Difference between revisions of "Creating Walther MA271 Fall2020 topic14: Uses of Feynman's Technique" - Rhea

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-=Application of Feynman's Technique=
-Even though the main concept of this technique has been covered in the [[Walther MA271 Fall2020 topic14: What is Feynman's Technique | What is Feynman's Technique]] page, there are many extensions of this technique and its uses. One of these applications is brought up by "Mu Prime Math" in his video about Feynman's technique.
-The general idea of the method is to set an integral equal to the function of an arbitrary variable in order to differentiate with respect to that variable and receive an integral that is easier to integrate. However, instead of using Feynman’s to set F(a) equal to the integral in the problem and differentiating it to get an easier integral, we can set F(a) equal to an easier integral and differentiate it to get the integral in the problem.
-For example, let's take the integral in the video:
-<center><math> \int_{0}^{1}(\ln{x}) dx</math></center>
-While there are other methods of solving this integral, it is also possible to use Feynman's technique to get the answer.
-However, if we just try and use Feynman's technique normally (by setting a function F(a) equal to <math> \int_{0}^{1}(\ln{ax}) dx</math>) we get a relationship that isn't particularly easy to define. As was explained, Feynman's technique circles back to itself, where we need to be able to find at least one value of F(a), which doesn't seem easy given its current definition.
-In order to solve this, we need to apply Feynman's method in a unique way. Just like we said above, rather than defining our given integral as a function, we should set the function equal to another integral (one that is easier to integrate). In order to choose the correct integral, we should choose a function that, when derived, contains our desired value of <math> \int_{0}^{1}(\ln{x}) dx</math>
-Since we are deriving in terms of "a", the only functions that come to mind which result in <math> \ln{x} </math>, are <math> x^{a} </math> and <math> a*\ln{x} </math>
-The issue, however, is that we want a function that is easy to integrate. Because of this, we will use <math> x^{a} </math>
-To proceed, we integrate F(a) with respect to "a":
-<center><math> F(a) = \int_{0}^{1}(x^{a}) dx \\
-F'(a) = \int_{0}^{1}(x^{a}\ln{x}) dx </math></center>
-As we can see, the derivative F'(a) can be set equal to our original integral if we set "a" equal to 0. Thus, F'(0) should get us the answer to the integral <math> \int_{0}^{1}(\ln{x}) dx </math>
-With this information, we can calculate F(a) very easily by just evaluating the integral, hence why we chose the function that was easier to integrate.
-<center><math> F(a) = \int_{0}^{1}(x^{a}) dx \\
-                    = \frac{x^{a+1}}{a+1}\Big|_{0}^{1} \\
-                    = \frac{1}{a+1}</math></center>
-The last step to solve this integral would be to take the derivative of F(a) and then set "a" equal to 0. We can find very simply that <math> F'(a) = -\frac{1}{(a+1)^{2}} </math>
-From before, we know that <math> F'(a) = \int_{0}^{1}(x^{a}\ln{x}) dx</math>, which means that:
-<center><math> F'(0) = \int_{0}^{1}(\ln{x}) dx = -\frac{1}{0+1}</math></center>
-As a result, our final solution gives us the answer that <math> \int_{0}^{1}(\ln{x}) dx = -1 </math>
-[[ Walther MA271 Fall2020 topic14 | Back to Feynman Integrals]]
-[[Category:MA271Fall2020Walther]]

Difference between revisions of "Creating Walther MA271 Fall2020 topic14: Uses of Feynman's Technique" - Rhea

Latest revision as of 20:20, 5 December 2020

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