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eduard
2 years ago
14

PLSASE HELP DUE TODAY WILL MARK YOU!!!

Mathematics
1 answer:
Xelga [282]2 years ago
4 0

Answer:

hjgjhgjhghjg

Step-by-step explanation:

ghjjgjesw54t32r3yggtfucgkyoughrhhfgh

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There are 1,000 grams in a kilogram. 
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In the vector space <img src="https://tex.z-dn.net/?f=P_2%28%5Cmathbb%7BR%7D%29" id="TexFormula1" title="P_2(\mathbb{R})" alt="P
tatuchka [14]

We could express any vector p\in P_2(\mathbb R) as

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16a+4b+c=0\implies c=-16a-4b

from which we can show that any such vector is a linear combination of some other vectors:

u=ax^2+bx+(-16a-4b)=a(x^2-16)+b(x-4)

More explicitly, we've shown that u is a linear combination of the vectors x^2-16 and x-4; in other words, u\in\mathrm{span}\{x^2-16,x-4\}, and in fact these two vectors form a basis for U. But this set does not span all of P_2(\mathbb R) because there's no combination of these two vectors that can be used to obtain a constant. We want the transformation to be usable for any vector in P_2(\mathbb R), so we need to add an additional vector extend the basis. We can do this simply by appending 1 into the spanning set. (Do check that the vectors remain linearly independent.)

Now we want the transformation to map those polynomials p(x) for which p(4)=0 to the zero vector. We know which vectors belong to the basis of U, so we need

f(x^2-16)=0

f(x-4)=0

f(1)=1

where the choice of the assignment for f(1) is arbitrary, so long as it's non-zero.

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3 years ago
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Answer:

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