- #1
Mandelbroth
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- 23
I'm in a high school pre-calculus class and a statistics class. For the latter, we are given z-tables for some of our tests. I don't like these z-tables.
Thus, I decided that a more direct approach (fundamental theorem of calculus) would be more accurate and, more importantly, more fun. My teacher kind of looked at the work and said "looks right to me". I think it's right (empirically it works and Wolfram says so), but I have no basis to see if I made a mistake or not (I'm basically just using my dad's old copy of Calculus and Analytic Geometry by Edwards and Penney and random resources on the internet).
To be clear, I'm looking for if I made a mistake in the derivation, not the answer. I tend to be good manipulating numbers in an erroneous way but still getting the right answer, but I want to make sure that I can do this correctly before I move on to something more difficult.
\int \frac{e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}}{\sigma \sqrt{2\pi}} dx = \frac{1}{\sigma \sqrt{2\pi}}\int e^{\frac{-z^2}{2}} dx
Letting z = (x-μ)/σ (fittingly), it follows that dz = \frac{dx}{\sigma} \Rightarrow dx = \sigma dz.
∴\frac{1}{\sigma \sqrt{2\pi}}\int e^{\frac{-z^2}{2}} dx = \frac{1}{\sqrt{2\pi}} \int e^{-\frac{z^2}{2}} dz = \frac{1}{\sqrt{2\pi}} \int \sqrt{e^{-z^2}} dz = \frac{1}{\sqrt{\pi}} \int_{0}^{z/\sqrt{2}} e^{-t^2} dt.
Finally, we get \displaystyle \frac{1}{\sqrt{\pi}} \int_{0}^{(\frac{x-\mu}{\sigma\sqrt{2}})} e^{-t^2} dt + C= \int \frac{e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}}{\sigma \sqrt{2\pi}} dx.
Thus, I decided that a more direct approach (fundamental theorem of calculus) would be more accurate and, more importantly, more fun. My teacher kind of looked at the work and said "looks right to me". I think it's right (empirically it works and Wolfram says so), but I have no basis to see if I made a mistake or not (I'm basically just using my dad's old copy of Calculus and Analytic Geometry by Edwards and Penney and random resources on the internet).
To be clear, I'm looking for if I made a mistake in the derivation, not the answer. I tend to be good manipulating numbers in an erroneous way but still getting the right answer, but I want to make sure that I can do this correctly before I move on to something more difficult.
\int \frac{e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}}{\sigma \sqrt{2\pi}} dx = \frac{1}{\sigma \sqrt{2\pi}}\int e^{\frac{-z^2}{2}} dx
Letting z = (x-μ)/σ (fittingly), it follows that dz = \frac{dx}{\sigma} \Rightarrow dx = \sigma dz.
∴\frac{1}{\sigma \sqrt{2\pi}}\int e^{\frac{-z^2}{2}} dx = \frac{1}{\sqrt{2\pi}} \int e^{-\frac{z^2}{2}} dz = \frac{1}{\sqrt{2\pi}} \int \sqrt{e^{-z^2}} dz = \frac{1}{\sqrt{\pi}} \int_{0}^{z/\sqrt{2}} e^{-t^2} dt.
Finally, we get \displaystyle \frac{1}{\sqrt{\pi}} \int_{0}^{(\frac{x-\mu}{\sigma\sqrt{2}})} e^{-t^2} dt + C= \int \frac{e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}}{\sigma \sqrt{2\pi}} dx.
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