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Question Number 6610 by Temp last updated on 05/Jul/16

If you have a real continuous function f(x),  If you take a distance between  f(x) and f(x+1), what is the average  distance within a≤x≤b?

$$\mathrm{If}\:\mathrm{you}\:\mathrm{have}\:\mathrm{a}\:\mathrm{real}\:\mathrm{continuous}\:\mathrm{function}\:{f}\left({x}\right), \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{take}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{between} \\ $$$${f}\left({x}\right)\:\mathrm{and}\:{f}\left({x}+\mathrm{1}\right),\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{average} \\ $$$$\mathrm{distance}\:\mathrm{within}\:{a}\leqslant{x}\leqslant{b}? \\ $$

Commented by Temp last updated on 05/Jul/16

let the distance be:  d=(√(∣f(x+1)−f(x)∣^2 +∣(x+1)−x∣^2 ))  d=(√(∣f(x+1)−f(x)∣^2 +1))

$$\mathrm{let}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{be}: \\ $$$${d}=\sqrt{\mid{f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)\mid^{\mathrm{2}} +\mid\left({x}+\mathrm{1}\right)−{x}\mid^{\mathrm{2}} } \\ $$$${d}=\sqrt{\mid{f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)\mid^{\mathrm{2}} +\mathrm{1}} \\ $$

Commented by Yozzii last updated on 05/Jul/16

d_(avg) =(1/(b−a))∫_a ^b d dx  d_(avg) =(1/(b−a))∫_a ^b (√((f(x+1)−f(x))^2 +1)) dx

$${d}_{{avg}} =\frac{\mathrm{1}}{{b}−{a}}\int_{{a}} ^{{b}} {d}\:{dx} \\ $$$${d}_{{avg}} =\frac{\mathrm{1}}{{b}−{a}}\int_{{a}} ^{{b}} \sqrt{\left({f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)\right)^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$$$ \\ $$

Commented by Temp last updated on 05/Jul/16

I thought so, thank you!

$$\mathrm{I}\:\mathrm{thought}\:\mathrm{so},\:\mathrm{thank}\:\mathrm{you}! \\ $$$$ \\ $$

Commented by Temp last updated on 05/Jul/16

is it (f(x+1)−f(x))^2   or  ∣f(x+1)−f(x)∣^2   ??

$$\mathrm{is}\:\mathrm{it}\:\left({f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)\right)^{\mathrm{2}} \\ $$$${or} \\ $$$$\mid{f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)\mid^{\mathrm{2}} \\ $$$$?? \\ $$

Commented by Yozzii last updated on 05/Jul/16

They are expressions equal in value  so either notation is appropriate.   I was taught that formula using parentheses.

$${They}\:{are}\:{expressions}\:{equal}\:{in}\:{value} \\ $$$${so}\:{either}\:{notation}\:{is}\:{appropriate}.\: \\ $$$${I}\:{was}\:{taught}\:{that}\:{formula}\:{using}\:{parentheses}. \\ $$

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