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- Thread starter pnaik2008
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- #2

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Here's a http://en.wikipedia.org/wiki/Dirac_delta_function" [Broken].

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- #3

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But the S* is linearly Independent, How can we prove that?

- #4

HallsofIvy

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Use the **definition** of "linearly independent". If [itex]a_1\phi_1(f)+ a_2\phi_2(f)+ a_3\phi_3(f)= 0[/itex] for **all** f, must [itex]a_1, a_2, a_3[/itex] all be 0?

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- #5

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what is the values of [tex]\phi_{0}[/tex] [tex]\phi_{1}[/tex] [tex]\phi_{-1}[/tex] ???

- #6

HallsofIvy

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a phi

Since P

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- #7

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What was the values of a b c ?

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HallsofIvy

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- #10

HallsofIvy

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I've already explained that in my response #6. af(-1)+ bf(0)+ cf(1)= 0 for **all** polynomials of order 2 or less. Take f to be x^{2}, x, and 1 and you get three equations to solve for a, b, and c.

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I've already explained that in my response #6. af(-1)+ bf(0)+ cf(1)= 0 forallpolynomials of order 2 or less. Take f to be x^{2}, x, and 1 and you get three equations to solve for a, b, and c.

Since it has taken already too much time, I will give a shot.

If f = x^2, then

f(0) = 0

f(-1) = 1

f(1) = 1

=> a*0 + b*1 + c*1 = 0

If f = x, then

f(0) = 0

f(-1) = -1

f(1) = 1

=> a*0 + b*(-1) + c*1 = 0

If f = 1, then

f(0) = 1

f(-1) = 1

f(1) = 1

=> a*1 + b*1 + c*1 = 0

So we have three equations

b + c = 0 (1)

-b + c = 0 (2)

a + b + c = 0 (3)

We get from (2)

b = c

Thus, we get from (1)

2b = 0

b = 0 (4)

Putting that to (1)

c = 0 (5)

Putting (4) and (5) to (3), we get

c = 0

Thus,

a + b + c = 0

This means that S* is linearly independent.

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I leave the rest for the questioner.

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- #12

HallsofIvy

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So you are taking af(0)+ bf(-1)+ cf(1)= 0? That's not what I had before but perfectly legal.Since it has taken already too much time, I will give a shot.

If f = x^2, then

f(0) = 0

f(-1) = 1

f(1) = 1

=> a*0 + b*1 + c*1 = 0

You mean a*0+ b*(-1)+ c*1= -b+ c= 0, right?If f = x, then

f(0) = 0

f(-1) = -1

f(1) = 1

=> a*0 + b*1 + c*1 = 0

No, -b+ c= 0.If f = 1, then

f(0) = 1

f(-1) = 1

f(1) = 1

=> a*1 + b*1 + c*1 = 0

So we have three equations

b + c = 0

b + c = 0

The three equationsa + b + c = 0

Simplifying

b + c = 0 (1)

a + b + c = 0 (2)

Substituting the equation (1) to (2)

a = 0

By trial and error, we know that

b = 0

c = 0

Thus,

a + b + c = 0

This means that S* is linearly independent.

However, the correct equations are -b+ c= 0, b+ c= 0, a+ b+ c= 0. Adding the first two equations, 2c= 0 so c= 0. Putting c= 0 into the first equation -b+ 0= 0 so b= 0. Putting b= c= 0 into the third equation, a+ b+ c= a+ 0+ 0= a= 0. Since a= b= c= 0, S* is linearly independent.

But I would have liked to have seen pnaik2008 at least

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I leave the rest for the questioner.

- #13

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So you are taking af(0)+ bf(-1)+ cf(1)= 0? That's not what I had before but perfectly legal.

You mean a*0+ b*(-1)+ c*1= -b+ c= 0, right?

I fixed the mistake.

I can't stand in waiting for b).

My attempt is:

The basis of S of (P_3, R) is (x^2, x, 1).

According to the definition of dual basis, for example the dual basis of the vector (1 0) is (1 0)^T.

Thus, the dual basis of (x^2, x, 1) is (x^2, x, 1).

- #14

HallsofIvy

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\(\displaystyle f

\(\displaystyle \phi_{-1}(f_{-1})= f_{-1}(-1)= 1,

\(\displaystyle \phi_{0}(f_{-1})= f_{-1}(0)= 0,

\(\displaystyle \phi_1(f_{-1})= f_{-1}(1)= 0

\(\displaystyle f

\(\displaystyle \phi_{0}(f_0)= f_{0}(-1)= 0,

\(\displaystyle \phi_{0}(f_0)= f_0(0)= 1,

\(\displaystyle \phi_1(f_0)= f_0(1)= 0

\(\displaystyle f

\(\displaystyle \phi_{0}(f_1)= f_{1}(-1)= 0,

\(\displaystyle \phi_{0}(f_1)= f_1(0)= 0,

\(\displaystyle \phi_1(f_1)= f_1(1)= 1

Since we are talking about polynomials of degree 2 or less, this reduces to

1) Find [itex]f_{-1}(x)= ax^2+ bx+ c[/itex] satisfying

a) [itex]f_{-1}(-1)= a- b+ c= 1[/itex]

b) [itex]f_{-1}(0)= c= 0[/itex]

c) [itex]f_{-1}(1)= a+ b+ c= 0[/itex]

2) Find [itex]f_{0}(x)= ax^2+ bx+ c[/itex] satisfying

a) [itex]f_{0}(-1)= a- b+ c= 0[/itex]

b) [itex]f_{0}(0)= c= 1[/itex]

c) [itex]f_{0}(1)= a+ b+ c= 0[/itex]

3) Find [itex]f_{1}(x)= ax^2+ bx+ c[/itex] satisfying

a) [itex]f_{1}(-1)= a- b+ c= 0[/itex]

b) [itex]f_{1}(0)= c= 0[/itex]

c) [itex]f_{1}(1)= a+ b+ c= 1[/itex]\)\)\)\)\)\)\)\)\)

- #15

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Fixing the formatting of HallOfIvy:

Reformatting what HallsOfIvy says

[tex]\begin{tabular}{ | f_-1(x) | f_0(x) | f_1(x) | }

& f_{-1}(x) & f_0(x) & f_1(x) \\

\hline

\phi_{-1} & 1 & 0 & 0 \\

\phi_0 & 0 & 1 & 0 \\

\phi_1 & 0 & 0 & 1 \\

ax^2 + bx + c: x=-1 & a -b + c = 1 & c = 0 & a + b + c = 0 \\

ax^2 + bx + c: x =0 & - - 0 & - - 1 & - - 0 \\

ax^2 + bx + c: x=1 & - - 0 & - - 0 & - - 1 \\

\end{tabular}[/tex]

I did not write variables to the last two rows again.

I am not sure about the #4 line.

a -b + c =1? # should it be 0.

Could you, please, hallOfIvy more explain your results?

[tex]f_{-1}(x)[/tex] such that

[tex]\phi_{-1}(f_{-1})= f_{-1}(-1)= 1, [/tex]

[tex]\phi_{0}(f_{-1})= f_{-1}(0)= 0,[/tex]

[tex]\phi_1(f_{-1})= f_{-1}(1)= 0[/tex]

[tex]f_{0}(x)[/tex] such that

[tex]\phi_{-1}(f_0)= f_{0}(-1)= 0, [/tex]# I fixed -1 here for phi

[tex]\phi_{0}(f_0)= f_0(0)= 1,[/tex]

[tex]\phi_1(f_0)= f_0(1)= 0[/tex]

[tex]f_{1}(x)[/tex] such that

[tex]\phi_{-1}(f_1)= f_{1}(-1)= 0, [/tex]# I fixed -1 here for phi

[tex]\phi_{0}(f_1)= f_1(0)= 0,[/tex]

[tex]\phi_1(f_1)= f_1(1)= 1[/tex]

Since we are talking about polynomials of degree 2 or less, this reduces to

1) Find [itex]f_{-1}(x)= ax^2+ bx+ c[/itex] satisfying

a) [itex]f_{-1}(-1)= a- b+ c= 1[/itex]# I am not sure should this be 1

b) [itex]f_{-1}(0)= c= 0[/itex]

c) [itex]f_{-1}(1)= a+ b+ c= 0[/itex]

2) Find [itex]f_{0}(x)= ax^2+ bx+ c[/itex] satisfying

a) [itex]f_{0}(-1)= a- b+ c= 0[/itex]

b) [itex]f_{0}(0)= c= 1[/itex]

c) [itex]f_{0}(1)= a+ b+ c= 0[/itex]

3) Find [itex]f_{1}(x)= ax^2+ bx+ c[/itex] satisfying

a) [itex]f_{1}(-1)= a- b+ c= 0[/itex]

b) [itex]f_{1}(0)= c= 0[/itex]

c) [itex]f_{1}(1)= a+ b+ c= 1[/itex]

Reformatting what HallsOfIvy says

[tex]\begin{tabular}{ | f_-1(x) | f_0(x) | f_1(x) | }

& f_{-1}(x) & f_0(x) & f_1(x) \\

\hline

\phi_{-1} & 1 & 0 & 0 \\

\phi_0 & 0 & 1 & 0 \\

\phi_1 & 0 & 0 & 1 \\

ax^2 + bx + c: x=-1 & a -b + c = 1 & c = 0 & a + b + c = 0 \\

ax^2 + bx + c: x =0 & - - 0 & - - 1 & - - 0 \\

ax^2 + bx + c: x=1 & - - 0 & - - 0 & - - 1 \\

\end{tabular}[/tex]

I did not write variables to the last two rows again.

I am not sure about the #4 line.

a -b + c =1? # should it be 0.

Could you, please, hallOfIvy more explain your results?

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