How to determine whether a function is even or odd in case the function has discontinuity at the origin?
$ sqrt {(1+a^2/x^2)} =>frac1 xsqrt{(x^2+a^2)}$
The first expression is even (i.e remains same when we put -x in place of x), while the second one is odd. What am I doing wrong while going from the first to the second expression? The expression has came as a result of integration of $intfrac{dx}{(x^2+a^2)^{3/2}}$ from -L to L. If I take the second expression the integration is non-zero, which I think is correct from the plot of $frac{1}{(x^2+a^2)^{3/2}}$ . For the other expression it is zero.
I am not asking for the result of the inegration. My questions are 1. How both the expressions are not same? 2. In general if there is a function which is discontinuous at the origin, how to find whether it is even or odd. Thank you for any help.
even-and-odd-functions
asked Nov 29 at 18:59
Siddhartha
3613
add a comment |
$ sqrt {(1+a^2/x^2)} =>frac1 xsqrt{(x^2+a^2)}$
The first expression is even (i.e remains same when we put -x in place of x), while the second one is odd. What am I doing wrong while going from the first to the second expression? The expression has came as a result of integration of $intfrac{dx}{(x^2+a^2)^{3/2}}$ from -L to L. If I take the second expression the integration is non-zero, which I think is correct from the plot of $frac{1}{(x^2+a^2)^{3/2}}$ . For the other expression it is zero.
I am not asking for the result of the inegration. My questions are 1. How both the expressions are not same? 2. In general if there is a function which is discontinuous at the origin, how to find whether it is even or odd. Thank you for any help.
even-and-odd-functions
asked Nov 29 at 18:59
Siddhartha
3613
2
$sqrt {1+frac{a^2}{x^2}}=frac{1}{|x|}sqrt{x^2+a^2}$, which is even either way.
– Shubham Johri
Nov 29 at 19:03
1
Being even/odd means that $f(t)=pm f(-t)$ for $t>0$. How does this have anything to do with a discontinuity at $0$?
– Federico
Nov 29 at 19:03
@Federico yes, there is no relation between being discontinuous at the origin and being even/odd. I understand, thanks. I got confused seeing the plots... for even and odd both cases f(0) give same value, if f(0) is finite. But in 1/x or 1/x^2 cases for even and for odd the plots shows different things at the origin.. but still then, even and odd, can be easily defined as you mentioned. Thanks!
– Siddhartha
Nov 29 at 19:17
1
@ShubhamJohri Thanks! I got it. $sqrt{x^2}=vert x vert $ always.
– Siddhartha
Nov 29 at 19:29
add a comment |
$ sqrt {(1+a^2/x^2)} =>frac1 xsqrt{(x^2+a^2)}$
The first expression is even (i.e remains same when we put -x in place of x), while the second one is odd. What am I doing wrong while going from the first to the second expression? The expression has came as a result of integration of $intfrac{dx}{(x^2+a^2)^{3/2}}$ from -L to L. If I take the second expression the integration is non-zero, which I think is correct from the plot of $frac{1}{(x^2+a^2)^{3/2}}$ . For the other expression it is zero.
I am not asking for the result of the inegration. My questions are 1. How both the expressions are not same? 2. In general if there is a function which is discontinuous at the origin, how to find whether it is even or odd. Thank you for any help.
even-and-odd-functions
asked Nov 29 at 18:59
Siddhartha
3613
$ sqrt {(1+a^2/x^2)} =>frac1 xsqrt{(x^2+a^2)}$
The first expression is even (i.e remains same when we put -x in place of x), while the second one is odd. What am I doing wrong while going from the first to the second expression? The expression has came as a result of integration of $intfrac{dx}{(x^2+a^2)^{3/2}}$ from -L to L. If I take the second expression the integration is non-zero, which I think is correct from the plot of $frac{1}{(x^2+a^2)^{3/2}}$ . For the other expression it is zero.
I am not asking for the result of the inegration. My questions are 1. How both the expressions are not same? 2. In general if there is a function which is discontinuous at the origin, how to find whether it is even or odd. Thank you for any help.
even-and-odd-functions
even-and-odd-functions
asked Nov 29 at 18:59
Siddhartha
3613
asked Nov 29 at 18:59
Siddhartha
3613
asked Nov 29 at 18:59
Siddhartha
3613
asked Nov 29 at 18:59
Siddhartha
3613
asked Nov 29 at 18:59
Siddhartha
3613
3613
2
$sqrt {1+frac{a^2}{x^2}}=frac{1}{|x|}sqrt{x^2+a^2}$, which is even either way.
– Shubham Johri
Nov 29 at 19:03
1
Being even/odd means that $f(t)=pm f(-t)$ for $t>0$. How does this have anything to do with a discontinuity at $0$?
– Federico
Nov 29 at 19:03
@Federico yes, there is no relation between being discontinuous at the origin and being even/odd. I understand, thanks. I got confused seeing the plots... for even and odd both cases f(0) give same value, if f(0) is finite. But in 1/x or 1/x^2 cases for even and for odd the plots shows different things at the origin.. but still then, even and odd, can be easily defined as you mentioned. Thanks!
– Siddhartha
Nov 29 at 19:17
1
@ShubhamJohri Thanks! I got it. $sqrt{x^2}=vert x vert $ always.
– Siddhartha
Nov 29 at 19:29
add a comment |
2
$sqrt {1+frac{a^2}{x^2}}=frac{1}{|x|}sqrt{x^2+a^2}$, which is even either way.
– Shubham Johri
Nov 29 at 19:03
1
Being even/odd means that $f(t)=pm f(-t)$ for $t>0$. How does this have anything to do with a discontinuity at $0$?
– Federico
Nov 29 at 19:03
@Federico yes, there is no relation between being discontinuous at the origin and being even/odd. I understand, thanks. I got confused seeing the plots... for even and odd both cases f(0) give same value, if f(0) is finite. But in 1/x or 1/x^2 cases for even and for odd the plots shows different things at the origin.. but still then, even and odd, can be easily defined as you mentioned. Thanks!
– Siddhartha
Nov 29 at 19:17
1
@ShubhamJohri Thanks! I got it. $sqrt{x^2}=vert x vert $ always.
– Siddhartha
Nov 29 at 19:29
2
2
$sqrt {1+frac{a^2}{x^2}}=frac{1}{|x|}sqrt{x^2+a^2}$, which is even either way.
– Shubham Johri
Nov 29 at 19:03
$sqrt {1+frac{a^2}{x^2}}=frac{1}{|x|}sqrt{x^2+a^2}$, which is even either way.
– Shubham Johri
Nov 29 at 19:03
1
1
Being even/odd means that $f(t)=pm f(-t)$ for $t>0$. How does this have anything to do with a discontinuity at $0$?
– Federico
Nov 29 at 19:03
Being even/odd means that $f(t)=pm f(-t)$ for $t>0$. How does this have anything to do with a discontinuity at $0$?
– Federico
Nov 29 at 19:03
@Federico yes, there is no relation between being discontinuous at the origin and being even/odd. I understand, thanks. I got confused seeing the plots... for even and odd both cases f(0) give same value, if f(0) is finite. But in 1/x or 1/x^2 cases for even and for odd the plots shows different things at the origin.. but still then, even and odd, can be easily defined as you mentioned. Thanks!
– Siddhartha
Nov 29 at 19:17
@Federico yes, there is no relation between being discontinuous at the origin and being even/odd. I understand, thanks. I got confused seeing the plots... for even and odd both cases f(0) give same value, if f(0) is finite. But in 1/x or 1/x^2 cases for even and for odd the plots shows different things at the origin.. but still then, even and odd, can be easily defined as you mentioned. Thanks!
– Siddhartha
Nov 29 at 19:17
1
1
@ShubhamJohri Thanks! I got it. $sqrt{x^2}=vert x vert $ always.
– Siddhartha
Nov 29 at 19:29
@ShubhamJohri Thanks! I got it. $sqrt{x^2}=vert x vert $ always.
– Siddhartha
Nov 29 at 19:29
add a comment |
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2
$sqrt {1+frac{a^2}{x^2}}=frac{1}{|x|}sqrt{x^2+a^2}$, which is even either way.
– Shubham Johri
Nov 29 at 19:03
1
Being even/odd means that $f(t)=pm f(-t)$ for $t>0$. How does this have anything to do with a discontinuity at $0$?
– Federico
Nov 29 at 19:03
@Federico yes, there is no relation between being discontinuous at the origin and being even/odd. I understand, thanks. I got confused seeing the plots... for even and odd both cases f(0) give same value, if f(0) is finite. But in 1/x or 1/x^2 cases for even and for odd the plots shows different things at the origin.. but still then, even and odd, can be easily defined as you mentioned. Thanks!
– Siddhartha
Nov 29 at 19:17
1
@ShubhamJohri Thanks! I got it. $sqrt{x^2}=vert x vert $ always.
– Siddhartha
Nov 29 at 19:29