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Difference between revisions of "User:Mathfantasia"

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We wish to find <math>\int^2_1 x^2dx</math>.  Recall that the formula for integration is <math>\int^b_a f(x)dx = \sum_{i=1}^n (height)(width) = \sum_{i=1}^n f(x_i)((b-a)/2)</math>.  So <math>x_i = a + \frac{(b-a)i}{n} = 1 + \frac{i}{n}</math>.  Width = <math> \frac{(b-a)i}{n} = \frac{i}{n}</math>.  <math>f(x_i) = (1 + \frac{i}{n})^2 = 1 + \frac{2i}{n} +  
 
We wish to find <math>\int^2_1 x^2dx</math>.  Recall that the formula for integration is <math>\int^b_a f(x)dx = \sum_{i=1}^n (height)(width) = \sum_{i=1}^n f(x_i)((b-a)/2)</math>.  So <math>x_i = a + \frac{(b-a)i}{n} = 1 + \frac{i}{n}</math>.  Width = <math> \frac{(b-a)i}{n} = \frac{i}{n}</math>.  <math>f(x_i) = (1 + \frac{i}{n})^2 = 1 + \frac{2i}{n} +  
−
\frac{i^2}{n^2}</math> .  The integral = <math>(\sum_{i=1}^n (1 + \frac{i}{n})^2 = 1 + \frac{2i}{n} + \frac{i^2}{n^2})\frac{1}{n} = \sum_{i=1}^n (1 + \frac{i}{n^2})^2 = 1 + \frac{2i}{n^2} + \frac{i^2}{n^3}) = \frac{1}{n}\sum_{i=1}^n 1 + \frac{2}{n^2}\sum_{i=1}^n + i \frac{1}{n^3}\sum_{i=1}^n i^2  </math>
+
\frac{i^2}{n^2}</math> .  The integral = <math>(\sum_{i=1}^n (1 + \frac{i}{n})^2)\frac{1}{n} = (1 + \frac{2i}{n} + \frac{i^2}{n^2})\frac{1}{n} = \sum_{i=1}^n (1 + \frac{i}{n^2})^2 = 1 + \frac{2i}{n^2} + \frac{i^2}{n^3}) = \frac{1}{n}\sum_{i=1}^n 1 + \frac{2}{n^2}\sum_{i=1}^n + i \frac{1}{n^3}\sum_{i=1}^n i^2  </math>

Revision as of 16:23, 16 August 2011

My Contributions:

Divisibility rules: for 7

Rule 3: "Tail-End divisibility." Note. This only tells you if it is divisible and NOT the remainder. Take a number say 12345. Look at the last digit and add or subtract a multiple of 7 to make it zero. In this case we get 12370 or 12310. Lop off the ending 0's and repeat. 1237 ==> 123 ==> 130 ==> 13 NOPE. Works in general with numbers that are relatively prime to the base (and works GREAT in binary).


Contributions On This Page Only:

Estimating Compound Interest

Look up and understand the "Rule of 72." My short synopsis here says that certain small percentages, say n%, will double in "72 divided by n" years.

I will now estimate $2^x$ between x=0 and x=1 by forming segments between the previous x values and x=.5. So for x<.5 we use 1 + .8x and otherwise 1.2x + .8.

I will now give an example of this method!

6% compounded yearly for 4 years with principal 20000. We see that 6% interest will make the principal double every 12 years (by the law of 72). But we only need 4 years, so with no more estimation we would need to know $2^{1/3}$. We estimate this with the above linear equations. So we want to multiply the principal, 20000, by ($1 + .8/3$). We end up getting 25336. And there ya go! Actual answer: 25250 (less than 1% off ... actually .3%)

LOL needed a place to latex on the go. Putting up some calculus notes.

We wish to find $\int^2_1 x^2dx$. Recall that the formula for integration is $\int^b_a f(x)dx = \sum_{i=1}^n (height)(width) = \sum_{i=1}^n f(x_i)((b-a)/2)$. So $x_i = a + \frac{(b-a)i}{n} = 1 + \frac{i}{n}$. Width = $\frac{(b-a)i}{n} = \frac{i}{n}$. $f(x_i) = (1 + \frac{i}{n})^2 = 1 + \frac{2i}{n} + \frac{i^2}{n^2}$ . The integral = $(\sum_{i=1}^n (1 + \frac{i}{n})^2)\frac{1}{n} = (1 + \frac{2i}{n} + \frac{i^2}{n^2})\frac{1}{n} = \sum_{i=1}^n (1 + \frac{i}{n^2})^2 = 1 + \frac{2i}{n^2} + \frac{i^2}{n^3}) = \frac{1}{n}\sum_{i=1}^n 1 + \frac{2}{n^2}\sum_{i=1}^n + i \frac{1}{n^3}\sum_{i=1}^n i^2$

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