Your videos take me back to my college days. And I mean over 50 years back. Back then, I had to pass an advanced math exam to qualify for my PhD in Aerospace Engineering. Today, I can enjoy your videos just for fun. This video was definitely fun to watch!
I used the error function extensively during my engineering studies. As I recall the Erf table that I used was “coarse” and required a bit of interpolations. I ended up creating my own set of tables, first using a programmable calculator and then a computer (those were just starting to be widely used). My professor was delighted, and subsequently he distributed copies of my tables for his students until his retirement.
@@literallydeadpool The function was used for estimates of doping impurities ranges in PN junctions. As I recall the integrations were performed on a printing desk calculator (HP-97?)and the thermal printer strips were pasted to form a table and then xeroxed. Subsequently I repeated the process on a “microcomputer” using BASIC, and the results were better presented and printed. A decade or so later in the mid/late 1980s I stumbled upon my old results, and decided to repeat the solution on the much faster HP-28 calculator (the first of its kind to utilize symbolic maths) using its wireless printer.
I love your content! I thought I hated math in high school, but later in life I found I just didn’t learn it the right way. I love how explain things that are so complex (to me) in a simple way that enables me to understand.
My approach was to consider the area under the Gaussian in the window from 0 to 1. Then shift the Gaussian curve right by one unit (u=x-1) and shift it left by one unit (u=x+1), and consider the area in the 0-to-1 window. Shifting right picks up the mirror image of the original integral, so equal area. Shifting left moves the right-side’s ‘tail’ into the window, with clearly less area under the tail.
You can make your last argument more rigorous by comparing the values of the integrand function of I_3 and I_2 within the interval of integration [0, 1]: e^-((x+1)^2) is always strictly less than e^-(x^2) for all x in [0, 1]. Therefore, when integrating over this interval, I_3 must be less than I_2.
I studied the error function extensively during my graduate school days. Mostly, the problems involved transient heat transfer and the various solutions for the diffusion equation, (Fick's law).
This error function method looks a bit overkill to my eyes. In the range x = 0 to 1, functions f(x) = e^-x² and g(x) = e^-(x-1)² cover the exact same area! We are just integrating the area in opposite directions! However, for h(x) = e^-(x+1)² clearly h(x) ≤ f(x) in the range x = 0 to 1, so integrating it must give smaller area than for f(x) and g(x).
This is a great video about the error function But I think I have another way to solve the problem without using it. first let's look at (x -1) ^ 2 on [0, 1] by u substition it is equal to u^2 on [-1, 0] but square function is even so (-u)^2 = u^2 or u^2 on [-1,0] is equal to u^2 on [0,1] therefore (x-1) ^ 2 = x^2 on the segment [0,1] I1 = I2 for x on [0,1] x + 1 > x (x+1) ^2 > x^2 because x >= 0 -((x+1) ^2) < -(x^2) exp(-((x+1) ^2)) < exp( -(x^2) ) because exp(x) is an increasing function Integrating doesn't change the order, so I1 = I2 > I3
The first two integrals should be equal, and both should be greater than the last integral, right? Easy problem for solving in your head by linear substitution.
There are a good calculus practice! But IMHO the easiest way to show it is graph method. We’re using the same region in all integrals, but the graph of e^(-x^2) is shifted by -1, 0 and 1 as well.
Fun with erf(x) and I note this was pitched as "error function application" but total overkill for the question. Eight minutes in, you get around to showing by change of variables that the first and second integrals are equal since e^(-x^2) is even and you are integrating it -1 to 0 or 0 to 1 respectively - no need for all the flipping limits and signs. The actual value is irrelevant for the moment. Similarly after the corresponding change of variable for the third integral, comparing the integrals of e^(-x^2) from 0 to 1 versus 1 to 2 respectively is easy since e^(-x^2) is monotonically decreasing for x>0 so the first two integrals are greater than e^(-1) while the third one is less than that value (we are just integrating over unit distance in each case after all). All the appeal to the shape of the graph of erf(x) and differences of magnitudes of erf(2) and erf(1) is unnecessary embroidery and involves a rather a unsatisfactory appeal to the shape of the graph of erf(x) after all this analysis.
Boss, just put x=1 in all three integrals, and within micro seconds u will come to know which one is bigger. The question only asks us to arrange in ascending order which can be fulfilled easily. Nowadays exams provide only few seconds or hardly a minute per question.
Another way to do this (without Error function) is by expanding all exponents and noting that all three integrands are: (I) exp(-x^2) * exp(2x - 1) (II) exp(-x^2) (III) exp(-x^2) * exp(-2x-1) They are all positive, so integrals will be positive. Since 0
Your videos take me back to my college days. And I mean over 50 years back. Back then, I had to pass an advanced math exam to qualify for my PhD in Aerospace Engineering. Today, I can enjoy your videos just for fun. This video was definitely fun to watch!
I used the error function extensively during my engineering studies. As I recall the Erf table that I used was “coarse” and required a bit of interpolations. I ended up creating my own set of tables, first using a programmable calculator and then a computer (those were just starting to be widely used). My professor was delighted, and subsequently he distributed copies of my tables for his students until his retirement.
interesting, what tools did you use and how did you store the numbers
@@literallydeadpool The function was used for estimates of doping impurities ranges in PN junctions. As I recall the integrations were performed on a printing desk calculator (HP-97?)and the thermal printer strips were pasted to form a table and then xeroxed. Subsequently I repeated the process on a “microcomputer” using BASIC, and the results were better presented and printed. A decade or so later in the mid/late 1980s I stumbled upon my old results, and decided to repeat the solution on the much faster HP-28 calculator (the first of its kind to utilize symbolic maths) using its wireless printer.
I love your content! I thought I hated math in high school, but later in life I found I just didn’t learn it the right way.
I love how explain things that are so complex (to me) in a simple way that enables me to understand.
I am so happy that I found this channel. It helps me to solve problems I never even knew existed. Fantastic professor, and fantastic content. Thanks!
That's a beautifully succinct and intuitive demonstration of the error function!
My approach was to consider the area under the Gaussian in the window from 0 to 1.
Then shift the Gaussian curve right by one unit (u=x-1) and shift it left by one unit (u=x+1), and consider the area in the 0-to-1 window.
Shifting right picks up the mirror image of the original integral, so equal area.
Shifting left moves the right-side’s ‘tail’ into the window, with clearly less area under the tail.
You’re an excellent teacher.
You can make your last argument more rigorous by comparing the values of the integrand function of I_3 and I_2 within the interval of integration [0, 1]: e^-((x+1)^2) is always strictly less than e^-(x^2) for all x in [0, 1]. Therefore, when integrating over this interval, I_3 must be less than I_2.
I studied the error function extensively during my graduate school days. Mostly, the problems involved transient heat transfer and the various solutions for the diffusion equation, (Fick's law).
I learn stg new every single time i tune into sir's channel..
GREAT WORK OF A WELL TRAINED PROF.
This error function method looks a bit overkill to my eyes. In the range x = 0 to 1, functions f(x) = e^-x² and g(x) = e^-(x-1)² cover the exact same area! We are just integrating the area in opposite directions! However, for h(x) = e^-(x+1)² clearly h(x) ≤ f(x) in the range x = 0 to 1, so integrating it must give smaller area than for f(x) and g(x).
Indeed, the error function is simply a way to tidy up notation and describe exactly what you said
exactly, answer is instant
@pojuantsalo3475 - Oops - missed your comment before posting same point. Cheers.
@@tassiedevil2200 Don't worry about it. 🙂
Very thankful to your contents Sir!
You should be proud of yourself.
This is a great video about the error function
But I think I have another way to solve the problem without using it.
first let's look at
(x -1) ^ 2 on [0, 1]
by u substition it is equal to
u^2 on [-1, 0]
but square function is even so (-u)^2 = u^2
or u^2 on [-1,0] is equal to u^2 on [0,1]
therefore (x-1) ^ 2 = x^2 on the segment [0,1]
I1 = I2
for x on [0,1]
x + 1 > x
(x+1) ^2 > x^2 because x >= 0
-((x+1) ^2) < -(x^2)
exp(-((x+1) ^2)) < exp( -(x^2) ) because exp(x) is an increasing function
Integrating doesn't change the order, so
I1 = I2 > I3
The first two integrals should be equal, and both should be greater than the last integral, right? Easy problem for solving in your head by linear substitution.
There are a good calculus practice! But IMHO the easiest way to show it is graph method. We’re using the same region in all integrals, but the graph of e^(-x^2) is shifted by -1, 0 and 1 as well.
Great concept , Great Video
Fun with erf(x) and I note this was pitched as "error function application" but total overkill for the question. Eight minutes in, you get around to showing by change of variables that the first and second integrals are equal since e^(-x^2) is even and you are integrating it -1 to 0 or 0 to 1 respectively - no need for all the flipping limits and signs. The actual value is irrelevant for the moment. Similarly after the corresponding change of variable for the third integral, comparing the integrals of e^(-x^2) from 0 to 1 versus 1 to 2 respectively is easy since e^(-x^2) is monotonically decreasing for x>0 so the first two integrals are greater than e^(-1) while the third one is less than that value (we are just integrating over unit distance in each case after all). All the appeal to the shape of the graph of erf(x) and differences of magnitudes of erf(2) and erf(1) is unnecessary embroidery and involves a rather a unsatisfactory appeal to the shape of the graph of erf(x) after all this analysis.
Boss, just put x=1 in all three integrals, and within micro seconds u will come to know which one is bigger. The question only asks us to arrange in ascending order which can be fulfilled easily. Nowadays exams provide only few seconds or hardly a minute per question.
wearing my favourite cap in this video!!
Thanks
Thank you 😊
I think you can use differentuation to prove the curve is concave or convex at fhat range.
Another way to do this (without Error function) is by expanding all exponents and noting that all three integrands are:
(I) exp(-x^2) * exp(2x - 1)
(II) exp(-x^2)
(III) exp(-x^2) * exp(-2x-1)
They are all positive, so integrals will be positive. Since 0
Input
integral_0^1 e^(-(x - 1)^2) dx = integral_0^1 e^(-x^2) dx
Result
True
Can u solve it by interpolation?
what chalk do you use?
Can you please do a full course on proof writing and also real analysis? 😭
Integral_0^1 e^(-(x-1)^2) dx >Integral_0^1 e^(-(x+1)^2 dx
Cool👍
I
1>I
3
No changes in integral value because you are only shifting the function. 1 second answer
I
1=I
2
0-1=-1
Took me 15 seconds to
X^2+6x-4^x=0 x=2 but how
Lambert W function ?