Root of 2 and area

  • Thread starter Monocles
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Main Question or Discussion Point

Ok, so doing a physics problem the other day I had to find a point c on a graph such that there was an equal amount of area between 0 and c and c and a point b > c. I found that for the graph f(x) = kx, the point was b over the square root of 2. This intrigued me so I looked for a generalization. Does this look correct? I'm still a low level math student (only in Calculus 2) so I don't know anything about the correct way to solve this stuff. Is this a proof as well? If not, how would I prove it?

[tex]f(x) = kx^n[/tex]


[tex]\int_{0}^{c} kx^n dx = \int_{c}^{b} kx^n dx[/tex]


[tex]k\int_{0}^{c} x^n dx = k\int_{c}^{b} x^n dx[/tex]


[tex]\left[ \frac{x^{n+1}}{n+1} \right]_{0}^{c} = \left[ \frac{x^{n+1}}{n+1} \right]_{c}^{b}[/tex]


[tex]c^{n+1} = b^{n+1} - c^{n+1}[/tex]


[tex]2c^{n+1} = b^{n+1}[/tex]


[tex]c = \frac{b}{\sqrt[n+1]{2}}[/tex]


thanks!
 

Answers and Replies

Seems correct to me. Explicit calculation is one way of proving things, so this can be considered a proof (if you supplement it with some text and a statement of what you have actually proven, assumptions ...) For example, must your "n" be a natural number, can it be negative, rational, complex..?
 
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OK, I decided to try and generalize it a bit more, but I hit a wall.

Maybe this should go in homework help section at this point? I don't know.

But anyways, I tried to generalize the same thing as the above, but with:


[tex]\sum_{i=1}^n k_i x^i[/tex]


I won't bother writing out all my work, but here is what I come up with:


[tex]\sum_{i=1}^n c^{i+1} = \frac{1}{2} \sum_{i=1}^n b^{i+1}[/tex]

Is there any way to simplify this further? It doesn't seem very useful in this form... (keep in mind I'm only in Calc 2 so if it would require Real Analysis or something like that its over my head at the moment)
 
Assuming this is correct ( I doubt it, what did you do with the factors [itex]\frac{1}{i+1}[/itex] coming from the integration of [itex]x^i[/itex]...? How did you cancel the k's...?), have you ever heard the term "geometric series"?
 
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Oops yeah you're right, I messed up getting rid of k.

Here is what I had right before that:


[tex]2 \left( \sum_{i=1}^{n} \frac{k_i c^{i+1}}{i+1} \right) = \sum_{i=1}^{n} \frac{k_i b^{i+1}}{i+1}[/tex]


Which I turned (incorrectly) into:


[tex]2 \left( \sum_{i=1}^{n} \frac{k_i}{i+1} + \sum_{i=1}^{n} c^{i+1} \right) = \sum_{i=1}^{n} \frac{k_i b^{i+1}}{i+1}[/tex]


I don't think I would be allowed to do this, would I? If I can, I can still eliminate k, but expanding those series makes it look like thats wrong too.


[tex]2 \left( \sum_{i=1}^{n} \frac{k_i}{i+1} \times \sum_{i=1}^{n} c^{i+1} \right) = \sum_{i=1}^{n} \frac{k_i b^{i+1}}{i+1}[/tex]


Edit: Yes I know about geometric series, I am having trouble seeing the connection to this though. I am getting pretty sleepy, though.
 
[tex]2 \left( \sum_{i=1}^{n} \frac{k_i c^{i+1}}{i+1} \right) = \sum_{i=1}^{n} \frac{k_i b^{i+1}}{i+1}[/tex]


Which I turned (incorrectly) into:


[tex]2 \left( \sum_{i=1}^{n} \frac{k_i}{i+1} + \sum_{i=1}^{n} c^{i+1} \right) = \sum_{i=1}^{n} \frac{k_i b^{i+1}}{i+1}[/tex]


I don't think I would be allowed to do this, would I?
Indeed, this is wrong.
Consider the case n=2:
[tex]\frac{k_1c^2}{2}+\frac{k_2c^3}{3}\neq\frac{k_1}{2}+\frac{k_2}{3}+c^2+c^3[/tex]

Do you any reason to belive this...?

[tex]
\sum_{i=1}^n{c^i}
[/tex]

by definition IS a geometric series. So if you're familar with the term and know in principle how to find a more explicit expression for them, where is your problem in applying it to this sum?
 

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