A curve y=f(x) passes through origin and slope of the tangent at any point (x, y) of the curve is (x4+2xy-1)/(1+x2), then find the value of the greatest integer which is less than or equal to f(-1)?
Non calculus solution (not a rigorous proof)
At(0,0), slope is -1
At (-ε, ε) slope is:
(ε^4-2ε^2-1)/(1+ε^2)
(-2ε^2-1)/(1+ε^2)
(-2ε^2-1)(1-ε^2) / ((1+ε^2)(1-ε^2))
(-1-1ε^2+2ε^4) / (1-ε^4)
-1-ε^2 = -(1+ε^2)
The curve passes through (-2ε, 2ε+ε^2)
So following the function from (0,0) up and to the left, the slope will be getting more negative.
Expect the function to pass through (-1, 1+k). (where k>0)
There is a good chance that k is small and the final answer is (provided k<1) that int(f(-1)) = 1
Computer estimate.
Take small x steps and compute y based on calculated slope.
(x,y) near (-1.0000000000000007, 1.141438461555864)
f(-1) near 1.141438461555864, possibly √2
greatest integer of f(-1) confirmed to be 1
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steps = 1000
x=[0]
y=[0]
for step in range(steps):
slope = (x[step]**4 + 2*x[step]*y[step] - 1) / (1 + x[step]**2)
x.append(x[step]-1/steps)
y.append(y[step] - slope/steps)
print(x[-1], y[-1])
print( int(y[-1]))
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Posted by Larry
on 2024-10-30 13:51:39 |
Probable solution
