Why is f-sub-xy equal to f-sub-yx? - Atheist Nexus2014-11-29T09:20:56Zhttp://www.atheistnexus.org/forum/topics/why-is-fsubxy-equal-to-fsubyx?groupUrl=mathematics&commentId=2182797%3AComment%3A1946029&x=1&feed=yes&xn_auth=noDoes someone have a link with…tag:www.atheistnexus.org,2012-05-08:2182797:Comment:19460292012-05-08T15:47:08.117ZRick Taylorhttp://www.atheistnexus.org/profile/RickTaylor
<p>Does someone have a link with instructions on how to include subscripts, deltas, and other nice math notation in one's reply? I could write out a proof of f_xy=f_yx (assuming continuity of the derivatives) if I knew how to do that.</p>
<p>Does someone have a link with instructions on how to include subscripts, deltas, and other nice math notation in one's reply? I could write out a proof of f_xy=f_yx (assuming continuity of the derivatives) if I knew how to do that.</p> The above proof can't work, b…tag:www.atheistnexus.org,2012-05-07:2182797:Comment:19445582012-05-07T20:51:22.555ZRick Taylorhttp://www.atheistnexus.org/profile/RickTaylor
<p>The above proof can't work, because f_xy is not always equal to f_yx. You need an additional assumption. For example, if f_xy and f_yx are both continuous functions in an open set, then f_xy=f_yx is true on the set (weaker assumptions are possible).</p>
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<p>The usual proof considers the limit as both h and k go to zero of f(a+h,b+h) -f(a,b+h)-f(a+h,b)+f(a,b), and shows that this limit is equal to both f_xy(a,b) and f_yx(a,b). It uses the assumption various partial derivatives are…</p>
<p>The above proof can't work, because f_xy is not always equal to f_yx. You need an additional assumption. For example, if f_xy and f_yx are both continuous functions in an open set, then f_xy=f_yx is true on the set (weaker assumptions are possible).</p>
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<p>The usual proof considers the limit as both h and k go to zero of f(a+h,b+h) -f(a,b+h)-f(a+h,b)+f(a,b), and shows that this limit is equal to both f_xy(a,b) and f_yx(a,b). It uses the assumption various partial derivatives are continuous, and the mean value theorem.</p>