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02 1 PointThe difference between the Hermite polynomial approximation p2n+1 of a function f and that function always goes to zero as n _ O_TrueFalse...

Question

02 1 PointThe difference between the Hermite polynomial approximation p2n+1 of a function f and that function always goes to zero as n _ O_TrueFalse

02 1 Point The difference between the Hermite polynomial approximation p2n+1 of a function f and that function always goes to zero as n _ O_ True False



Answers

True or False The polynomial function $f(x)$ has a factor $x+2$ if and only if $f(2)=0 .$ Justify your answer.

So the topic of this video is to to explore what happens to tailor paul Newman expansions. The further away you get from the center point and the question asked is whether the error between the original function and the linear function is always greater than the error between the function and the second order. And while it might be the case when you're near the center point, that is not true when you are far away. And to give an example, simply use any arbitrary function here, I'll use co sign X. This first degree order of coastline. Ex well, this is simply just one and the second degree. Well this is one minus X squared over two factorial. And if you were to sketch a graph or you find that if this is cosign X, you had the tangent line and then you have the parabola. Well, simply pick a super far away value. Say x equals two pi right here. What you get is well, The air between the function and the linear approximation is zero, But the air between the function and the quadratic is actually non zero sum positive value. And, well, we know that positive numbers are greater than zero, which contradicts what we assume the first place, and therefore the statement is false that the second degree order approximation is always better than the linear approximation.

Pressure is asking that that the given statement is to force and the statement is a polynomial function off degree and wit rial ovations has exactly and complex CEOs at most and off them are really zeros. So we need to find out that this segment is true or force. Hey, this is statement is too, because it is the result off the fundamental todo a zebra and defector cheery. That's why we can say that this is the correct statement and we can find it out on the fundamental curable and zebra and defector tear. So this is the complete answer of this pressure.

Problem. Number 11. The statement is through. Assume that April packs is equal to a explosive. Be then, if dash offense is able to a ah, the our land, the are then on a turban. C and D is an A to integration from sea to the off square root off. One lost a squared E X, which is equal to its bare root off. One plus is where T minus sea. The approximation is and equal to submission from cape with 12 a sequel to end over square root of one plus a square in the X. Okay, just equal toe. Want us a square eight minus C. So the approximation gives the exact our clips.

In this problem, we will cover local linearity. So to solve this problem, we have to refer back to the tangent plane approximation, which I have written here in green and we see that we will have to double check to make sure that this tangent plane approximation is correct. So we have to find the partial derivatives With respect to X&Y at the .01, and also the value of the function at 01. So we'll begin by finding f. 01. And if we plug that in, we will get that. This is just one Now to find the partial derivative of respective X. And that's just taking the derivative of the function holding the variable. Why fixed? So keeping Y in front we have E. To the X. Squared. And we also want to multiply by the derivative of the exponents itself. So two X. And that gives us two x. Y. Each of the X square. If we were to plug in the .01 into this partial derivative, we would get that. The whole thing is zero. And now we move on lastly to the partial derivative with respect to Y. And that's just take more function, the derivative of the function holding the X. Variable fixed now. And when we take the derivative with respect to why we're just going to get E. To the X. Square. And if we were to plug in the .01 into this partial derivatives, we would get one because each of the zero is just one. So now using our information, we can write out our approximation. So we have F X. Y is approximately equal. So one plus. Since the partial derivative respect the X. Is zero, there's not gonna be any X terms in this equation. So we can move right along to the Y. And we see that the partial derivative is one. So we'll just have why my just one here and of course the positive one and negative one cancels out. So this can be simplified to F. Of X. Y approximately equal. So why? That means that the statement given and blue is true.


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