Is there a way to integrate the log of a function, f(x) in Matlab without defining eg l = log(f(x)?
I have the following code:
x = 0:0.001:2.5;
gamma_l = @(x) 2*x;
And I want to integrate the following:
integral( log(gamma_l), 0 , 0.6 )
But it gives me the error:
Undefined function 'log' for input arguments of type
'function_handle'.
I know that I could just define:
gamma_l_l = @(x) log(2*x);
integral( gamma_l_l, 0 , 0.6 )
Because it works in this way. However, I would like to know why the first case does not work. And if there is a way to integrate the function without defining a new function.
matlab integration
asked Nov 13 '18 at 16:05
TeconTecon
183
add a comment |
I have the following code:
x = 0:0.001:2.5;
gamma_l = @(x) 2*x;
And I want to integrate the following:
integral( log(gamma_l), 0 , 0.6 )
But it gives me the error:
Undefined function 'log' for input arguments of type
'function_handle'.
I know that I could just define:
gamma_l_l = @(x) log(2*x);
integral( gamma_l_l, 0 , 0.6 )
Because it works in this way. However, I would like to know why the first case does not work. And if there is a way to integrate the function without defining a new function.
matlab integration
asked Nov 13 '18 at 16:05
TeconTecon
183
1
The first case does not work becauselog()requires numbers as inputs, and you are not giving it a number, you are giving it a function handle. The logarithm of a function handle is not defined.
– Ander Biguri
Nov 13 '18 at 16:07
Thanks, @Ander! So do you think that the most efficient (only) way is to define another function? Isn't there a way to define composite function inside the integral?
– Tecon
Nov 13 '18 at 16:11
1
Gnovice's answer is how you should handle this ;)
– Ander Biguri
Nov 13 '18 at 16:32
add a comment |
I have the following code:
x = 0:0.001:2.5;
gamma_l = @(x) 2*x;
And I want to integrate the following:
integral( log(gamma_l), 0 , 0.6 )
But it gives me the error:
Undefined function 'log' for input arguments of type
'function_handle'.
I know that I could just define:
gamma_l_l = @(x) log(2*x);
integral( gamma_l_l, 0 , 0.6 )
Because it works in this way. However, I would like to know why the first case does not work. And if there is a way to integrate the function without defining a new function.
matlab integration
asked Nov 13 '18 at 16:05
TeconTecon
183
I have the following code:
x = 0:0.001:2.5;
gamma_l = @(x) 2*x;
And I want to integrate the following:
integral( log(gamma_l), 0 , 0.6 )
But it gives me the error:
Undefined function 'log' for input arguments of type
'function_handle'.
I know that I could just define:
gamma_l_l = @(x) log(2*x);
integral( gamma_l_l, 0 , 0.6 )
Because it works in this way. However, I would like to know why the first case does not work. And if there is a way to integrate the function without defining a new function.
matlab integration
matlab integration
asked Nov 13 '18 at 16:05
TeconTecon
183
asked Nov 13 '18 at 16:05
TeconTecon
183
asked Nov 13 '18 at 16:05
TeconTecon
183
asked Nov 13 '18 at 16:05
TeconTecon
183
asked Nov 13 '18 at 16:05
TeconTecon
183
183
1
The first case does not work becauselog()requires numbers as inputs, and you are not giving it a number, you are giving it a function handle. The logarithm of a function handle is not defined.
– Ander Biguri
Nov 13 '18 at 16:07
Thanks, @Ander! So do you think that the most efficient (only) way is to define another function? Isn't there a way to define composite function inside the integral?
– Tecon
Nov 13 '18 at 16:11
1
Gnovice's answer is how you should handle this ;)
– Ander Biguri
Nov 13 '18 at 16:32
add a comment |
1
The first case does not work becauselog()requires numbers as inputs, and you are not giving it a number, you are giving it a function handle. The logarithm of a function handle is not defined.
– Ander Biguri
Nov 13 '18 at 16:07
Thanks, @Ander! So do you think that the most efficient (only) way is to define another function? Isn't there a way to define composite function inside the integral?
– Tecon
Nov 13 '18 at 16:11
1
Gnovice's answer is how you should handle this ;)
– Ander Biguri
Nov 13 '18 at 16:32
1
1
The first case does not work because
log() requires numbers as inputs, and you are not giving it a number, you are giving it a function handle. The logarithm of a function handle is not defined.– Ander Biguri
Nov 13 '18 at 16:07
The first case does not work because
log() requires numbers as inputs, and you are not giving it a number, you are giving it a function handle. The logarithm of a function handle is not defined.– Ander Biguri
Nov 13 '18 at 16:07
Thanks, @Ander! So do you think that the most efficient (only) way is to define another function? Isn't there a way to define composite function inside the integral?
– Tecon
Nov 13 '18 at 16:11
Thanks, @Ander! So do you think that the most efficient (only) way is to define another function? Isn't there a way to define composite function inside the integral?
– Tecon
Nov 13 '18 at 16:11
1
1
Gnovice's answer is how you should handle this ;)
– Ander Biguri
Nov 13 '18 at 16:32
Gnovice's answer is how you should handle this ;)
– Ander Biguri
Nov 13 '18 at 16:32
add a comment |
1 Answer
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Your variable gamma_l is an anonymous function, and the log function is not designed to accept function handles as an input. Instead, you need to define a second anonymous function that evaluates gamma_l for a given value, then passes the numeric result to log, like so:
result = integral(@(x) log(gamma_l(x)), 0, 0.6);
answered Nov 13 '18 at 16:08
gnovicegnovice
116k13231335
add a comment |
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Your variable gamma_l is an anonymous function, and the log function is not designed to accept function handles as an input. Instead, you need to define a second anonymous function that evaluates gamma_l for a given value, then passes the numeric result to log, like so:
result = integral(@(x) log(gamma_l(x)), 0, 0.6);
answered Nov 13 '18 at 16:08
gnovicegnovice
116k13231335
add a comment |
Your variable gamma_l is an anonymous function, and the log function is not designed to accept function handles as an input. Instead, you need to define a second anonymous function that evaluates gamma_l for a given value, then passes the numeric result to log, like so:
result = integral(@(x) log(gamma_l(x)), 0, 0.6);
answered Nov 13 '18 at 16:08
gnovicegnovice
116k13231335
add a comment |
Your variable gamma_l is an anonymous function, and the log function is not designed to accept function handles as an input. Instead, you need to define a second anonymous function that evaluates gamma_l for a given value, then passes the numeric result to log, like so:
result = integral(@(x) log(gamma_l(x)), 0, 0.6);
answered Nov 13 '18 at 16:08
gnovicegnovice
116k13231335
Your variable gamma_l is an anonymous function, and the log function is not designed to accept function handles as an input. Instead, you need to define a second anonymous function that evaluates gamma_l for a given value, then passes the numeric result to log, like so:
result = integral(@(x) log(gamma_l(x)), 0, 0.6);
answered Nov 13 '18 at 16:08
gnovicegnovice
116k13231335
edited Nov 13 '18 at 16:14
answered Nov 13 '18 at 16:08
gnovicegnovice
116k13231335
answered Nov 13 '18 at 16:08
gnovicegnovice
116k13231335
answered Nov 13 '18 at 16:08
gnovicegnovice
116k13231335
116k13231335
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The first case does not work because
log()requires numbers as inputs, and you are not giving it a number, you are giving it a function handle. The logarithm of a function handle is not defined.– Ander Biguri
Nov 13 '18 at 16:07
Thanks, @Ander! So do you think that the most efficient (only) way is to define another function? Isn't there a way to define composite function inside the integral?
– Tecon
Nov 13 '18 at 16:11
1
Gnovice's answer is how you should handle this ;)
– Ander Biguri
Nov 13 '18 at 16:32