how to know if my ideas didnt tought before?

[spkarra]

Tomer,
Few months ago, I thought they were new and posted some of my conjectures.
In this forum, the members responded in a professional manner.

So, my advice is "Give a try", who knows what you had in your mind might be a new concept.
Good Luck,
Sastry Karra

[blob100]

Originally Posted by spkarra

Tomer,
Few months ago, I thought they were new and posted some of my conjectures.
In this forum, the members responded in a professional manner.

So, my advice is "Give a try", who knows what you had in your mind might be a new concept.
Good Luck,
Sastry Karra

Hello Karra.
I can't understand what do you talk about.
My conjectures were trivially false and unreadable.

Tomer.

[blob100]

The fifth problem (1):
V1=(x1,y1),
V2=(x2,y2).
C is the angle between the two vectors.
A is the angle between V1 and X.
B is the angle between V2 and X.

cosC=cos(|A-B|)=cosAcosB+sinAsinB=
(x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))=G.
G is an easy way to show the whole phrase.

We try to prove:
|V1||V2|cosC=x1x2+y1y2.
((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2))=|V1||V2|.

cosC|V1||V2|=G|V1||V2|=(x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))(((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2)))=x1x2+y1y2.

[R.D. Silverman]

Originally Posted by blob100

The fifth problem (1):
V1=(x1,y1),
V2=(x2,y2).
C is the angle between the two vectors.
A is the angle between V1 and X.
B is the angle between V2 and X.

cosC=cos(|A-B|)=cosAcosB+sinAsinB=
(x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))=G.
G is an easy way to show the whole phrase.

We try to prove:
|V1||V2|cosC=x1x2+y1y2.
((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2))=|V1||V2|.

cosC|V1||V2|=G|V1||V2|=(x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))(((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2)))=x1x2+y1y2.

Please make this a little more readable.....

There are several ways to prove the result. This may be one of them,
but your notation is all "jammed together".

BTW, This is a very important result.

[blob100]

Originally Posted by R.D. Silverman

Please make this a little more readable.....

There are several ways to prove the result. This may be one of them,
but your notation is all "jammed together".

BTW, This is a very important result.

Yes, I know this result is important,
And my proof is not readable,
The problem I found aiming my writing was that geometical pictures can't be drawn here, and more then it, the phrases aren't written with the popular mathematical form (as you, me and everyone would write on a sheet of paper, I mean, not as on the computer, forum).

I am able to write it on a word page and send it to you by e-mail.
I can try explaining what is written.

Thanks Tomer.

[blob100]

Originally Posted by blob100

The fifth problem (1):
V1=(x1,y1),
V2=(x2,y2).
C is the angle between the two vectors.
A is the angle between V1 and X.
B is the angle between V2 and X.

cosC=cos(|A-B|)=cosAcosB+sinAsinB=
(x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))=G.
G is an easy way to show the whole phrase.

We try to prove:
|V1||V2|cosC=x1x2+y1y2.
((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2))=|V1||V2|.

cosC|V1||V2|=G|V1||V2|=(x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))(((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2)))=x1x2+y1y2.

|V1|=(x1,y1),
|V2|=(x2,y2).
C is the angle between the two vectors (you denoted it as theta).
A is the angle between V1 and X.
B is the angle between V2 and X.

cosC=cos(|A-B|)=cosAcosB+sinAsinB=
(x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))=
(x2/|V2|)(x1/|V1|)+(y1/|V1|)(y2/|V2|)=G.
G is an easy way to show the whole phrase.
_______________________________
We try to prove:
|V1||V2|cosC=x1x2+y1y2.
((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2))=|V1||V2|.
_______________________________
cosC|V1||V2|=G|V1||V2|=((x2/|V2|)(x1/|V1|)+(y1/|V1|)(y2/|V2|))|V1||V2|=
x1x2+y1y2.

[R.D. Silverman]

Originally Posted by blob100

Yes, I know this result is important,
And my proof is not readable,
The problem I found aiming my writing was that geometical pictures can't be drawn here, and more then it, the phrases aren't written with the popular mathematical form (as you, me and everyone would write on a sheet of paper, I mean, not as on the computer, forum).

I am able to write it on a word page and send it to you by e-mail.
I can try explaining what is written.

Thanks Tomer.

|
|
|
| /(x1, y1)
| /
| / .(x2,y2)
| / .
| / .
| / .
|/a . b
----------------------------------

Let L1 = length of the segment labelled (x1,y1)
L2 = '' (x2, y2)


Then cos(b) = x2/L2
cos(a + b) = x1/L1

What then is cos(a)?

[axn]

Originally Posted by R.D. Silverman

|
|
|
|      /(x1, y1)
|     /
|    /      .(x2,y2)
|   /     .
|  /    .
| /  . 
|/a . b
----------------------------------

Looks better with code tag

[blob100]

Originally Posted by R.D. Silverman

|
|
|
| /(x1, y1)
| /
| / .(x2,y2)
| / .
| / .
| / .
|/a . b
----------------------------------

Let L1 = length of the segment labelled (x1,y1)
L2 = '' (x2, y2)


Then cos(b) = x2/L2
cos(a + b) = x1/L1

What then is cos(a)?

I can't understand what is drawn here.
I think, after some changes I gave, you will understand easily what I wrote there.

[blob100]

Originally Posted by axn

Looks better with code tag

1) cos(a)=cos(a+b)cos(b)+sin(a+b)sin(b).
sin(a+b)=x1/|V1|
sin(b)=x2/|V2|
cos(a+b)=y1/|V1|
cos(b)=y2/|V2|
2) cos(a)=(x1/|V1|)(x2/|V2|)+(y1/|V1|)(y2/|V2|).
If we product cos(a) with |V1||V2| we may get by (2):
cos(a)|V1||V2|=x1x2+y1y2.

[R.D. Silverman]

Originally Posted by blob100

1) cos(a)=cos(a+b)cos(b)+sin(a+b)sin(b).
sin(a+b)=x1/|V1|
sin(b)=x2/|V2|
cos(a+b)=y1/|V1|
cos(b)=y2/|V2|
2) cos(a)=(x1/|V1|)(x2/|V2|)+(y1/|V1|)(y2/|V2|).
If we product cos(a) with |V1||V2| we may get by (2):
cos(a)|V1||V2|=x1x2+y1y2.

This works nicely.