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Generating general formula for multiple derivatives of f^n(x), given F(x)= -3/x
- Thread starter nicmun
- Start date
I'm having trouble coming up with a general formula for f^n(x)
F(x)= -3/x
F'(x)= 3/x^2
F''(x)= -6x/x^4
F'''(x)= 18x^4/x^8
From this I can get the denominator is x^(n^2), but how do I get the formula for the numerator?
Well, I'd begin by noting that your expression for the denominator is not correct. If that were true, the denominators would be {x, x4, x9, x16, ...} Additionally, your expressions for the derivatives are technically right, but it might help to see what's really going on if you write it in a different form. You would have:
\(\displaystyle f(x) = \dfrac{-3}{x}\)
\(\displaystyle f^{(1)}(x) = \dfrac{3}{x^2}\)
\(\displaystyle f^{(2)}(x) = \dfrac{-6}{x^3}\)
\(\displaystyle f^{(3)}(x) = \dfrac{18}{x^4}\)
That aside, let's look at each of the numerators and how they relate to the previous one. We can clearly see that if n is even, then the numerator is negative, but if n is odd, the numerator is positive. This tells us the expression for the numerator must involve (-1)n+1 in some way. Then we can look at the absolute values of the numerators, and see that 3 = 3 * 1, 6 = 3 * 2, 18 = 6 * 3, etc. Are you seeing a pattern here? What is the fourth derivative of f(x)? Does it follow this pattern?
Let's start with notation.I'm having trouble coming up with a general formula for f^n(x)
F(x)= -3/x
F'(x)= 3/x^2
F''(x)= -6x/x^4
F'''(x)= 18x^4/x^8 But this is wrong. The coefficient is + 24.
From this I can get the denominator is x^(n^2), but how do I get the formula for the numerator?
\(\displaystyle n = 0 \implies f_n(x) = -\ \dfrac{3}{x}.\)
\(\displaystyle n > 0 \implies f_{n}(x) = \text {derivative of } f_{n-1}(x).\)
\(\displaystyle n = 1 \implies f_1(x) = \text {derivative of } f_0(x) = \left ( \dfrac{d}{dx}\ \left (-\ \dfrac{3}{x} \right ) \right ) =\\
\dfrac{3}{x^2} = 3 * 1! * (-\ 1)^2 * x^{- 2} = 3 * n! * (-\ 1)^{(n+1)} * x^{-(n+1)} .\)
\(\displaystyle \therefore \exists \text { at least one integer } k \text { such that }\\
k \ge 1 \text { and } f_k(x) = 3k! * (-\ 1)^{(k+1)} * x^{-\ (k+1)}.\)
\(\displaystyle \text {CASE I: } k \text { is odd} \implies k + 1 \text { is even, } k + 2 = (k + 1) + 1 \text { is odd,}\)
\(\displaystyle \text {and } f_k = 3k! * (-\ 1)^{(k+1)} * x^{-\ (k+1))} = 3k! * x^{-\ (k+1))}.\)
\(\displaystyle \therefore f_{k+1} = \left ( \dfrac{d}{dx}\ 3k! * x^{-\ (k + 1)} \right ) =\\
= 3k!\{- (k + 1)\}k! * x^{\{-(k+1) - 1\}} = 3(k + 1)k! * (-\ 1) * x^{\{-(k+1) - 1\}} =\\
3(k + 1)! * (-\ 1)^{\{(k + 1)+ 1\}} * x^{-\{(k+1) + 1\}}.\)
\(\displaystyle \text {CASE II: } k \text { is even} \implies k + 1 \text { is odd, } k + 2 = (k + 1) + 1 \text { is even,}\)
\(\displaystyle \text {and } f_k = 3k! * (-\ 1)^{(k+1)} * x^{-\ (k+1))} = (-\ 1) * 3k! * x^{-\ (k+1))}.\)
\(\displaystyle \therefore f_{k+1} = \left ( \dfrac{d}{dx}\ (-\ 1) * 3k! * x^{-\ (k + 1)} \right ) =\\
= 3(-\ 1)\{-\ (k + 1)\}k! * x^{\{-(k+1) - 1\}} = 3(k + 1)k! * x^{-\ (k+1) - 1\}} =\\
3(k + 1)! * (-\ 1)^{\{(k + 1)+ 1\}} * x^{-\ \{(k+1) + 1\}}.\)
\(\displaystyle \text {THUS, by induction, } n > 0 \implies f_k(x) = 3n! * (-\ 1)^{(n+1)} * x^{-\ (n+1)}.\)
Well, I'd begin by noting that your expression for the denominator is not correct. If that were true, the denominators would be {x, x4, x9, x16, ...} Additionally, your expressions for the derivatives are technically right, but it might help to see what's really going on if you write it in a different form. You would have:
\(\displaystyle f(x) = \dfrac{-3}{x}\)
\(\displaystyle f^{(1)}(x) = \dfrac{3}{x^2}\)
\(\displaystyle f^{(2)}(x) = \dfrac{-6}{x^3}\)
\(\displaystyle f^{(3)}(x) = \dfrac{18}{x^4}\)
That aside, let's look at each of the numerators and how they relate to the previous one. We can clearly see that if n is even, then the numerator is negative, but if n is odd, the numerator is positive. This tells us the expression for the numerator must involve (-1)n+1 in some way. Then we can look at the absolute values of the numerators, and see that 3 = 3 * 1, 6 = 3 * 2, 18 = 6 * 3, etc. Are you seeing a pattern here? What is the fourth derivative of f(x)? Does it follow this pattern?
Whatever the pattern is I'm not getting it and it's driving me insane I've been racking my brain all day, I'm sure it's obvious but I've looked at it too long at this point. The fourth derivative would be 72. 3*24 = 72. So it's 3 * something but I don't see how you go from 1 to 2 to 6 to 24.
Whatever the pattern is I'm not getting it and it's driving me insane I've been racking my brain all day, I'm sure it's obvious but I've looked at it too long at this point. The fourth derivative would be 72. 3*24 = 72. So it's 3 * something but I don't see how you go from 1 to 2 to 6 to 24.
Actually, you had the third derivative correct to begin with. I'm not sure why JeffM said the coefficient should be 24, as that's not correct. The third derivative of \(\displaystyle \dfrac{-3}{x}\) is \(\displaystyle \dfrac{18}{x^4}\). WolframAlpha agreeshttps://www.wolframalpha.com/input/?i=third+derivative+of+-3/x. The pattern you're looking for then becomes:
- 3
- 3 = 3 * 1
- 6 = 3 * 2 = 3 * (1 * 2)
- 18 = 6 * 3 = 3 * (1 * 2 * 3)
- 72 = 18 * 4 = 3 * (1 * 2 * 3 * 4)
- etc.
Are you familiar with the factorial? That will prove immensely helpful here.