I tried converting to exponential, e^(1+x^2) -x -1 = 0 Then finding a contradiction but didn't find one. Then i differentiated to find a minima > 0. 2xe^(1+x^2) -1 = 0 but didn't find a solution as the term 2xe^(1+x^2) doesn't equal to 1 at any point x. Again even if there was any it still would have been a local minima.
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2I’m voting to close this question because it is not a computer programming question. It is a mathematics question. – Raymond Chen Oct 21 '20 at 05:08
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i gave tag as math and transcendental equations? – Aditya Jha Oct 21 '20 at 05:10
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1This is a computer programming site. Perhaps you want https://math.stackexchange.com/ – Raymond Chen Oct 21 '20 at 05:12
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@AdityaJha Read the tag descriptions to see why math-only questions are off-topic on SO. Also, `log(1+x) <= x < 1 + x^2`. – dxiv Oct 21 '20 at 06:02
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Let f(x) = 1 + x^2 - log(1 + x)
Conditional: x > -1
f'(x) = 2x -1/(1+x)
f'(x) = 0 => x = (sqrt(3) - 1) / 2
f"(x) = 2 + 1/(1+x)^2, always greater than 0
So f(sqrt(3) - 1) / 2) must be the minimum value of f
f(sqrt(3) - 1) / 2) is approximate 0.8
So f(x) always greater than 0.8
Therefore, f(x) = 0 has no solution.
Therefore log(1+x) and (1+x^2) do not intersect.
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