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Here we have $f(x)=x^{3}$, if we substitute x with 2, we will get $f(2)=8$. Now if we took the derivative of $f(x)=x^{3}$, we will get $f'(x)=3x^{2}$, and $f'(2)=12$. We can tell from this that a change in x, will change $f(x)$ 12 times as fast as x is changing.

$f(2.001)=8.012006001$

$f(2.002)=8.024024008$

$f(2.003)=8.036054027$

I can see the pattern here, it is changing 12 times as x is changing, however why is it when I try with for example 2.1 I don't get the same pattern?

$f(2.1)=9.261$

$f(2.2)=10.648$

$f(2.3)=12.167$

Friedrich
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  • Because the derivative is constantly changing perhaps? The derivative is only $12$ at $x=2$, not at $x>2$. I'm also not quite sure what exactly you mean with $* \ *$ –  Mar 13 '18 at 17:59
  • It's to show the patterns. f(2.001) = 8.012006001. The ** are added to separate the parts of the pattern. – Preston Millward Mar 13 '18 at 18:02
  • @vrugtehagel when you substitute "2.001" you get 8.012 ( not exactly ), and when you change it to 2.002 you get 8.024.. it is changing by 0.012, but when substituting 2.1 and making a change you don't get that change by 12 times. – Friedrich Mar 13 '18 at 18:05
  • It was meant to show the pattern as @Preston said, however I removed it so it doesn't cause any confusion – Friedrich Mar 13 '18 at 18:06

3 Answers3

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You actually do get the same pattern, it's just that the numbers in the pattern have all started to flow together. $$f(2.001) = 8 + .001 * 12 + 6 * (.001)^2 + (.001)^3$$ $$f(2.002) = 8 + .002 * 12 + 6 * (.002)^2 + (.002)^3$$ $$...$$ $$f(2.1) = 8 + .1 * 12 + 6 * (.1)^2 + (.1)^3 = 9.261$$

Preston Millward
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  • I had never noticed this before. This question is really interesting. – Preston Millward Mar 13 '18 at 18:14
  • This was very elegant and very interesting, thank you very much :) – Friedrich Mar 13 '18 at 18:44
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    The pattern you're observing is called the Taylor series for the given function. The derivative is only an approximation, but the Taylor series is exact -- and the Taylor series for a polynomial is in fact equal to the polynomial itself:

    $$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3$$

    Here, $f(2)=8$, $f'(2)=12$, $f''(2)=12$, and $f'''(2)=6$.

     – symplectomorphic Mar 13 '18 at 18:48
  • That's sweet. Thanks! – Preston Millward Mar 13 '18 at 19:09
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$9.261 - 8 = 1.261 = 12.61*0.1 \approx 12*0.1$

$10.648 - 9.261 = 1.387=13.87*0.1 \approx 12*0.1$

So you do get the same pattern.

....

$f(x)$ is increasing $12$ times faster than $x$ only at exactly $x=2$. At just a tiny smidgen past $x=2$ it will be increasing at a faster rate then $12$.

So the distance travel from $x=2$ to $x=2 + a$ will have to be more than $12a$ because $f(x)$ would be traveling faster then $12$ for all the time $x \in (2, 2+a]$.

On the other hand, at $x=2+a$ we have $f(x)$ is increasing at $3(4+4a + a^2)=12 +12a + 3a^2$. wheres for all $x \in [2,2+a)$, $f(x)$ was increasing at rate slower than $12 + 12a + 3a^2$. So the distance travelled will be less than $a(12+12a+3a^2)$.

So the total distance $d$ will be $12a < d < 12a+12a^2+3a^2$.

For very small values of $a$ (particularly in comparison to $x$) the margin of error is small but for large values it is not.

$f(2) = 8; f(2.01)= 8.120601$

$f'(2) = 12; f'(2.01) = 12.1203$

So $f(2) + 12*.01 < f(2.01) < f(2) + 12.1203*0.01\implies 8.12 < 8.120601 < 8.121203$.

But $f(2.1) = 9.261$ and $f'(2.1) = 3(2.1)^2 = 13.21$ so

$f(2) + 12*.1 < f(2.1) < f(2) + 13.21*1 \implies 9.2 < 9.261 < 9.321$.

All is as expected.

fleablood
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Linear approximations are just tangent line approximations and they are local.

That is they work the best if your change in $x$ is very small.

In your case, your change of $0.001$ is good but the change of $0.1$ is not very good.

As you noticed, for $2.2$ and $2.3$ the estimates are even worse.

For $2.3$ your estimate is $8+12(.3)=11.6$ while the exact value is $2.3 ^3 = 12.2$ an error of $.6$ which is rather large.

Mohammad Riazi-Kermani
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