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Let 0 = V/3. show that(a)O(az...

Question

Let 0 = V/3. show that(a)O(az

let 0 = V/3. show that (a) O(az



Answers

Show that for $V$ less than zero, $I_{\text {net }} \approx-I_{0}$

Problem tells us that the magnitude of the vector is zero. And to show that the vector components will be 00 So the Man Utd zero that means the square root of the X component and the white compelling the some of those two. But after total equal zero. So if we square about signs, um, we know that X squared plus y squared has to be equal to zero. And the only way that can happen because x squared and why squared? Neither one can be negative because of just the definition of squaring. Um, the only way that they can equal zero is if x equals y, and they both equal zero, which implies that the two components will be zero. Okay, Thank you, very.

In this question, we will be talking about linear operators and inner product spaces. The question asks us to prove that the identity operator and the zero operator satisfy the property of being their own a joints or being self a joint. Recall that if we have a vector space V. Mm and X and Y are too arbitrary vectors envy, then we call the operator B. Um Or we say that the operator V has an ad joint denoted B. Star. If the inner product of the image of X under B. And why is the same as the inner product of Yeah. X. And the image of the star? The image of why under B. Star for any X. And Y envy. Mhm. So this must be true for any vectors X. And walk. Mhm. Well, we want to show that if we put the identity operator here or if we put the identity operator here and here, this is satisfied, meaning that the I. Star or Okay, I've got any operator satisfies this. I start property And same thing with zero. So consider the image of X under eye. Yeah. Which is just X. If we take the inner product of this image and why we get the inner product of X and Y. Which we immediately recognize as being the inner product of X. And the image of Why under the identity, meaning that I satisfies this definition of being B. Star for the operator B. Which is also I in this case and says this is true for any vector X and Y. I equals the ad right of I. Mhm. Mhm. Now, for zero of X. And why? Yeah, we know. Uh Oh well, we'll consider the inner product of O. Of X and Y. And we know that the image of X under Missouri mapping is just the zero vector. Which uh huh. By definition of inner product. Inter products Gives a zero in our product. And using the same logic that is also the inner product of X. And the zero vector. Okay. And that zero vector can also be written as the image of Why? Under the zero mapping, showing that The zero mapping is also its own Android. And that brings us to the end of this video. Thank you for watching.

You're s three. Your, um now riveted of the absolute value invective function with objective function is non zero sheerness. That Artie is not equal to zero elector, then review t of the absolute value of our team is equal to one over. The absolute value are this is actually a normal value. I yorn hands rt died, it are. And then the hints were given that norm or we were You are tee are even the archaea tonight with you? This means that the one R T is equal to the swelling or dotted with itself is nice. You know he's working from the left side. We have really with respective team of warm Are you? No, we have by year five in particular, Change rule objective functions and real value functions that this is equal to Yorn are finally fine Can give you another a better way to do this would be recognize that about well actually use a visit differentiation. So you know that have been says no are he's weird equal to r t are tee surviving visit differentiation You differentiate such a respected t get to are tee we need to of one party is equal to one right side point. Our prime a T or was rt are prime a t. This is the same as our teeth. Out of our way. Que our prime a tea out of it are not right, are you are kind of so again, you have a normal bar. 90 years since 90 So I go outside. Like to know Get that you Oh, are is eat what? To one over the normal Hard to. All right, we got our are you not our find it too. And this is exactly

We need to show that in the negative X is approximately equal to one minus X. Near X equals zero. So we have E to the negative X is our F of X. And to show that it is approximately need to tangent line approximation. So this is F A plus F prime at a times X minus A. So we need to find F at zero because they told us near X equals zero, say zero, so F at zero is E to the negative X. Either the negative X Eating The zero is 1. Uh, F prime of X is equal to negative E to the negative X. So again, f prime at zero. This would be a negative one times X. So this is equal to one minus X.


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