Maths, help me with it

Jintor

Jintor

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This isn't even my problem, but I can't freaking figure this out! Forgotton so much over the last three months.

Picture4-1.png


I have no idea what's going on anymore. Somebody explain it. Hopefully with pictures.
 
I'll give it a shot.

So, you know that

f(x) = 1 / x^2

And this also means that

f(x+h) = 1 / (x+h)^2

When we substitute in the original equation (you better write it out):

(f(x+h) - f(x)) / h
= ( (1/(x+h)^2) - (1/x^2) ) / h
= h/(x+h)^2 - h/x^2
= h/(x^2+2xh+h^2) - h/x^2

And to be able to easily bring together the two fractions, we multiply top and bottom of the first fraction with x^2

= (hx^2)/(x^2(x^2+2xh+h^2)) - h/x^2
= (hx^2)/(x^2(x^2+2xh+h^2)) - (h(x^2+2xh+h^2))/(x^2(x^2+2xh+h^2))
= ((hx^2) - (hx^2+2xh^2+h^3)))/(x^2(x^2+2xh+h^2))
= (hx^2-hx^2-2xh^2-h^3)/(x^2(x^2+2xh+h^2))
= (2xh^2+h^3)/(x^4+2x^3h+h^2x^2))
 
This is ridiculously easy though. It's only substituting and then simplifying the fraction.
 
Damn, I'm always late to math threads.
 
There are 3 kinds of people in this world, those who can count and those who can't.
 
Im worried that I was about to say 'what about the 3rd kind of person?'
 
Insano said:
I'll give it a shot.

So, you know that

f(x) = 1 / x^2

And this also means that

f(x+h) = 1 / (x+h)^2

When we substitute in the original equation (you better write it out):

(f(x+h) - f(x)) / h
= ( (1/(x+h)^2) - (1/x^2) ) / h
= h/(x+h)^2 - h/x^2 (That's not correct.)
= h/(x^2+2xh+h^2) - h/x^2

And to be able to easily bring together the two fractions, we multiply top and bottom of the first fraction with x^2

= (hx^2)/(x^2(x^2+2xh+h^2)) - h/x^2
= (hx^2)/(x^2(x^2+2xh+h^2)) - (h(x^2+2xh+h^2))/(x^2(x^2+2xh+h^2))
= ((hx^2) - (hx^2+2xh^2+h^3)))/(x^2(x^2+2xh+h^2))
= (hx^2-hx^2-2xh^2-h^3)/(x^2(x^2+2xh+h^2))
= (2xh^2+h^3)/(x^4+2x^3h+h^2x^2))
Click to expand...

[f(x+h) - f(x)] / h
=[(1/(x+h)^2) - (1/x^2)] / h
=[(1/(x+h)^2) - (1/x^2)] * 1/h
=1/h * [(1/(x+h)^2) - (1/x^2)]

1/h * {{(x^2)- [(x+h)^2]} / {x^2[(x+h)^2]}}
1/h * {{(x^2)- [x^2 + h^2 + 2xh]} / {x^2[(x+h)^2]}}
1/h * {(x^2 - x^2 - h^2 - 2xh)} / {x^2[(x+h)^2]}}
1/h * {( - h^2 - 2xh) / {x^2[(x+h)^2]}}
1/h * {- h(h+2x) / {x^2[(x+h)^2]}} (The h's cancel out.)
[-1(h+2x)] / {x^2[(x+h)^2]}
(-h-2x) / {x^2[(x+h)^2]}
(-h-2x) / {x^2 [x^2 + h^2 + 2xh]}
(-h-2x) / {x^4 + (x^2*h^2) + (2x^3*h)}
 
There are only 10 types of people in the world: Those who understand binary, and those who don't
 
**** math in the ass...

i forgot everything except adding and subtracting.
 
I should relearn calculus. You can do calculus on anything and get lots of interesting info.
 
vanillacrazycake said:
[f(x+h) - f(x)] / h
=[(1/(x+h)^2) - (1/x^2)] / h
=[(1/(x+h)^2) - (1/x^2)] * 1/h
=1/h * [(1/(x+h)^2) - (1/x^2)]

1/h * {{(x^2)- [(x+h)^2]} / {x^2[(x+h)^2]}}
1/h * {{(x^2)- [x^2 + h^2 + 2xh]} / {x^2[(x+h)^2]}}
1/h * {(x^2 - x^2 - h^2 - 2xh)} / {x^2[(x+h)^2]}}
1/h * {( - h^2 - 2xh) / {x^2[(x+h)^2]}}
1/h * {- h(h+2x) / {x^2[(x+h)^2]}} (The h's cancel out.)
[-1(h+2x)] / {x^2[(x+h)^2]}
(-h-2x) / {x^2[(x+h)^2]}
(-h-2x) / {x^2 [x^2 + h^2 + 2xh]}
(-h-2x) / {x^4 + (x^2*h^2) + (2x^3*h)}
Click to expand...
Aw, crap. You're right :O

I quickly noted these lines down this morning, my mistake. The irony is that I had to do it quickly, because I had a math lesson.
 
Ha-ah. Huh. Hmmm.

This makes a lot more sense.

The only thing I didn't do, apparently, was to take the 1/h outside the initial equasion, and to work with the numerator properly. : |

Thanks a bunch you guys.
 
Jintor said:
This isn't even my problem, but I can't freaking figure this out! Forgotton so much over the last three months.

Picture4-1.png


I have no idea what's going on anymore. Somebody explain it. Hopefully with pictures.
Click to expand...

Hrmmm, seems to me like this is the statement of the fundamental theorem of calculus. Although it is not the question you are asking, you can state that the whole thing approaches the derivative of 1/x^2 as h approaches 0. That would be -2/x^3. So without doing any real maths I can tell you that at the lim of h->0 your answer simplifies to -2/x^3. I am guessing that is the second part of your question.

For the actual maths part you can use this to check the solution (or you could use the solution to determine the derivative of f(x))
=(-2x-h)/(x^4+2hx^3+h^2*x^2)

You can see that this solution simplifies to -2/x^3 (the derivative of f(x)) when h approaches 0.
 
Dan said:
Hrmmm, seems to me like this is the statement of the fundamental theorem of calculus. Although it is not the question you are asking, you can state that the whole thing approaches the derivative of 1/x^2 as h approaches 0. That would be -2/x^3. So without doing any real maths I can tell you that at the lim of h->0 your answer simplifies to -2/x^3. I am guessing that is the second part of your question.

For the actual maths part you can use this to check the solution (or you could use the solution to determine the derivative of f(x))
=(-2x-h)/(x^4+2hx^3+h^2*x^2)

You can see that this solution simplifies to -2/x^3 (the derivative of f(x)) when h approaches 0.
Click to expand...
I love you dan
 
Just when I left school behind, it appears on my internets! Arrghh!
 
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