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Hi Ronak ... I have solved Your Physics questions , I found all of them very Interesting . I really much impressed from your Problem making skills . I really appreciate awesomeness of your Problems ! Great work .. !
I'am regret that I joined brilliant very late , My relative brother recommended this site to me recently ...
There are so many good questions of maths and physics , and I found that :: You and Jatin yadav and Deepanshu and David Mattingly and Josh silverman Posted so many self made questions .. I'am really amazed with your skills !

@Nishu Sharma
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I know it's pretty late to type this but thanks for the appreciation, I see you are from jaipur also, in which area of jaipur are you from, I am from jaipur too. @Nishu sharma

@Ronak Agarwal
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Hi , Actually My home is at jaipur , but I live in kota alone , near by to my aunt , My family lives in Jaipur. My home in backyard of Vidhan-sabha .. where do you live ?

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

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## Comments

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TopNewest@Brian Charlesworth @Kunal Joshi @Ronak Agarwal @Raghav Vaidyanathan @Calvin Lin

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@Nishu sharma

The answer is:

$\displaystyle \large \large e^{\frac{x^2}{4}}$

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Yes ! But I gave this as a Challange , So Everyone should post thier approaches .. Like karthik did !

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Well, it's very easy. And Raghav has given exactly the correct answer considering it to be very easy.

$\displaystyle {I}_{1} = \int_{0}^{x}{{e}^{-{z}^{2}+zx}}$

Completing the square,

$\displaystyle {I}_{1} = \int_{0}^{x}{{e}^{{x}^{2}/4}{e}^{-{z}^{2}+zx+{x}^{2}/4}}$

$\displaystyle {I}_{1} = {e}^{{x}^{2}/4}\int_{0}^{x}{{e}^{-{(z-x/2)}^{2}}}$

Using u-substitution and a little bashing up, we get

$\displaystyle {I}_{1} = {e}^{{x}^{2}/4}\int_{-x/2}^{x/2}{{e}^{-{u}^{2}}}$

And hence, $\displaystyle {I}_{1} = 2{e}^{{x}^{2}/4}\int_{0}^{x/2}{{e}^{-{u}^{2}}}$

By using error function now,

$\displaystyle {I}_{1} = {e}^{{x}^{2}/4}\sqrt{\pi}\text{erf}(x/2)$

Now, $\displaystyle {I}_{2} = \int_{0}^{x}{{e}^{-{(z/2)}^{2}}}$

Using v-substitution and some 'using of facts',

$\displaystyle {I}_{2} = 2\int_{0}^{x/2}{{e}^{-{v}^{2}}}$

$\displaystyle {I}_{2} = \sqrt{\pi}\text{erf}(x/2)$

Therefore, $\displaystyle \frac{{I}_{1}}{{I}_{2}} = {e}^{{x}^{2}/4}$

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Kartik there was no need to uneccasarily bring the error function into play here.

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Yes, you can convert one integral directly into the other without erf().

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Yeah, I agree. That was a mistake!

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Hi Ronak ... I have solved Your Physics questions , I found all of them very Interesting . I really much impressed from your Problem making skills . I really appreciate awesomeness of your Problems ! Great work .. ! I'am regret that I joined brilliant very late , My relative brother recommended this site to me recently ... There are so many good questions of maths and physics , and I found that :: You and Jatin yadav and Deepanshu and David Mattingly and Josh silverman Posted so many self made questions .. I'am really amazed with your skills !

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@Nishu sharma

I know it's pretty late to type this but thanks for the appreciation, I see you are from jaipur also, in which area of jaipur are you from, I am from jaipur too.Log in to reply

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Also, never miss out $dz$ inside integral. It is a bad habit.

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Yeah. But it is tedious. :P

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Nicely done , but there is more Smarter approach for this .. Think about it Kartik sharma

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Can anyone explain this https://brilliant.org/discussions/thread/extremely-weird-integration/?ref_id=946409

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