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anonymous
 one year ago
help please
anonymous
 one year ago
help please

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0find the dual basis of {(1,0,0),(0,1,0),(0,0,1) and explain

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0notice that the 1 shifts each time.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01, 0, 0 0,1,0 0,0,1 thats binary code

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the first column is 2^0 second column is 2^1 third column is 2 ^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you know how to count in tens. you know the first column is 09 the second column is 10  19 etc., remember the abacus? it is similar for the binary system

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll draw you a picture!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439037967941:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i don,'t still get why 2^ of numbers here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0another picture in base 10 (decimal system, which you are more used to) dw:1439038148182:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There are only 2 numbers 1 and zero that is why it is base 2 or 2^

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so how come the numbers they raised to? 2^n

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0look up positional number systems in wikipedia it should explain it more to you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The value that a digit taks on depends on its position, as I have shown

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for example, 110 in decimal system is 1 x 100 + 1 x 10 + 1 x 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, you did for base 2 but why base 10?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just showed you the same principle applies in base 2 as it applies in base 10 (decimal) which you were more familiar with

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but we are not using base 10 right since you said that only numbers are involved . ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the value of the number (e.g. 001) depends on itsa place or position) is what I am saying

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So of 1 is placed in the units column in the decimal system then it can only take the value between 0 and 9

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0your question was about the dual basis of this, I have offered an answer in relation to that. i.e that the digits given are base 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No. It said find the dual basis of this, and explain. you have found the dual basis. reproduce my table i n relation to 2^0, "^1 etc, and then state that the value of oo1,010 depends on their position.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you get 7 if you add the numbers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, so the table with the position is the dual basis?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for something to have a "dual basis" it must have two possible outcomes.i.e. binary. you show that as follows: dw:1439039911382:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you will get 100% trust me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now you are asked to explain: so here goes:( the value of the number zero or 1 depends on the position) that's it!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thanks . i have a question on inner produte. can i ask?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0compute the I2 inner product of x and y where x=1,1/2,1,3,1/4 and y=1/3,1/4,1/5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's not what the dual basis is. if you have a real inner product space \(V,\langle\cdot,\cdot\rangle\) then an ordered basis \(\{b_i\}\) induces a basis for the dual space \(V^*\) of linear maps \(V\to R\), such that $$b^i(v)=\langle b_i,v\rangle$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is equivalent to the idea of the transpose, where we identify column vectors with \(v\in V\) and so row vectors with \(u\in V^*\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what is the duel basis of my question sir?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in other words, for an orthonormal basis \(\{e_i\}\), like you have, our dual basis looks like \(e^i\), so that \(e^i(e_i)=1\) and \(e^i(e_j)=0\) for \(i\ne j\). in other words, our dual basis lets us pick out things parallel to the corresponding original basis vectors

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so a dual basis for \(\{(1,0,0),(0,1,0),(0,0,1)\}\) looks like \(\{\delta_{1j},\delta_{2j},\delta_{3j}\}\) where \(\delta_{ij}\) denotes the Kronecker delta: $$\delta_{ij}=\left\{\begin{matrix}1&\text{if }i=j\\0&\text{otherwise}\end{matrix}\right.$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now in \(\ell^2(\mathbb N)\) we're talking about a sequence space and we have an inner product like: $$\langle x,y\rangle=\sum_{k=0}^\infty x_ky_k$$ for real sequences \(x,y\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\langle\left(1,\frac12,1,3,\frac14,0,0,0,\dots\right),\left(\frac13,\frac14,\frac15,0,0,0,\dots\right)\rangle=1\cdot\frac13+\frac12\cdot\frac14+1\cdot\frac15=\dots$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you compute \(1/3+1/8+1/5\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(40+15+24)/120=79/120
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