If #x# is nearly equal to one, then find the approximate value of #{mx^m-nx^n}/{m-n}#?

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Answer 1 · 2018-02-14T14:12:24

#1#. Note that Cesareo R. gives a much better approximation.
Another route to his approximation is given below.

If #x~~1#, then both #x^m# and #x^n# are close to #1#.

Better Approximation

We can use the equation of the tangent line at #a=1# to get the linearization (or linear approximation) at #1# of #f(x) = (mx^m-nx^n)/(m-n)#

We'll get

#y = L(x) = f(a)+f'(a)(x-a)#.

So we need:

#y = L(x) = f(1)+f'(1)(x-1)#.

We find #f(1) = 1# and

#f'(x) = (m^2x^(m-1)-n^2x^(n-1))/(m-n)# so #f'(1) = (m+n)#.

Giving linearization at #1#:

#y = L(x) = 1 + (m+n)(x-1)#.

Answer 2 · 2018-02-14T14:19:28

See below.

Making #x = 1+epsilon# we have

# {mx^m-nx^n}/{m-n} = (m(1+epsilon)^m-n(1+epsilon)^n)/(m-n)#

but

#(1+epsilon)^n = 1+n epsilon + O(epsilon^2)# so

# {mx^m-nx^n}/{m-n} = (m ( 1+m epsilon + O(epsilon^2))-n( 1+n epsilon + O(epsilon^2)))/(m-n)#

then

#{mx^m-nx^n}/{m-n} approx (m ( 1+m epsilon )-n( 1+n epsilon ))/(m-n) = 1+(m+n)epsilon #

but

#1+(m+n)epsilon = 1+(m+n)(1+epsilon)-(m+n) = 1+(m+n)(x-1)#

and finally

# {mx^m-nx^n}/{m-n} approx 1+(m+n)(x-1)#