K9-47G Posted September 30, 2005 Share Posted September 30, 2005 This is the problem in my book. If g(x)=x/e^x, find g^(n)(x). I don't really understand what the problem is asking me to find. It is in the differentiation section of the book, if that helps at all. I think it may be asking for a formula... By the way, the n in the formula represents how many times to take the derivative of g(x). Link to comment Share on other sites More sharing options...

Dave Posted September 30, 2005 Share Posted September 30, 2005 What it's looking for is an explicit formula (in terms of n and x) for the n'th derivative of g. How do you approach it? The way I would consider is to work out the first few terms. Also, consider writing g as [imath]g(x) = xe^{-x}[/imath]. Once you've worked out the first few terms and figured out what the pattern is, make a guess at what you think the formula is. You can then prove it using induction (which is an easy proof). Link to comment Share on other sites More sharing options...

K9-47G Posted September 30, 2005 Author Share Posted September 30, 2005 So would it be g^n(x)=g(x)+n(e^-x)..... I'm assuming that the derivative of e^-x is e^-x.... Link to comment Share on other sites More sharing options...

mezarashi Posted September 30, 2005 Share Posted September 30, 2005 So would it be g^n(x)=g(x)+n(e^-x)..... I'm assuming that the derivative of e^-x is e^-x.... Don't just assume that. The derivative of exp(-x) is -exp(-x). Notice that to differentiate g(x) = xexp(-x) you will need to use the product rule. Using that, write out: g'(x) g''(x) g'''(x) and so on. Enough so that you see a pattern. For example if I had the function f(x) = exp(ax) Then f'(x) = a exp(ax) f''(x) = a^2 exp(ax) f'''(x) = a^3 exp(ax) I can conclude that f(n)(x) = a^n exp(ax). Got it? Link to comment Share on other sites More sharing options...

K9-47G Posted September 30, 2005 Author Share Posted September 30, 2005 well I did write it out, this is what I got. g'(x)= xe^-x+e^-x g"(x)= xe^-x+2e^-x g'''(x)= xe^-x+3e^-x so I concluded that the formula would be g^n(x)= g(x)+n(e^-x). I'm just not sure if my math is right. Link to comment Share on other sites More sharing options...

Dave Posted September 30, 2005 Share Posted September 30, 2005 That's not entirely correct. For example: [imath]g'(x) = e^{-x} - xe^{x}[/imath] [imath]g''(x) = -2e^{-x} +xe^{x}[/imath] You must make sure to observe your negative signs Also, you might well see a pattern, but that certainly isn't a proof. Whilst it's an easy inductive proof, it is required to make your observation concrete. Link to comment Share on other sites More sharing options...

mezarashi Posted September 30, 2005 Share Posted September 30, 2005 I'm just not sure if my math is right. Right off, I disagree with your g'(x). Refer to what i said about the derivative of exp(-x). Edit: or what dave just posted right now Link to comment Share on other sites More sharing options...

Dave Posted September 30, 2005 Share Posted September 30, 2005 so I concluded that the formula would be g^n(x)= g(x)+n(e^-x). I'm just not sure if my math is right. I don't think this is a good form for the answer. You want an explicit formula in terms of x, and whilst you have one there, it's much neater to write down something like [imath]g^{(n)}(x) = nx + b[/imath] or whatever. Link to comment Share on other sites More sharing options...

K9-47G Posted September 30, 2005 Author Share Posted September 30, 2005 so e^-x derived is -e^-x.... We haven't covered that yet. Now I'm getting g'(x)=e^-x-xe^-x g''(x)=-2e^-x+xe^-x g'''(x)= 3e^-x-xe^-x and so on, but I have no idea how to make an explicit formula out of that because the negatives are alternating. Link to comment Share on other sites More sharing options...

Dave Posted September 30, 2005 Share Posted September 30, 2005 Note that [imath](-1)^{n+1}[/imath] is -1 when n is even, and 1 when n is odd. It's a common thing to find. Obviously if you haven't covered stuff like the chain rule and all that, you don't need to go on and prove it by induction. I had assumed that you were at some kind of 1st year university level maths or something Link to comment Share on other sites More sharing options...

K9-47G Posted September 30, 2005 Author Share Posted September 30, 2005 The inductive proof is: show true for n=1, assume true for n=k and show true for n=k+1, right? Link to comment Share on other sites More sharing options...

mezarashi Posted September 30, 2005 Share Posted September 30, 2005 so e^-x derived is -e^-x.... We haven't covered that yet. You should have convered: d/dx ( exp(ax) ) = a exp(ax)' date=' where a is a constant. Now let a = -1. and so on, but I have no idea how to make an explicit formula out of that because the negatives are alternating. (-1)^n alternates between positive and negative for increasing integer values of n. Link to comment Share on other sites More sharing options...

Dave Posted September 30, 2005 Share Posted September 30, 2005 That's the idea. It's blatently easy because you can just differentiate from the statement of n=k being true. Link to comment Share on other sites More sharing options...

BobbyJoeCool Posted September 30, 2005 Share Posted September 30, 2005 isn't [math]g^n (x)[/math] the same as saying [math](g(x))^n[/math]? In which case the derivative of the function for any n can be given using the simple diferentiaition rules... [math]g(x)^n=(xe^{-x})^n[/math] Genera power rule states... (or the power rule with the chain rule) [math]g(x)=(a)u^b, g'(x)=(ab)u^{b-1}u'[/math] so we use this on the function... with.. [math]u=xe^{-x}[/math] and u'= to the derivative of u... using the product rule... [math]-xe^{-x}+e^{-x}=(1-x)e^{-x}[/math] [math]g'(x)^n=n(xe^{-x})^{n-1}(1-x)e^{-x}[/math] or would that be way off and too simple? EDIT: all those funky f's I had were changed to g's... opps. EDIT mark II: I just noticed that KG said to take the deriviatves and not Dave... so this is kinda pointless isn't it... Link to comment Share on other sites More sharing options...

K9-47G Posted September 30, 2005 Author Share Posted September 30, 2005 So would my formula be g^n(x)= -n(-1^n)e^-x+(-1^n)xe^-x. I know there must be an easier way to write that. Link to comment Share on other sites More sharing options...

K9-47G Posted September 30, 2005 Author Share Posted September 30, 2005 How do you format your math work to look bold and easier to read? Link to comment Share on other sites More sharing options...

Dave Posted September 30, 2005 Share Posted September 30, 2005 isn't [math]g^n (x)[/math] the same as saying [math](g(x))^n[/math']?[/math] Generally, we interpret g^{n} as: [math]\underbrace{g \circ \cdots \circ g}_{\text{n times}}[/math] The circle indicates composition of the function. This is mainly because mathematicians are lazy, and in general we're more interested in composition than multiplication. We use [imath]g^{(n)}[/imath] to denote the n'th derivative. EDIT mark II: I just noticed that KG said to take the deriviatves and not Dave... so this is kinda pointless isn't it... Pretty much Link to comment Share on other sites More sharing options...

BobbyJoeCool Posted September 30, 2005 Share Posted September 30, 2005 How do you format your math work to look bold and easier to read? [math]\LaTeX[/math] Link to tutorial thread And I misunderstood the question... so my answer is wrong... Link to comment Share on other sites More sharing options...

Dave Posted September 30, 2005 Share Posted September 30, 2005 So would my formula be g^n(x)= -n(-1^n)e^-x+(-1^n)xe^-x. I know there must be an easier way to write that. Yes, but you can simplify that quite a bit. Notice that by factoring the -1 out, you get the much nicer [imath]g^{(n)}(x) = (-1)^{n+1} ( ne^{-x} - xe^{-x})[/imath]. Link to comment Share on other sites More sharing options...

BobbyJoeCool Posted September 30, 2005 Share Posted September 30, 2005 This is mainly because mathematicians are lazy' date=' and in general we're more interested in composition than multiplication.[/quote'] isn't that a contradiciton of terms? lazy, but more interested in a harder way of doing things? But yes, math people are lazy, who else would come up with [math]\Delta x[/math] to represent Change in x? Link to comment Share on other sites More sharing options...

Dave Posted September 30, 2005 Share Posted September 30, 2005 isn't that a contradiciton of terms? lazy, but more interested in a harder way of doing things? Well, I suppose we just don't want to bother writing down loads of compositions, so we come up with a shorthand way of doing it. Multiplication just isn't as useful as composition Link to comment Share on other sites More sharing options...

K9-47G Posted September 30, 2005 Author Share Posted September 30, 2005 [math] g^{(n)}(x)=-n(-1^n)e^{-x}+(-1^n)xe^{-x} [/math] Link to comment Share on other sites More sharing options...

Dave Posted September 30, 2005 Share Posted September 30, 2005 By the way, it's not [imath]-1^n[/imath] since this is generally interpreted as [imath]-(1^n) = -1[/imath]. You'll want to write [imath](-1)^n[/imath]. Link to comment Share on other sites More sharing options...

K9-47G Posted September 30, 2005 Author Share Posted September 30, 2005 ok, thanks. Link to comment Share on other sites More sharing options...

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