Prove Lagrange Polynomials Basis of $\mathbb{R}_{n}[X]

[God's Pen]

hello everyone
for $$i=1,2,...,(n+1)$$ let $$P_{i}(X)=\frac{\prod_{1\leq j\leq n+1,j\neq i}(X-a_j)}{\prod_{1\leq j\leq n+1,j\neq i}(a_i-a_j)}$$
prove that $$(P_1,P_2,...P_{n+1})$$ is basis of $$\mathbb{R}_{n}[X]$$.
i already have an answer but i don't understand some of it.
...
we have $$B=\{P_1, \cdots , P_{n+1} \} \subset \mathbb{R}_n[x]$$ and $$\dim_{\mathbb{R}} \mathbb{R}[x]=n+1.$$ so, in order to show that $$B$$ is a basis for $$\mathbb{R}_n[x],$$ we only need

to show that $$B$$ is linearly independent.for any $$i$$ we have $$P_i(a_i)=1$$ and $$P_j(a_i)=0,$$ for $$i \neq j.$$ now, to show that $$B$$ is linearly independent, suppose that $$\sum_{j=1}^{n+1}c_j P_j(x)=0,$$ for some
$$c_j \in \mathbb{R}.$$ put $$x=a_i$$ to get $$c_i=0. \ \Box$$
...
what i don't understand,
why should we have $$P_j(X)$$ and not $$P_i(X)$$
why we had to find $$P_j(a_i)$$ ?
and why we have $$\sum_{j=1}^{n+1}$$ and not $$\sum_{i=1}^{n+1}$$ ?
thank you so much.

 

[krishna mohan]

Hmmm..er...I don't know if I have understood your question properly ...but let me try to answer...


Your question is..why have we used j instead of i in the summation? Well..as far as I can see...i and j are just dummy variables ..it does not matter whether you use i or j or k...

You can very well use i if you want...but then you have to always remember that i denotes a general element...then..to prove the coefficients as zero, you can't put x=a_(i) as i is now reserved for the general element..you have to use some other letter in that case...


Now, as for the question as to why we had to find P_(j)(a_i)...by putting x=a_(i), you are reducing all elements of the sum to zero except the term with index j=i. This term is
c_i * P_(i)(a_i)=c_i.

 

[God's Pen]

Hmmm..er...I don't know if I have understood your question properly ...but let me try to answer...


Your question is..why have we used j instead of i in the summation? Well..as far as I can see...i and j are just dummy variables ..it does not matter whether you use i or j or k...

You can very well use i if you want...but then you have to always remember that i denotes a general element...then..to prove the coefficients as zero, you can't put x=a_(i) as i is now reserved for the general element..you have to use some other letter in that case...


Now, as for the question as to why we had to find P_(j)(a_i)...by putting x=a_(i), you are reducing all elements of the sum to zero except the term with index j=i. This term is
c_i * P_(i)(a_i)=c_i.

thank you.
but if i and j are just dummy variables and it doesn't matter which one we put.
how could we have $$P_i(a_i)=1$$ and $$P_j(a_i)=0$$ ??
?

 

[krishna mohan]

Welll...first thing you should realize is that even if you use i or j or k or a or b or c, the two expressions are fundamentally different...


$$P_{i}(a_{i})$$ has the the index for $$P$$ and $$a$$ as the same..

$$P_{j}(a_{i})$$ has the the index for $$P$$ and $$a$$ as different..of course, the assumption here being that, whatever i and j are, they are not equal..i.e if we put i=1, then j cannot be 1...


See the definition of $$P_{i}(X)$$...In the numerator, there is a product...the product contains all terms of the form $$X-a_{j}$$ such that j is not equal to i...

Like, if i=3, then the product contains $$(X-a_{1}),(X-a_{2}),(X-a_{4}),(X-a_{5}) etc$$ but not $$(X-a_{3})$$...


Then, you can see why putting $$X=a_{3}$$ won't make the expression to zero..but $$X=a_{j}$$... such that j is not three...will reduce the whole term to zero...

 

[God's Pen]

Welll...first thing you should realize is that even if you use i or j or k or a or b or c, the two expressions are fundamentally different...


$$P_{i}(a_{i})$$ has the the index for $$P$$ and $$a$$ as the same..

$$P_{j}(a_{i})$$ has the the index for $$P$$ and $$a$$ as different..of course, the assumption here being that, whatever i and j are, they are not equal..i.e if we put i=1, then j cannot be 1...


See the definition of $$P_{i}(X)$$...In the numerator, there is a product...the product contains all terms of the form $$X-a_{j}$$ such that j is not equal to i...

Like, if i=3, then the product contains $$(X-a_{1}),(X-a_{2}),(X-a_{4}),(X-a_{5}) etc$$ but not $$(X-a_{3})$$...


Then, you can see why putting $$X=a_{3}$$ won't make the expression to zero..but $$X=a_{j}$$... such that j is not three...will reduce the whole term to zero...

i think I'm getting somewhere,thank u so mush for your help.