How can we prove B(x,y) = B(y,x)?

Win_odd Dhamnekar

Win_odd Dhamnekar

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How to show that beta function defined by [math]B(x,y)=\displaystyle\int_0^1t^{(x-1)}(1-t)^{(y-1)} dt ,[/math] for x > 0, y > 0 satisfies the relation B( x, y ) = B( y, x ) for x > 0, y > 0

My attempt to answer:

My hp 50g calculator proved the above beta function for ( x, y ) and ( y, x ). Is there any other way to prove it?
 
Dhamnekar Winod said:
How to show that beta function defined by [math]B(x,y)=\displaystyle\int_0^1t^{(x-1)}(1-t)^{(y-1)} dt ,[/math] for x > 0, y > 0 satisfies the relation B( x, y ) = B( y, x ) for x > 0, y > 0

My attempt to answer:

My hp 50g calculator proved the above beta function for ( x, y ) and ( y, x ). Is there any other way to prove it?
Click to expand...
Did you try swapping x and y in the definition, and manipulating the integral to show it is equal to B(x,y)? (Hint: try a substitution.)

A calculator, of course, can't prove anything.

Incidentally, in your title you used the symbol ß, which is actually the German "eszett", or "sharp S". If you were aiming at a Greek beta, you can just use B, which is the capital beta! Making it lower case would no longer name the beta function.
 
Dhamnekar Winod said:
How to show that beta function defined by [math]B(x,y)=\displaystyle\int_0^1t^{(x-1)}(1-t)^{(y-1)} dt ,[/math] for x > 0, y > 0 satisfies the relation B( x, y ) = B( y, x ) for x > 0, y > 0

My attempt to answer:

My hp 50g calculator proved the above beta function for ( x, y ) and ( y, x ). Is there any other way to prove it?
Click to expand...

You'll have to ask your calculator :) (From the "couldn't resist" list)
 
I would think that the method your calculator used is the only way to prove this.
 
Symmetrical Property of the Beta Function: B(x,y) = B(y, x)
\(\displaystyle =B(x,y) = \displaystyle\int_0^1 t^{x-1}(1-t)^{y-1}dt\)
Let t= 1 - u ⇒ u = 1 - t so dt = -du

at t = 0 ⇒ u = 1 and at t = 1 ⇒ u = 0
\(\displaystyle B(x,y) = \displaystyle\int_1^0 (1-u)^{x-1} u^{y-1}-du\)

\(\displaystyle B(x,y) = -\displaystyle\int_1^0 u^{y-1} (1-u)^{x-1} du = \displaystyle\int_0^1 u^{y-1} (1-u)^{x-1}du = B(y,x)\)
 
Why are you suddenly subtracting du and then multiplying by du?
 
Dhamnekar Winod said:
- du indicates du is negative.
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Whether du is negative or not, why are you subtracting du?
For the record, if x<0, then -x is NOT negative. That is, putting a negative sign in front of a number does not always make that number negative. For example, -7 is a number and if I put a negative sign in front of it then it will be come positive.
 
Steven G said:
Whether du is negative or not, why are you subtracting du?
For the record, if x<0, then -x is NOT negative. That is, putting a negative sign in front of a number does not always make that number negative. For example, -7 is a number and if I put a negative sign in front of it then it will be come positive.
Click to expand...
All he needed was parenthesis. (-du). That's all you needed to say.

-Dan
 
I disagree. This is a forum where we want students to think rather than just give answers.
 
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