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Derivatives

Derivatives

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Acolyte
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Loughborough

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Suppose f and g are two infinitely differentiable functions on the reals, and suppose that, for some real A, k>0, n>0, and for all x>A:
f^(n)(x) - g^(n)(x) > k (where ^(n) means nth derivative.)

Show that there exists a number B so that for all x>B, f(x) - g(x) > 0. Deduce that, if f is e^qx for q positive, and g is a polynomial, there exists a B for which f>g for all x>B.

Once again I ask mathematical types not to call out the answer until others have had a chance to think about it. This one's off the top of my head, so I apologise if people find it too easy or hard, or if I've got the problem wrong.

Note to the non-mathematical: a derivative is a 'rate of change', and the derivative of an nth derivative is a (n+1)th derivative. So the second derivative of f is 'rate of change of rate of change of f', and so on.

r
CHAOS GHOST!!!

Elsewhere

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Originally posted by Acolyte
Suppose f and g are two infinitely differentiable functions on the reals, and suppose that, for some real A, k>0, n>0, and for all x>A:
f^(n)(x) - g^(n)(x) > k (where ^(n) means nth derivative.)

Show that there exists a number B so that for all x>B, f(x) - g(x) > 0. Deduce that, if f is e^qx for q positive, and g is a polynomial, there exists a B for w ...[text shortened]... ivative. So the second derivative of f is 'rate of change of rate of change of f', and so on.
This is clever, but isn't it only of interest to 'mathematical types' or maybe Thomas Malthus 😉?

Acolyte
Now With Added BA

Loughborough

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Originally posted by Acolyte
Suppose f and g are two infinitely differentiable functions on the reals, and suppose that, for some real A, k>0, n>0, and for all x>A:
f^(n)(x) - g^(n)(x) > k (where ^(n) means nth derivative.)

Show that there exists a number B so that for all x>B, f(x) - g(x) > 0. Deduce that, if f is e^qx for q positive, and g is a polynomial, there exists a B for w ...[text shortened]... ivative. So the second derivative of f is 'rate of change of rate of change of f', and so on.
Just noticed that f and g need only be n times differentiable. There are a few variations on this theme, and I was messing around with them after posting, hence the edits.

Acolyte
Now With Added BA

Loughborough

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Hint (and I don't care who answers it now): the magic word is Induction.

f

my head

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Originally posted by Acolyte
Suppose f and g are two infinitely differentiable functions on the reals, and suppose that, for some real A, k>0, n>0, and for all x>A:
f^(n)(x) - g^(n)(x) > k (where ^(n) means nth derivative.)

Show that there exists a number B so that for all x>B, f(x) - g(x) > 0. Deduce that, if f is e^qx for q positive, and g is a polynomial, there exists a B for w ...[text shortened]... ivative. So the second derivative of f is 'rate of change of rate of change of f', and so on.
must...learn....cal..cu...lus....

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