Exercise
Prove that for x and y that are close to zero the following holds
1+x−y1≈1−x+y
Proof
Since the equation we need to prove has the estimator sign, we probably need to use the linear approximation formula. Since there are two variables, we will use the 2-variable linear approximation formula
f(x,y)≈f(x0,y0)+fx′(x0,y0)⋅(x−x0)+fy′(x0,y0)⋅(y−y0)
Therefore, we need to define the following
x,y,x0,y0,f(x)
And place them in the formula. x and y will be the numbers that appear in the question, and the points
x0,y0
will be the closest values to x and y respectively, which are easy for calculations.
In our exercise, we will define
x0=0,y0=0
The function will be the right side in the equation that we need to prove
f(x,y)=1+x−y1
In the formula above we see the function partial derivatives. Hence, we calculate them.
fx′(x,y)=(1+x−y)2−1
fy′(x,y)=(1+x−y)2−1⋅(−1)=
=(1+x−y)21
We put all the data in the formula and get
1+x−y1≈f(0,0)+fx′(0,0)⋅(x−0)+fy′(0,0)⋅(y−0)=
=1+0−01+(1+0−0)2−1⋅x+(1+0−0)21⋅y=
=1−1⋅x+1⋅y=
=1−x+y
Hence, we get
1+x−y1≈1−x+y
As required.
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