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Let E = Q[z]/(r* 2) _ Prove that E is field.Find the prineipal representative of [3x' 528 + 62? + 1] in E Find the multiplicative inverse of 21 + 1] in E....

Question

Let E = Q[z]/(r* 2) _ Prove that E is field.Find the prineipal representative of [3x' 528 + 62? + 1] in E Find the multiplicative inverse of 21 + 1] in E.

Let E = Q[z]/(r* 2) _ Prove that E is field. Find the prineipal representative of [3x' 528 + 62? + 1] in E Find the multiplicative inverse of 21 + 1] in E.



Answers

Verify that $ \text{div} \, \textbf{E} = 0 $ for the electric field $ \textbf{E(x)} = \dfrac{\varepsilon Q}{| \textbf{x} |^3} \textbf{x} $.

Okay. And this problem, we have an electric diable and were asked, were given its potential, which is a potential The disciple is equal to one over four pi absolute knots He which is the disciple moment times the coast I have data for fate is this angle road that the disciple that's all over r squared were asked to find the electric field the relationship between the electric field the electric potential is that the electric field is equal to the negative radiant of the mention Fletcher potential Expanding it out is DD x v extraction plus D d v D y the Y direction and so forth. So jump into it. So to start, we can break up this Grady in into each individual section we confined the electric field, the ex direction, which is simply negative. DVD X. So we're taking an X derivative of that expression one over for pie. Absalon nods. He goes on data over R squared. Okay, so we can now express this our distance in terms of x and y, which will be useful to us. So we'll have number four by a cyanide. You, Josephine think he co signed data over route exported plus y squared. We can also express the coastline data in terms of our X's and y's. Some fly even more so we have the same constant. I'm gonna leave. We'll be there before by absolute nods and little bit of algebra Report. We can replace the coastline data over Route X squared plus y squared to be X over X squared plus y squared to three abs Wisconsin Out. Take the derivative. We will end up with kind of a nasty expression. To be honest. Negative p over four pi absalon nuts Now bear with me x squared plus y squared resulted three abs Mine is excellent. Three abs x word plus y squared to the 1/2 comes to your ex This is all over x squared plus y squared cubed. So, like I said, kind of gross derivative. Simplifying it down more. We have pee over four pi absalon nods Times two x squared minus y squared all over X squared plus y squared to the five halves. And now we can go back to our data's and ours if we wish. Did you pee over for pie? Absolute knots to co sign squared data mine a sine squared data. Nice trick identity right there over R cubed. And that's it for the ex direction for the wind direction. We're taking a negative derivative. And why of the potential, which will simply after pulling out a conscious being negative p x for four ply Absolutely not times negative three halves X squared. That's why I squared the negative five halves as to why, once again we have. It's a somewhat difficult derivative. I'm skipping some steps because it will take way too long. We included all of them to do some simplification to X squared minus y squared over X squared plus y squared the five hands. Yeah, if we wish, How end up being Oh, it's miss something still like to take a step back. Okay, actually, we have three x y over x squared plus y squared to the five halves, and that simplifies down when we go back to signs and co signs. The didn't close on data signed data over are cute. All right, so it's noticed something here. Notice that the electric field of a di poll is actually given. It actually falls off as the cube of the radius away from the dye pole, usually with electric field, it falls off as R squared. This is private. Interesting result it's worth noting probably was writing down the disciple field is not proportional to the one of the art inverse squared. It's actually over are cute and that's it.

Electric fields. For a Dybul, that's a plus charge, and that's a minus. Charge. The distance between them is, let's say this is do or two and this is deal. We're too. And then we are trying to find the electric field over here for the pasture. The electrified will go in this direction for the minus charge electric visually in this direction. So if we have, if we have this distance to be our and this angles to be Ted up, then that daughter electric field will be to E signed it off. So electric field, you do or die polled. It was two e sign did, uh, it was two times. He's one way for by absolutely not. So these are the Chargers class Q and minus cube. So you over doesn't a You were s where times scientists are scientifically is. Do you over do over A. So we have whatever. Four by it's the not you d over a cube, and that is about one of our four by it's a not be over our cube because this is B and is almost equal to our for small

In this question we will be dealing with exponential equations. The question asks us to find the inverse of this function here. E to the power of X -E. to the power of -X divided by two and stated. The domains and ranges of these two functions F and the inverse of a Okay, yeah, we start with the domain of this function F. The domain is defined as the set of all input values for which the function is defined or has an output in other words. So what is the set of all X values for which ffx exists? Well, we know we can have E to the power of X. That is either the power of X exists for any real number X. And similarly the same thing is true for each of the power of minus x. In fact, that's just a reflection of dysfunction each other part books and so all real numbers X are valid. Next we have the range. The range is a bit more tricky to figure out, but uh we can still kind of reason mathematically there is. We noticed that F x is a large negative number or even just uh unknown small negative number in a manner of speaking Something less than -2. For example, then this number becomes quite a bit bigger than this number. In fact, already, when you get to well, okay, X equals negative thought. Just for the sake of example, that's About 0.007 And effects is -5. This number is pretty huge. And so um When X is less than -5, Well, when access wasn't in a good five, first of all, this is even bigger than what I just said. This is even smaller than 0.007. And so, in a sense, this term is dominating and the function is just it's almost exactly the same as the function With this being zero. So without this term. Yeah, in the opposite case of X is greater than five, for example, this becomes very big and this becomes very small, this is the dominant term and the function becomes almost exactly the same as this function. So, yeah, this for negative numbers or at least for substantial negative numbers. And this for substantial positive numbers, approximate the function F And we know that this function takes on all negative values, negative real values. Because of the negative sign on this being positive, mistakes are positive values because each of the power of X is always positive. And so we can deduce from this that F of X can take on any negative number or any positive number. And wow, that's not a very specific or very precise or rigorous uh statement it will do for the sake of this question. So we can say the F can take on all real numbers. Well, I mentioned positive and negative but we can easily see that if X zero Than these two terms are equal to each other and this denominator is zero. So the function can take on the value zero as well. Okay, so we have our domain and range which are both the set of all real numbers. Now we want to find the inverse of the function F. This inverse function is the function that takes F of X as an input in our prospects for any X in the domain. So any real number X. Specifically this now uh this is what the inverse of F does and that's how we can define it. But it's a bit awkward, awkward to work with mathematically. Excuse me to uh to be able to manipulate the equation more easily. I'll just call this ffx. Why? So X is my output and why? Sorry? X is my input. And why is my output for the function F. The function of inverse is the function that takes. Why as an input, That is the output of F. And gives X as an output, which is the input of that. Oh, if re functions are opposites not to find offenders, I'll just express X as a function of why? So reverse this equation. So As you might imagine first we multiplied by two, then we want some way to get rid of this X. Or to get X. So that it's not only experiment, we can't take the law algorithm because we have a difference of two exponential. And so one thing I will do is rewrite this minus X term. Each of the power of minus X as a fraction and then clear the denominator when we do that. Uh we noticed something interesting here, we have a to the power of X squared here we have a multiple of each of the power of X. And here we have a constant term that reminds us of a quadratic equation because we have a square term. Uh the problem with an experiment of one or the constant and in fact this is a quadratic equation except the variable in a sense is not a variable. It's the expression each of the power of X. And so we can just say let A equals E to the power of X. Give this a variable name so that it looks like a quadratic. So when we do that we have to we must be careful because each of the power of X is always positive. And so we can't assume A can take on any real value. Like we do for most variables of quadratic equations but we have to say that A is positive. Okay, so again, don't lose track of what our goal is. Our goal is to isolate X and write it in terms of a function of water. So since a contains X, we want to isolate A and write it as a function of why? So that we can later on substitute A equals B to the power of X To isolate A. We use the quadratic formula. So hopefully if you're doing pretty calculus, you're familiar with the quadratic formula already and you might even be yeah familiar with the form that allows you to Right the reduced form directly without having to reduce by two later. If this coefficient is even like in our case here anyways this is what we get why plus or minus the square root of Y squared plus one. Remember that A represents each of the power of X. So it would be wrong to say it is a negative value. And so if the minus here possibility with the minus is negative, we have to rule that out because that's not possible in the context of the previous uh equation. Which is this? No. How do we know which one is greater out of these two to tell if is negative or not? Well it's actually quite simple. We know that square root of y squared plus one must be greater than the square root of y squared right? Because here we have just added one and the square root function is an increasing function. So wow, adding 1 to the argument. What's going on? Increasing the argument of? The square root is going to increase the value of the whole like the square root of that number. And what is the square root of Y squared? That's just the absolute value of why? In other words the value of why? Regardless of its sign. So that's greater if y is positive and that implies at this expression is greater than why. And if what was negative, it implies that this expression is greater than negative Y. Which is the positive counterpart of the negative number. Why? And so we have that. This must be greater than why, no matter what what it is, no matter what real value, meaning that this is a negative expression. And so this is not possible. Plus is the only possibility how we can substitute B to the power of X and rewrite this in the log arrhythmic form to solve for X. There is our inverse function function that takes the output of S as an input and outputs a number that was input it into X into F. Originally since we usually right functions in this form with the letter X always representing the input. We can write the inverse function like this. Fn versa. Becks. Is this expression in terms of that input? X. Yeah. Yeah. And now we found the inverse function. Are we done not quite remember that we're talking about domain and range as well. However, it should be much easier to find the domain and range in this case because Uh huh. Since I said the domain is the set of all possible inputs of the domain of F. For example, is the set of all possible inputs in F. And the input of F is always the output of F in verse. And the range is the set of all outputs, then we know that the range of FN verse is the domain of F, and the domain of F inverse is just the range of F and both domain and range. For F, we're all real numbers, so the same is true for offenders. And this concludes this video.

We know the M E is the medium. Therefore, m E is gonna be 1/2. Do you see a friend is also the median. Therefore we know FN is also 1/2 D c. Therefore, we noto um and is a B plus D c over too. Therefore, this indicates that we have e f equals seven minus dese malice M e minus half an. Therefore, we have a B minus D c over too.


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