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Question 1170642: I need help on this question:
Show that, for small values of x^2, (1-2x^2)^-2 -(1+6x^2)^2/3≈kx^4.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Let's rewrite the given expression into the form f(x) - g(x)
where,
f(x) = (1-2x^2)^(-2)
g(x) = (1+6x^2)^(2/3)
This will allow us to find the Taylor polynomial expansion for each function.
For that, we'll need the derivatives.
First Derivative:
f(x) = (1-2x^2)^(-2)
f ' (x) = -2*(1-2x^2)^(-3)*d/dx[1-2x^2]
f ' (x) = -2*(1-2x^2)^(-3)*(-4x)
f ' (x) = 8x*(1-2x^2)^(-3)
Don't forget the chain rule.
Second Derivative:
f ' (x) = 8x*(1-2x^2)^(-3)
f '' (x) = d/dx[8x*(1-2x^2)^(-3)]
f '' (x) = d/dx[8x]*(1-2x^2)^(-3) + 8x*d/dx[(1-2x^2)^(-3)]
f '' (x) = 8*(1-2x^2)^(-3) + 8x*(-3)(1-2x^2)^(-4)*d/dx[1-2x^2]
f '' (x) = 8*(1-2x^2)^(-3) + 8x*(-3)(1-2x^2)^(-4)*(-4x)
f '' (x) = 8*(1-2x^2)^(-3) + 96x^2(1-2x^2)^(-4)
f '' (x) = 8*(1-2x^2)(1-2x^2)^(-4) + 96x^2(1-2x^2)^(-4)
f '' (x) = (8-16x^2)(1-2x^2)^(-4) + 96x^2(1-2x^2)^(-4)
f '' (x) = (8-16x^2 + 96x^2)(1-2x^2)^(-4)
f '' (x) = (8+80x^2)(1-2x^2)^(-4)
Note the use of the power rule on line 3, then the chain rule on line 4
Third Derivative:
f '' (x) = (8+80x^2)(1-2x^2)^(-4)
f ''' (x) = d/dx[(8+80x^2)]*(1-2x^2)^(-4)+(8+80x^2)*d/dx[(1-2x^2)^(-4)]
f ''' (x) = (160x)*(1-2x^2)^(-4)+(8+80x^2)*(-4)*(1-2x^2)^(-5)*d/dx[1-2x^2]
f ''' (x) = (160x)*(1-2x^2)^(-4)+(8+80x^2)*(-4)*(1-2x^2)^(-5)*(-4x)
f ''' (x) = (160x)*(1-2x^2)^(-4)+(1280x^3 + 128x)*(1-2x^2)^(-5)
f ''' (x) = (160x)*(1-2x^2)*(1-2x^2)^(-5)+(1280x^3 + 128x)*(1-2x^2)^(-5)
f ''' (x) = (160x-320x^3)*(1-2x^2)^(-5)+(1280x^3 + 128x)*(1-2x^2)^(-5)
f ''' (x) = (160x-320x^3+1280x^3+128x)*(1-2x^2)^(-5)
f ''' (x) = (960x^3+288x)*(1-2x^2)^(-5)
Fourth Derivative:
f ''' (x) = (960x^3+288x)*(1-2x^2)^(-5)
f '''' (x) = d/dx[(960x^3+288x)]*(1-2x^2)^(-5)+(960x^3+288x)*d/dx[(1-2x^2)^(-5)]
f '''' (x) = (2880x^2+288)*(1-2x^2)^(-5)+(960x^3+288x)*(-5)*(1-2x^2)^(-6)*d/dx[1-2x^2]
f '''' (x) = (2880x^2+288)*(1-2x^2)^(-5)+(960x^3+288x)*(-5)*(1-2x^2)^(-6)*(-4x)
f '''' (x) = (2880x^2+288)*(1-2x^2)^(-5)+(19200x^4+5760x^2)*(1-2x^2)^(-6)
f '''' (x) = (2880x^2+288)*(1-2x^2)*(1-2x^2)^(-6)+(19200x^4+5760x^2)*(1-2x^2)^(-6)
f '''' (x) = (-5760x^4+2304x^2+288)*(1-2x^2)^(-6)+(19200x^4+5760x^2)*(1-2x^2)^(-6)
f '''' (x) = (-5760x^4+2304x^2+288+19200x^4+5760x^2)*(1-2x^2)^(-6)
f '''' (x) = (13440x^4+8064x^2+288)*(1-2x^2)^(-6)
We could keep going with even higher order derivatives, but this is sufficient for the problem. The reason why is because we only need to go as high as the x^4 term of the Taylor expansion, so that ties to the fourth derivative of f(x). Anything beyond this point will contribute some change but not much compared to the larger terms. The x^2 is already small, which means stuff like x^5,x^6,x^7,... are even smaller. That's why they don't play a large role.
Since we want x^2 to be small, this implies that x is small.
So the Taylor polynomial for f(x) will be centered at x = 0.
We have these functions
f(x) = (1-2x^2)^(-2)
f ' (x) = 8x*(1-2x^2)^(-3)
f '' (x) = (8+80x^2)(1-2x^2)^(-4)
f ''' (x) = (960x^3+288x)*(1-2x^2)^(-5)
f '''' (x) = (13440x^4+8064x^2+288)*(1-2x^2)^(-6)
Evaluate each function at x = 0
f(x) = (1-2x^2)^(-2)
f(0) = (1-2*0^2)^(-2)
f(0) = (1)^(-2)
f(0) = 1
f ' (x) = 8x*(1-2x^2)^(-3)
f ' (0) = 8*0*(1-2*0^2)^(-3)
f ' (0) = 0
f '' (x) = (8+80x^2)(1-2x^2)^(-4)
f '' (0) = (8+80*0^2)(1-2*0^2)^(-4)
f '' (0) = (8)(1)^(-4)
f '' (0) = (8)(1)
f '' (0) = 8
f ''' (x) = (960x^3+288x)*(1-2x^2)^(-5)
f ''' (0) = (960*0^3+288*0)*(1-2*0^2)^(-5)
f ''' (0) = (0)*(1-2*0^2)^(-5)
f ''' (0) = 0
f '''' (x) = (13440x^4+8064x^2+288)*(1-2x^2)^(-6)
f '''' (0) = (13440*0^4+8064*0^2+288)*(1-2*0^2)^(-6)
f '''' (0) = (288)*(1)^(-6)
f '''' (0) = (288)*(1)
f '''' (0) = 288
Let t(x) be the Taylor approximation for f(x) centered around x = 0
We can say
t(x) = f(0) + (f'(0)/(1!))*(x-0) + (f''(0)/(2!))*(x-0)^2 + (f'''(0)/(3!))*(x-0)^3 + (f''''(0)/(4!))*(x-0)^4
t(x) = 1 + (0/1)*(x-0) + (8/2)*(x-0)^2 + (0/6)*(x-0)^3 + (288/24)(x-0)^4
t(x) = 1 + 4x^2 + 12x^4
If you graphed f(x) and t(x) together, then zoomed in really close to x = 0, you should see that the graphs are very similar to one another. I recommend using graphing software that allows for different color functions and that allows you to toggle the function on or off, so you can see the two different functions contain nearly the same set of points. Desmos and Geogebra are two free such options.
As you move away from x = 0, f(x) and t(x) start to disagree with one another. So t(x) is only useful for small values of x.
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Follow the same basic steps for g(x).
I won't list those steps because I'll let you do that part.
This is what you should get (or you could get equivalent versions to these)
g(x) = (1+6x^2)^(2/3)
g ' (x) = 8x*(6x^2+1)^(-1/3)
g '' (x) = (16x^2+8)*(6x^2+1)^(-4/3)
g ''' (x) = (-64x^3-96x)*(6x^2+1)^(-7/3)
g '''' (x) = (640x^4+1920x^2-96)(6x+1)^(-10/3)
Evaluate each function at x = 0
g(x) = (1+6x^2)^(2/3)
g(0) = (1+6*0^2)^(2/3)
g(0) = 1
g ' (x) = 8x*(6x^2+1)^(-1/3)
g ' (0) = 8*0*(6*0^2+1)^(-1/3)
g ' (0) = 0
g '' (x) = (16x^2+8)*(6x^2+1)^(-4/3)
g '' (0) = (16*0^2+8)*(6*0^2+1)^(-4/3)
g '' (0) = 8
g ''' (x) = (-64x^3-96x)*(6x^2+1)^(-7/3)
g ''' (0) = (-64*0^3-96*0)*(6*0^2+1)^(-7/3)
g ''' (0) = 0
g '''' (x) = (640x^4+1920x^2-96)(6x+1)^(-10/3)
g '''' (0) = (640*0^4+1920*0^2-96)(6*0+1)^(-10/3)
g '''' (0) = -96
Let r(x) be the Taylor expansion of g(x) around x = 0
r(x) = g(0) + (g'(0)/(1!))*(x-0) + (g''(0)/(2!))*(x-0)^2 + (g'''(0)/(3!))*(x-0)^3 + (g''''(0)/(4!))*(x-0)^4
r(x) = 1 + (0/1)*(x-0) + (8/2)*(x-0)^2 + (0/6)*(x-0)^3 + (-96/24)*(x-0)^4
r(x) = 1 + 4x^2 - 4x^4
Like the relationship with t(x) and f(x), the connection with r(x) and g(x) is only really useful for small values of x.
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So going back to the original problem of finding f(x) - g(x)
That's approximately the same as t(x) - r(x) when x is small. So when x is centered around x = 0.
Both r and t are polynomials which can be quickly subtracted as such
t(x) - r(x) = ( 1 + 4x^2 + 12x^4 ) - ( 1 + 4x^2 - 4x^4 )
t(x) - r(x) = 1 + 4x^2 + 12x^4 - 1 - 4x^2 + 4x^4
t(x) - r(x) = 12x^4 + 4x^4
t(x) - r(x) = 16x^4
The "1+4x^2" terms basically canceled out leaving the x^4 terms to be combined.
We see that t(x)-r(x) is 16x^4
So f(x) - g(x) is approximately 16x^4
So in the end,
(1-2x^2)^(-2) - (1+6x^2)^(2/3)
is approximately equal the form kx^4 where k = 16. This only applies for small values of x^2.
You can use a graph to help visually confirm this.
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