Difference between revisions of "What is Feynman's Technique"

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~~=What is Feynman's Technique?=~~

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Feynman's Technique of integration utilizes parametrization and a mix of other different mathematical properties in order to integrate an integral that is can't be integrated through normal processes like u-substitution or integration by parts. It primarily focuses on setting a function equal to an integral, and then differentiating the function to get an integral that is easier to work with. A simple example would be an integral such as:

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~~<center><math> \int_{0}^{\infty}(e^{-x^2}*cos(2x)) dx</math></center>~~

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As we can see, there isn't any particular place that we can use u-substitution or integration by parts to produce a solution easily, but Feynman shows us how we can parameterize the integral as a function, focusing on the cosine factor of the integrand. By writing the integral as a function, we can change the expression to:

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~~<center><math> F(a) = \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx</math> (where a = 2)</center>~~

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This allows us to extract an x from the cosine segment of the integrand by differentiating with respect to a, making the left portion of the integrand <math>x*e^{-x^2}</math>, which is much easier to deal with than just <math>e^{-x^2}</math>

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~~From here, our differentiated equation is <math> F'(a) = \int_{0}^{\infty}(-x*e^{-x^2}*sin{(a*x)}) dx</math>, which we can then integrate using integration by parts.~~

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~~Doing so, however, would only get us:~~

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~~<center><math> \frac{sin{(a*x)}}{2e^{x^2}}\Big|_{0}^{\infty} - \frac{a}{2}\int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx</math></center>~~

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~~With this, we can see that the left side of the subtraction operation evaluates to 0, while the right side is just <math>-\frac{a}{2}F(a)</math>~~

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~~Thus, our result is <math>F'(a) = -\frac{a}{2}F(a)</math>~~

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Now, to proceed with the rest of the calculation, we need to express this equation in terms of differential fractions. By that, I mean writing it in the form <math> \frac{dF}{da} = (-\frac{a}{2})F</math> (source: blackpenredpen's explanation). With this, we can manipulate the equation to show <math>\frac{dF}{F} = -\frac{a}{2}da</math>

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After integrating both sides, we are left with <math>\ln{F} = -\frac{a^2}{4} + C</math>. By then considering ln as a logarithmic function of base "e", we can conclude that <math>F = e^{-\frac{a^2}{4}}*C_1</math>

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~~Finally, to solve for <math>C_1</math>, we substitute the original equation <math> F(a) = \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx</math> back into or new equation to get:~~

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~~<center><math>\int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx = e^{-\frac{a^2}{4}}*C_1</math></center>~~

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~~Since the function works for any number "a", we can choose the variable to be set as 0, giving us <math>\int_{0}^{\infty}(e^{-x^2}) dx = e^{-\frac{a^2}{4}}*C_1</math>~~

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Although we haven't covered it, this is the Gaussian integral, which, for the sake of this explanation, we will just take the value as <math>\frac{\sqrt{\pi}}{2}</math>(if you would like to know how this is calculated, there are probably solutions on YouTube or Google).

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~~Because of this, we now know that the value of <math>C_1</math> is the Gaussian integral, which gives us our last equation which we set equal to our original F(a):~~

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~~<center><math>F(a) = \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx = e^{-\frac{a^2}{4}}*\frac{\sqrt{\pi}}{2}</math></center>~~

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~~[[ Walther MA271 Fall2020 topic14 | Back to Feynman Integrals]]~~

Difference between revisions of "What is Feynman's Technique" - Rhea

Latest revision as of 20:35, 30 November 2020

Alumni Liaison

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-=What is Feynman's Technique?=
-Feynman's Technique of integration utilizes parametrization and a mix of other different mathematical properties in order to integrate an integral that is can't be integrated through normal processes like u-substitution or integration by parts. It primarily focuses on setting a function equal to an integral, and then differentiating the function to get an integral that is easier to work with. A simple example would be an integral such as:
-<center><math> \int_{0}^{\infty}(e^{-x^2}*cos(2x)) dx</math></center>
-As we can see, there isn't any particular place that we can use u-substitution or integration by parts to produce a solution easily, but Feynman shows us how we can parameterize the integral as a function, focusing on the cosine factor of the integrand. By writing the integral as a function, we can change the expression to:
-<center><math> F(a) = \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx</math> (where a = 2)</center>
-This allows us to extract an x from the cosine segment of the integrand by differentiating with respect to a, making the left portion of the integrand <math>x*e^{-x^2}</math>, which is much easier to deal with than just <math>e^{-x^2}</math>
-From here, our differentiated equation is <math> F'(a) = \int_{0}^{\infty}(-x*e^{-x^2}*sin{(a*x)}) dx</math>, which we can then integrate using integration by parts.
-Doing so, however, would only get us:
-<center><math> \frac{sin{(a*x)}}{2e^{x^2}}\Big|_{0}^{\infty} - \frac{a}{2}\int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx</math></center>
-With this, we can see that the left side of the subtraction operation evaluates to 0, while the right side is just <math>-\frac{a}{2}F(a)</math>
-Thus, our result is <math>F'(a) = -\frac{a}{2}F(a)</math>
-Now, to proceed with the rest of the calculation, we need to express this equation in terms of differential fractions. By that, I mean writing it in the form <math> \frac{dF}{da} = (-\frac{a}{2})F</math> (source: blackpenredpen's explanation). With this, we can manipulate the equation to show <math>\frac{dF}{F} = -\frac{a}{2}da</math>
-After integrating both sides, we are left with <math>\ln{F} = -\frac{a^2}{4} + C</math>. By then considering ln as a logarithmic function of base "e", we can conclude that <math>F = e^{-\frac{a^2}{4}}*C_1</math>
-Finally, to solve for <math>C_1</math>, we substitute the original equation <math> F(a) = \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx</math> back into or new equation to get:
-<center><math>\int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx = e^{-\frac{a^2}{4}}*C_1</math></center>
-Since the function works for any number "a", we can choose the variable to be set as 0, giving us <math>\int_{0}^{\infty}(e^{-x^2}) dx = e^{-\frac{a^2}{4}}*C_1</math>
-Although we haven't covered it, this is the Gaussian integral, which, for the sake of this explanation, we will just take the value as <math>\frac{\sqrt{\pi}}{2}</math>(if you would like to know how this is calculated, there are probably solutions on YouTube or Google).
-Because of this, we now know that the value of <math>C_1</math> is the Gaussian integral, which gives us our last equation which we set equal to our original F(a):
-<center><math>F(a) = \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx = e^{-\frac{a^2}{4}}*\frac{\sqrt{\pi}}{2}</math></center>
-[[ Walther MA271 Fall2020 topic14 | Back to Feynman Integrals]]