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Conjecture about the nth derivative of the function f(x)=e^(ax)

  • Thread starter Dustinsfl
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  • #1
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Make a conjecture about the nth derivative of the function f(x)=e^(ax). This conjecture should be written as a self contained proposition including an appropriate quantifier.

What is the last sentence saying to do. I know what a conjecture is but I am confused on what the book wants here.
 

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  • #2
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Make a conjecture about the nth derivative of the function f(x)=e^(ax). This conjecture should be written as a self contained proposition including an appropriate quantifier.

What is the last sentence saying to do. I know what a conjecture is but I am confused on what the book wants here.
To help you understand what "conjecture" means, consider the following small example. For example, let's pretend that I happen to notice that

[tex]1+2 = 3 = \frac{2 (2+1)}{2}[/tex]

[tex]1+2+3 = 6 = \frac{3( 3+1)}{2}[/tex]

[tex]1+2+3+4= 10 = \frac{4(4+1)}{2}[/tex]

then I might hypothesize that

[tex]1+2+3+\dotsm + (n-1) + n = \frac{n(n+1)}{2}[/tex]

Such an hypothesis (based on observed patterns) is a conjecture about the formula for the sum 1+2+3+...+(n-1)+n. It is an educated guess that appears to be true, but needs to be proved (or possibly disproved -- maybe our hypothesis is wrong).

So for your question, try playing around with all the derivatives (first, second, third, etc.) of the function e^(ax) and see if you can find any patterns. Then make a "conjecture" about the "nth derivative"
 
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So the self contained proposition doesn't mean to do anything special or different then a just making any conjecture?
 
  • #4
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So the self contained proposition doesn't mean to do anything special or different then a just making any conjecture?
I believe that the intention of this problem is that the conjecture you make has only to do with the function f(x) = e^(ax) and its derivatives, and nothing more (self-contained). Otherwise, we could make up just about any conjecture.
 

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