Question

show that e^x>1+x+(x^2/2), for all x>0 @MIT 18.02 Multiva…

Answer 1

If able, write out the maclaurin series for e^x. That shows this

Answer 2

depends on your definition of
\[e^x\]

Answer 3

what do you mean depending on the definition of e^x

Answer 4

Consider the function f(x) = e^x - (x^2/2 + x + 1). Note that for x = 0, f(x) = 0. Now if you can show that the derivative of f'(x) is always positive for x > 0, that will do the trick.

f'(x) = e^x - (x + 1)

Still a bit of a problem. But notice now that f'(x) = 0.
If we can show that (f'(x))' = f''(x) > 0 for x > 0, then that will do the trick.

f''(x) = e^x - 1.

As e^x > 1 for x > 0, it is indeed the case that f''(x) > 0 for x > 0.

Hence f"(x) > 0 for x > 0

And therefore f(x) > 0 for x > 0.

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It's possible now to generalize this solution to show that for \( x > 0 \)
\[ e^x > \sum_{i=0}^n x^i/i! \]

Answer 5

("Hence f"(x) > 0 for x > 0" should read "Hence f'(x) > 0 for x > 0")

Answer 6

I really like that proof. Nice