Is #e^x# the unique function of which derivative is itself? Can you prove it?

Question

Is #e^x# the uniq function of which derivative is itself? Can you prove it? #f'(x)=f(x)=>f(x)=e^x# , is this true everytime? Is there any second function like this?

Chloe Alexander · Accepted Answer

Almost. If #k# is any constant then #f(x) = k e^x# satisfies #f'(x) = f(x)#.

Uniqueness of #e^x# is given by requiring #f(x)# to be defined everywhere, and #f(0) = 1#.

Any function of the form #f(x) = ke^x# satisfies #f'(x) = f(x)# In order for #e^x# to be the only solution we need an extra condition such as #f(0) = 1# Suppose #f(x)# is a well behaved function from #RR# to #RR# with #f(0) = 1#. We can write the Maclaurin series for #f(x)# as: #f(x) = sum_(n=0)^oo a_n x^n" "# for some constants #a_0, a_1,...# Then: #1 = f(0) = a_0# and: #f'(x) = sum_(n=0)^oo (d/(dx) a_n x^n)# #color(white)(f'(x)) = sum_(n=0)^oo (n a_n x^(n-1))# #color(white)(f'(x)) = sum_(n=0)^oo (n+1) a_(n+1) x^n# Since we want #f'(x) = f(x)#, we have: #sum_(n=0)^oo (n+1) a_(n+1) x^n = sum_(n=0)^oo a_n x^n# So equating the coefficients, we find: #{ (a_0 = 1), (a_(n+1) = a_n/(n+1)" for "n >= 0") :}# So: #1/a_0 = 1/1 = 0!# #1/a_1 = 1/a_0 = 1!# #1/a_2 = 2/a_1 = 2!# #1/a_3 = 3/a_2 = 3!# and: #1/a_(n+1) = (n+1) (1/a_n) = (n+1)!# Hence: #f(x) = sum_(n=0)^oo x^n/(n!)# which is one of the definitions of #e^x# Then if #k# is any constant, we find: #d/(dx) k e^x = k d/(dx) e^x = k e^x# So: #f(x) = k e^x# is also a solution of #f(x) = f'(x)# Footnote There are some non well behaved functions that also satisfy #f'(x) = f(x)# For example: #f(x) = { ("undefined" " if " x=0), (abs(e^x)" if "x != 0) :}#