa...lore/curiosity/ It looks like we'll be able to track its progress using this application.

Here is a link to the raw images taken by the rover: http://mars.jpl.nasa...multimedia/raw/

Yes... Congratulations to the Mars Rover Curiosity team!!!!

‎1 hour and 31 minutes till we land on Mars!!! Finger are crossed, and I wish the Mars Rover Curiosity team the best of luck http://www.space.com...laboratory.html

Thank you. It brings a smile to my face knowing that my work actually proved to be useful . I do have a blog here at SFN where I posted a polished version of this thread. However, there must have been a software update because the page is not rendering correctly. Summer session is over and I have a three week break from my academic studies before fall session begins. With that being said, I will work towards updating my blog and post a link here once it is fixed. As for your questions, I have explored similar variations to the one that you suggested. So far, I have not found any other useful properties other than the ones I posted in this thread. On a separate note, I am curious as to how you used my equations to check your work. If you are not comfortable sharing your work publicly in this thread, then you can always send me a private message.

Adam, to understand why [math]c^2[/math] even appears in the equations, we will take a look at deriving time dilation. The following image illustrates a light clock at rest: This clock measures time according to: [math]\Delta t = \frac{2 L}{c}[/math] where [math]L[/math] is the distance separating the reflective plates and [math]c[/math] is the speed of light. This gives us: [math]\frac{\text{distance}}{\text{speed}}=\frac{\text{distance}}{\text{distance}/\text{time}} = \text{time}[/math] We have twice the distance, [math]2L[/math], because the light must travel to the top plate and be reflected back to the bottom plate in order to measure time. An observer in motion relative to the light clock will not see the path of light as a straight up and down path. Instead, they will see the following: As we can see from the above image, an observer moving at a speed of [math]v[/math] relative to the light clock will see the path of the light elongated to a distance [math]R[/math], which is equal to the hypotenuse of the triangle: Because the speed of light is the same regardless of one's frame of reference, an observer moving at a speed of [math]v[/math] relative to the light clock will see that the clock is measuring time according to: [math]\Delta \tau = \frac{2 R}{c}[/math] where [math]R=\sqrt{\left(\frac{v \Delta \tau}{2}\right)^2+L^2}[/math] as determined by the distance formula (Pythagorean theorem). Multiplying both sides of the equation by [math]c[/math] and squaring the result, we get: [math]c^2 \Delta \tau^2 = 2^2 R^2 = 4 \left( \frac{v^2 \Delta \tau^2}{4} +L^2\right)=v^2 \Delta \tau^2+4L^2[/math] Subtracting [math]v^2 \Delta \tau^2[/math] from both sides yield: [math]c^2 \Delta \tau^2 - v^2 \Delta \tau^2 = 4L^2[/math] Factoring [math]\Delta \tau^2[/math] on the left side yields: [math]\Delta \tau^2 \left(c^2 - v^2 \right) = 4L^2[/math] Now we can divide both sides by [math]\left(c^2 - v^2 \right)[/math]: [math]\Delta \tau^2 = \frac{4L^2}{ \left(c^2 - v^2 \right)}[/math] Factoring [math]c^2[/math] from the denominator on the right side and taking the square root of the result yields: [math]\Delta \tau = \frac{2L}{ c \sqrt(1 - \frac{v^2}{c^2})}[/math] Because [math]\Delta t = 2L/c[/math] we can substitute [math]\Delta t[/math] into the final result to yield the time dilation formula: [math]\Delta \tau = \frac{\Delta t}{ \sqrt(1 - \frac{v^2}{c^2})}[/math] As you can see from the above formula, [math]\Delta \tau = \gamma \Delta t[/math] where [math]\gamma = \frac{1}{ \sqrt(1 - \frac{v^2}{c^2})}[/math] is the Lorentz factor and it contains [math]v^2[/math] and [math]c^2[/math] (Gamma, [math]\gamma[/math], forms the basis for transformation in special relativity). Applying this to other equations, such as those for energy, yields:

I recently moved into a new apartment, and I finally have internet again. I'll give you guys another week on this one ; )

Nope that's not the answer : (

This can be a fun problem to solve if you like calculus. There is a cow tethered by a rope to a circular barn on the north side of a fence which extends eastward radially from the barn. The rope and the fence has a length equal to the circumference of the barn. Find the grazing area as a function of the radius of the barn.

Ok... Let's devise an experiment right here so that we can verify your claims. I'll create several images and send a PM including one of the images to anyone a referee who is willing to participate at a specified time (must be someone like Swansont or Cap'n). Then, at the specified time that is agreed upon by everyone, we will think about the image for 10 minutes (or whatever time you think is necessary for you to get a mental picture of the image). We will count the number of images that you get correct. I will count a hit even if you can only get part of the image correct. Are you game?

I never said that indeterminate forms cannot be evaluated. I simply demonstrated why the limit[math], \ \lim_{(x,y) \to (0,0)} \frac{y}{x} \ ,[/math] is undefined. Indeterminate forms, [math]\ \frac{0}{0}, \ \frac{\infty}{\infty}, \ \infty - \infty, \ 0 \times \infty, \ 0^0, \ \infty^0, \ 1^\infty \ ,[/math] can be evaluated using limits.

We can use L'Hopital's rule to verify the accuracy of your statement: [math]f(x)=8-x^3[/math] [math]g(x)=2-x[/math] such that: [math]F(x)=\frac{f(x)}{g(x)}[/math] where: [math]\lim_{x \to 2} \frac{f(x)}{g(x)} \ = \ \lim_{x \to 2} \frac{f'(x)}{g'(x)} \ = \ \lim_{x \to 2} \frac{-3\, x^2}{-1} \ = \ \frac{3 \times 2^2}{1} \ = \ 12[/math] Note: In most cases, L'Hopital's rule can be used to determine the limit of indeterminate forms - [math]\frac{0}{0}[/math] and [math]\frac{\infty}{\infty}[/math]. However, this is getting off-topic and it would be better to start a thread in the mathematics forum discussing indeterminate forms.