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Use (8.5) to show that the given functions are linearly independent.$$e^{i x}, e^{-i x}$$...

Question

Use (8.5) to show that the given functions are linearly independent.$$e^{i x}, e^{-i x}$$

Use (8.5) to show that the given functions are linearly independent. $$e^{i x}, e^{-i x}$$



Answers

Show that the $m$ functions $e^{r x}, x e^{r x}, \ldots, x^{m-1} e^{r x}$ are linearly independent on $(-\infty, \infty) .[$ Hint: Show that these functions are linearly independent if and only if $1, x, \ldots, x^{m-1}$ are linearly independent.]

In this video to go through the answer to question number 31 from chapter 9.4 so as to find that Ah, the determinant off this matrix and show this equal to zero. Yes, this is quite easy. It's two by two matrix. So it's the bottom. Right? Uh, which is two times model t time by the top left Just T square serves T t to t squared times model T minus bomb left, which is to tee times talk right, which is T Sizemore t. Which means that t squared my tea, which seek with zero. It's the next up. Castillo show that the factors which have written here is t wan t to, ah, linearly independence on for the values of tea on the real line. Okay, so, through that we want to assume that we can write it. Ah, right. Allan e, assume off these two vectors see? Want to see to being constants? You want to show that we assume that we can write it as a linear Cem zero and they show that this implies that C one and C two you are equal to zero. Okay, let's write that in component form. So go C one times t squared plus C two times T Monty Equal saver on dhe. See one times t plus C two times. Monty is equal to zero where I've divided the bomb equation by two. Okay, so now look, let's look at two different cases. It's the first, the case where t is less than zero. What he is less than hearing about t is minus T. So equations are gonna become C one minus C two times T squares equals zero on dhe minus C one plus C two times t is equal zero and t copies in a row. So therefore, we must have that C one is equal city. Get on for tea is greater than or equal to zero. The components off Theo equation become C one close C T times t squared because when t it's great and we could see a roti. Much of the city is equal to t the secret zero and then the bottom equation becomes C one plus c t times by t equals zero. Okay, so then this implies that C one is equal to minus ct. Okay, we're looking for two pet linearly at linear dependence on the whole of the real line. So both these equations C one equals minus C two on DSI one equal see two. They both need to be satisfied. If that both size fives, then we must have that C one equals CT equals zero. So, therefore

Use the Ron skin to show that the given functions are linearly independent on the given interval. Well, I need to write F one of X f two of X and of three of X. Now I have to take the derivative in the next line and now I have to take the derivative in the next line. If I take the determinant of this, I get to That's it. And so because that is not zero than it does mean that the set is linearly independent.

Okay. This question asks us to use the Ron ski in to show that the given values are linearly independent on the given interval of I. And so I'm just gonna go ahead and write the given values on the top row of the Ron skin. Um, the 1st 1 is the sign of X. This second is the co sign of X and the third is the tangent of X. Oops! The interval is negative. Pi over two two pi over to and so driven of of the sign is the co sign driven of the coastline is the sign opposite sign Um, derivative of the tangent is the Sikh int squared of X derivative of the coastline is the opposite of the sign Derivative of the opposite of the sign is the opposite of the co sign and derivative of the seek. It squared of X Um, that would be the seeking squared of X is one over the co sign of X squared, which would be the co sign of X to the negative to power, which would be negative, too. Co sign of axe times the derivative of the co sign of axe, which is the opposite of the sign of X. So that's going to be too sine x co sign X Silly me. I forgot to write my negative three power in here. So I've got a race. This so that would be to sign X over co sign Queued of X. And the reason why we need a negative three here is if it's not obvious, is we have to subtract one from negative to when you subtract one from negative to you get negative three. So to sign, axe over co sign Cubed of X. All right, so now we have to use the Ron skin. Okay, Um interesting. And I'm wondering how I'm going to do this with the given amount of space. Um, but first, I'm gonna erase my interval. I'm going to remember what that interval is. Then I have to do the determine it, which think is gonna be pretty crazy to dio. But let's do it anyway. I'm just rewriting the 1st 2 columns. All right, so the first term is going to be this times this times this this plus this times this times this plus this time this time this minus this times This times this minus this time this time. This minus this time this times This? Um uh, let's just look at some of these terms. The first term, the first red line is going to be negative Sine squared of X times two sign of X over co sign of three coast on the third car X, which will give us negative too. Sign cubed of X over coastline cubed of X which would be tangent cubed of X So that first term will end up being negative Two tangent cubed of X. Okay, let's look at the second term co sign of axe times one over co sine squared of X times Negative sign of X. So that's going to give us negative tangent of X. Look at the next one tangent of X times co sine squared of X negative. Verifying that I'm using my negatives correctly. Okay, that's going to be negative. Sign Exco sine x Okay, minus 10 X sine x sine x tan x sine x sine x Oh, my goodness. Um, that would be the sign queued of X over the coup sine X Let's do the next one. So this is gonna be a plus. So we've got sine X. We have co sign of X and we have one over cause and squared of X. So that would just be tangent of X. And interestingly enough, that cancels out. Okay, last one co sine squared of X is gonna be a minus in front of it, by the way, and I wanted to use blue co sine squared of X over coast on Cuba Vax That would be, um, won over co sign of X. This would be to tangent of X. So we're left with that. And so I am going to go to a graphing calculator and I'm going to put that function in the graphing calculator. Negative two tangent. Jude Cube of acts I didn't like that functions Tangent of X. Okay, there we go. Negative two tangent of X to the third hour. Spoonley Grete. Okay, rice minus sign of X. Could you sign of X minus fine of X to the third power over co sign of X minus two. Engine of X Okay. And going to set my interval to be the interval that we were given, which is negative. Pi over to two pi over two, actually I'm gonna set it to be negative part of pie. Just, uh see what it looks like. Interesting. My step is going to be pi over two, so I can see from negative pi over two two pi over two. It does have a value. However, it must not include negative pi over two or positive pi over two. Because at those values, it is undefined. And I could see that the interval waas negative pi over two too high over to not including the starting point or the ending point. And so, yes, it does have a value. It actually equals zero at one point, which is the origin. And by the way, it looks like this. No, this is gonna be negative. Pi over two. And this is positive pie over to thinking there is most likely a way to simplify that expression. Um, but nevertheless, that's what we got. So, um

Were given three functions and asked to use the Ron ski in to verify that the three functions are linearly independent. The three functions are f of X equals one F of X equals three acts and f of X equals X squared, minus one on the interval from negative infinity to infinity. Well, I take the derivative that's gonna be 03 and two x Take the driven of again 002 The determinant is just going to be one times three times to which is six that determine it is not zero and so the functions are


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