While solving a cubic equation we come across Tschirnhaus transformation. This transformation is used to express a cubic equation in depressed form. In Tschirnhaus transformation we substitute x by t−b/3a. But why it is −b/3a that is the unique value for it. In this post I will show you why such transformation is useful in case of a depressed cubic.
Look at the following figure.
By looking at the figure we may think the the curve is symmetrical about some x = a. But let us prove it. The expression of the cubic is f(x) = ax3 + bx2 + cx + d. Differentiating f(x) with respect to x we get f '(x) = 3ax2 + 2bx + c. This is the rate of change of curve. Differentiating again we get f ' '(x) = 6ax + 2b. This is rate of change of slope of the curve. This value is zero for x = −b/3a. f '(x) is quadratic in x so it has two roots or in other words we can express it as the product of two linear expressions. The sum of these two roots is −2b/3a. Let the two roots of the expression f '(x) be α and β. Then α + β = −2b/3a. As we can see that the sum of slopes is twice of the rate of change of slope. So the curve is symmetrical about the line x = −b/3a. On both sides of this line at equal distance the value of slope has the same modulus.
The curve given in the figure is y = x3 − 6x2 + 5x − 4. Here (−b/3a) = 6/3 = 2. Shifting the origin to X = x − 2 we get
y = (x+2)3 − 6(x+2)2 + 5(x+2) − 4
y = x3 + 8 + 6x2 + 12x − 6x2 − 24x − 24 + 5x + 10 v 4
y = x3 − 7x − 10
So we see that when we shift the origin to −b/3a we get a cubic whose graph is symmetrical about the axis x=0.
Look at the following figure.
By looking at the figure we may think the the curve is symmetrical about some x = a. But let us prove it. The expression of the cubic is f(x) = ax3 + bx2 + cx + d. Differentiating f(x) with respect to x we get f '(x) = 3ax2 + 2bx + c. This is the rate of change of curve. Differentiating again we get f ' '(x) = 6ax + 2b. This is rate of change of slope of the curve. This value is zero for x = −b/3a. f '(x) is quadratic in x so it has two roots or in other words we can express it as the product of two linear expressions. The sum of these two roots is −2b/3a. Let the two roots of the expression f '(x) be α and β. Then α + β = −2b/3a. As we can see that the sum of slopes is twice of the rate of change of slope. So the curve is symmetrical about the line x = −b/3a. On both sides of this line at equal distance the value of slope has the same modulus.
The curve given in the figure is y = x3 − 6x2 + 5x − 4. Here (−b/3a) = 6/3 = 2. Shifting the origin to X = x − 2 we get
y = (x+2)3 − 6(x+2)2 + 5(x+2) − 4
y = x3 + 8 + 6x2 + 12x − 6x2 − 24x − 24 + 5x + 10 v 4
y = x3 − 7x − 10
So we see that when we shift the origin to −b/3a we get a cubic whose graph is symmetrical about the axis x=0.
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