All you have to do is factor out sqrt[4x^2-1] into its base terms shown right there on the denominator and cancel those out, leaving only f(x)=4x. Plug in f(1), f(2), etc and you're done. It does seem kind of drawn out to go all the way to f(60) though...
All you have to do is factor out sqrt[4x^2-1] into its base terms shown right there on the denominator and cancel those out, leaving only f(x)=4x. Plug in f(1), f(2), etc and you're done. It does seem kind of drawn out to go all the way to f(60) though...
Hm. It's been a while since I used my math skills, and I even remember something similar, but
sqrt[4x^2-1] factors into sqrt[2x-1]*sqrt[2x+1], not sqrt[2x-1]+sqrt[2x+1], right? How did you cancel out (4x+AB)/(A+B) into 4x? I'm pretty sure the trick here has something to do with 4x being sqrt[2x-1]^2+sqrt[2x+1]^2, so the whole equation is (A^2+B^2+AB)/(A+B), but I can't quite put my finger on it. If we multiply both parts by (A-B), we get as far as (A^3-B^3)/2, but that doesn't really ease up the task of calculating f(1)+...+f(60). I always hated those kind of tasks, where you have to either take a lucky guess at how to solve it, or solve tons of similar tasks to memorize those tricks.
Hm. It's been a while since I used my math skills, and I even remember something similar, but
sqrt[4x^2-1] factors into sqrt[2x-1]*sqrt[2x+1], not sqrt[2x-1]+sqrt[2x+1], right? How did you cancel out (4x+AB)/(A+B) into 4x? I'm pretty sure the trick here has something to do with 4x being sqrt[2x-1]^2+sqrt[2x+1]^2, so the whole equation is (A^2+B^2+AB)/(A+B), but I can't quite put my finger on it. If we multiply both parts by (A-B), we get as far as (A^3-B^3)/2, but that doesn't really ease up the task of calculating f(1)+...+f(60). I always hated those kind of tasks, where you have to either take a lucky guess at how to solve it, or solve tons of similar tasks to memorize those tricks.
Ah, damn it.
Geh, you're right, forgot about the annoying + sign after 4x or this would've been too easy. Er... hmm. I tried squaring both sides and that didn't lead me anywhere; all I got was
I'd have to ask my old Calc professor if this is actually solvable, but considering how ridiculous it looks (up to f(60)? really?), the ridiculousness is probably part of the strip's joke only mathematicians would understand.
Are we solving for x? Since maybe we could just choose any random numbers between 3 and 60 which can replace all the x in the equation and see if it matches (and if there aren't any, lets just choose any random numbers aside between 3 and 60). So this make it solvable, but it's a freaking long task to solve it.
I'm just making a guess, my math skills are bit rusty anyway.
Are we solving for x? Since maybe we could just choose any random numbers between 3 and 60 which can replace all the x in the equation and see if it matches (and if there aren't any, lets just choose any random numbers aside between 3 and 60). So this make it solvable, but it's a freaking long task to solve it.
I'm just making a guess, my math skills are bit rusty anyway.
No, we're not solving for x. The board is not translated yet, but this seems to be a summation problem. As in, "add the results from F(1) to F(60)," or sigma of f(x), with i=1 and n=60 (http://en.wikipedia.org/wiki/Summation). Simplying f(x) would really help, which is what I was attempting to do, but it doesn't seem to be possible to in this case. Unless there's a Calculus pro lurking about here who says otherwise...
Anyway, the joke is that the teacher's being really mean making Homura add up to f(60), and with a problem not presented in her previous timelines.
Homu says she already knows the answer, so she's just playing around at not knowing to get Madoka to cheer for her.
You can't use calculus here, because calculus requires a continuous area. You can only solve the problem by just replacing X with all those different integers one after the other.
This problem is basically the worst possible thing to try and do with nothing but head math on a black board. It teaches nothing, really, even if you can do it. It's just stupidly long to have to add together all those different things even before you factor in the fact that it requires taking square roots of things nearly impossible to calculate by head math. (Numbers like square root of 13 being added to other square roots and then dividing out against another number with a square root.)
This is just plain the sort of thing you build computers to calculate for you. Stupid, repetitive math.
You can't use calculus here, because calculus requires a continuous area. You can only solve the problem by just replacing X with all those different integers one after the other.
Such problems generally imply to simplify f(x) to something trivial enough to do E[f(x)] manually. It's just that f(x) doesn't seem to be possible to simplify in this case.
Homu says she already knows the answer, so she's just playing around at not knowing to get Madoka to cheer for her.
You can't use calculus here, because calculus requires a continuous area. You can only solve the problem by just replacing X with all those different integers one after the other.
This problem is basically the worst possible thing to try and do with nothing but head math on a black board. It teaches nothing, really, even if you can do it. It's just stupidly long to have to add together all those different things even before you factor in the fact that it requires taking square roots of things nearly impossible to calculate by head math. (Numbers like square root of 13 being added to other square roots and then dividing out against another number with a square root.)
This is just plain the sort of thing you build computers to calculate for you. Stupid, repetitive math.
The answer is 886.65
So i did a quick program in C to calculate the sum of f(1) to f(60) as shown in the strip, and i got a different answer.(665.0). The code is as follows:
Show
#include <stdio.h>
#include <math.h>
#include <errno.h>
#include <fenv.h>
double funct (int x);
int main(void)
{
int x;
double y = 0.0;
for (x = 1; x < 61; x++)
y = y + funct(x);
printf("%f", y);
return 0;
}
double funct(int x){
return ((4*x+sqrt((4*(x*x))-1))/(sqrt((2*x)+1)+sqrt((2*x)-1)));
}
Homu says she already knows the answer, so she's just playing around at not knowing to get Madoka to cheer for her.
You can't use calculus here, because calculus requires a continuous area. You can only solve the problem by just replacing X with all those different integers one after the other.
This problem is basically the worst possible thing to try and do with nothing but head math on a black board. It teaches nothing, really, even if you can do it. It's just stupidly long to have to add together all those different things even before you factor in the fact that it requires taking square roots of things nearly impossible to calculate by head math. (Numbers like square root of 13 being added to other square roots and then dividing out against another number with a square root.)
This is just plain the sort of thing you build computers to calculate for you. Stupid, repetitive math.
The answer is 886.65
I'm late to the party, but I can't just let it slide that you guys think this is so impossible.
First of all, we reach the state that Type-kun described, i.e. (A^3-B^3)/2, where A=sqrt[2x+1] and B=sqrt[2x-1]. Now we note that when A is at (say) x=n, then it is equal to B at x=n+1, i.e. sqrt[2n+1] = sqrt[2(n+1)-1]. This then also means the same is true for A^3 and B^3, and thus each A^3 cancels with the B^3 in the next term. We have what is called a telescoping sum. Therefore, the answer is half of A^3 evaluated at x=60 minus B^3 evaluated at x=1, or (sqrt[2*60+1]^3-sqrt[2*1-1]^3)/2 =(sqrt[121]^3 - sqrt[1]^3)/2 =(11^3-1)/2 =(1331-1)/2 =1330/2 =665 exactly.
AS for whether this could be done on a blackboard, it's a lot of writing but not impossible I would say, and not tedious either.
TeacherTeacherAkemi, why don't you give this one a shot?I don't know how to solve this...Do your best!Is it really okay for me to solve the same problem... so many times...?I thought I'd let you try something different today.Homu?
