Trouble understanding a step in a proof, algebra with summation
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I have this line in a proof that I do not understand and would like some help understanding please.
$$sum_i=1^nx_i^2-nbarx^2 = sum_i=1^n(x_i^2-2x_ibarx+barx^2)$$
Can anybody enlighten me on this transformation please?
Thanks.
proof-explanation
add a comment |Â
up vote
0
down vote
favorite
I have this line in a proof that I do not understand and would like some help understanding please.
$$sum_i=1^nx_i^2-nbarx^2 = sum_i=1^n(x_i^2-2x_ibarx+barx^2)$$
Can anybody enlighten me on this transformation please?
Thanks.
proof-explanation
Is there any hypotheses on the $x_i$s?
â Bernard
Sep 5 at 11:14
what is $barx$?
â giannispapav
Sep 5 at 11:16
yes, $barx_i$ is the average.
â Bucephalus
Sep 5 at 11:17
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have this line in a proof that I do not understand and would like some help understanding please.
$$sum_i=1^nx_i^2-nbarx^2 = sum_i=1^n(x_i^2-2x_ibarx+barx^2)$$
Can anybody enlighten me on this transformation please?
Thanks.
proof-explanation
I have this line in a proof that I do not understand and would like some help understanding please.
$$sum_i=1^nx_i^2-nbarx^2 = sum_i=1^n(x_i^2-2x_ibarx+barx^2)$$
Can anybody enlighten me on this transformation please?
Thanks.
proof-explanation
proof-explanation
asked Sep 5 at 11:08
Bucephalus
455216
455216
Is there any hypotheses on the $x_i$s?
â Bernard
Sep 5 at 11:14
what is $barx$?
â giannispapav
Sep 5 at 11:16
yes, $barx_i$ is the average.
â Bucephalus
Sep 5 at 11:17
add a comment |Â
Is there any hypotheses on the $x_i$s?
â Bernard
Sep 5 at 11:14
what is $barx$?
â giannispapav
Sep 5 at 11:16
yes, $barx_i$ is the average.
â Bucephalus
Sep 5 at 11:17
Is there any hypotheses on the $x_i$s?
â Bernard
Sep 5 at 11:14
Is there any hypotheses on the $x_i$s?
â Bernard
Sep 5 at 11:14
what is $barx$?
â giannispapav
Sep 5 at 11:16
what is $barx$?
â giannispapav
Sep 5 at 11:16
yes, $barx_i$ is the average.
â Bucephalus
Sep 5 at 11:17
yes, $barx_i$ is the average.
â Bucephalus
Sep 5 at 11:17
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
If $bar x$ is the average $bar x = frac1n sum_i=1^n x_i$, then
$$sum_i=1^n 2x_i bar x = 2bar x sum_i=1^n x_i = 2n bar x ^2.
$$
Furthermore,
$$
sum_i=1^n bar x^2 = bar x^2 sum_i=1^n 1 = nbar x ^2
$$
Oh, nice, now I get it. Thanks @Kusma
â Bucephalus
Sep 5 at 11:21
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $bar x$ is the average $bar x = frac1n sum_i=1^n x_i$, then
$$sum_i=1^n 2x_i bar x = 2bar x sum_i=1^n x_i = 2n bar x ^2.
$$
Furthermore,
$$
sum_i=1^n bar x^2 = bar x^2 sum_i=1^n 1 = nbar x ^2
$$
Oh, nice, now I get it. Thanks @Kusma
â Bucephalus
Sep 5 at 11:21
add a comment |Â
up vote
1
down vote
accepted
If $bar x$ is the average $bar x = frac1n sum_i=1^n x_i$, then
$$sum_i=1^n 2x_i bar x = 2bar x sum_i=1^n x_i = 2n bar x ^2.
$$
Furthermore,
$$
sum_i=1^n bar x^2 = bar x^2 sum_i=1^n 1 = nbar x ^2
$$
Oh, nice, now I get it. Thanks @Kusma
â Bucephalus
Sep 5 at 11:21
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $bar x$ is the average $bar x = frac1n sum_i=1^n x_i$, then
$$sum_i=1^n 2x_i bar x = 2bar x sum_i=1^n x_i = 2n bar x ^2.
$$
Furthermore,
$$
sum_i=1^n bar x^2 = bar x^2 sum_i=1^n 1 = nbar x ^2
$$
If $bar x$ is the average $bar x = frac1n sum_i=1^n x_i$, then
$$sum_i=1^n 2x_i bar x = 2bar x sum_i=1^n x_i = 2n bar x ^2.
$$
Furthermore,
$$
sum_i=1^n bar x^2 = bar x^2 sum_i=1^n 1 = nbar x ^2
$$
answered Sep 5 at 11:16
Kusma
3,355218
3,355218
Oh, nice, now I get it. Thanks @Kusma
â Bucephalus
Sep 5 at 11:21
add a comment |Â
Oh, nice, now I get it. Thanks @Kusma
â Bucephalus
Sep 5 at 11:21
Oh, nice, now I get it. Thanks @Kusma
â Bucephalus
Sep 5 at 11:21
Oh, nice, now I get it. Thanks @Kusma
â Bucephalus
Sep 5 at 11:21
add a comment |Â
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Is there any hypotheses on the $x_i$s?
â Bernard
Sep 5 at 11:14
what is $barx$?
â giannispapav
Sep 5 at 11:16
yes, $barx_i$ is the average.
â Bucephalus
Sep 5 at 11:17