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Question. Can you help me prove $$ \exp(-x)=\frac{1}{\exp(x)} \quad ? $$

I need to prove this for a maths homework. What's given is that

  1. $ \displaystyle \exp(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!} $.
  2. $ \exp(1) = e $.
  3. $ \exp(x+y) = \exp(x) \exp(y)$.

I managed to prove other terms but I can't do this, I've tried for hours. Sorry that I don't know how to properly write down the equation 1).

Sangchul Lee
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lulu
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  • No other givens, like what $-x$ means? This is slightly a provocative way to say it, but these extra things are actually important in some cases... – abiessu Jan 13 '23 at 07:35
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    Also, it is recommended that the question you have should appear in the body of what you write, not just in the title. A good way to do this might be to present an approach you might have taken to the problem. – abiessu Jan 13 '23 at 07:37
  • I removed the inappropriate [tag:theorem-provers] tag – please avail yourself of the tag summaries when choosing tags. – joriki Jan 13 '23 at 07:41
  • Here's a tutorial and reference for formatting math on this site. – joriki Jan 13 '23 at 07:42
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    Is it given that $\rm{exp}(0)=1$? [If not it is not hard to show by plugging $x=0$ to the given series for $e^x.$] Once that is known, apply rule 3 in the case where $y=-x$ and you can derive what you need about $\rm{exp}(-x).$ – coffeemath Jan 13 '23 at 07:46
  • What happens if you multiply $e^x$ and $e^{-x}$ using the power series definition? – David Lui Jan 13 '23 at 08:02

1 Answers1

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The first relation provides $\exp(0) = 1$ by mere evaluation; the third one gives then : $$ 1 = \exp(0) = \exp(x-x) = \exp(x)\exp(-x) \verb+ +\Leftrightarrow\verb+ + \exp(-x) = \frac{1}{\exp(x)} $$

Abezhiko
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