The aim is to show a form which will update the div on the page with the result
I'm not sure, but may be you init form with the wrong id?
$('#formDiv').fadeIn('slow').ajaxForm(options)
When form id is feditform. And in the correct example located on http://plungjan.name/eetest/facy.html $('#feditform').ajaxForm(options);
feditform
$('#feditform').ajaxForm(options);