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Wednesday, November 28, 2007. E pi or pi e is greater. Lets consider e x and x e. When x is 0, e x = 1 and x e is 0. Now we know e x x e when x is 0. Lets find the mininum difference between the two. Now consider the equation. When the value of y is minimum, dy/dx = 0. D/dx (e x - x e) = 0. E x - e x (e-1) = 0. E x - e(x e)/x = 0. X(e x) - e(x e) = 0. Hence the minimum value of y is at x = e. The value of y at x=e is 0. Now that we proved e x - x e is positive at x = 0 and the minimum is 0 at x = e,.

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Wednesday, November 28, 2007. E pi or pi e is greater. Lets consider e x and x e. When x is 0, e x = 1 and x e is 0. Now we know e x x e when x is 0. Lets find the mininum difference between the two. Now consider the equation. When the value of y is minimum, dy/dx = 0. D/dx (e x - x e) = 0. E x - e x (e-1) = 0. E x - e(x e)/x = 0. X(e x) - e(x e) = 0. Hence the minimum value of y is at x = e. The value of y at x=e is 0. Now that we proved e x - x e is positive at x = 0 and the minimum is 0 at x = e,.

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Suriya | suriyaprakash.blogspot.com Reviews

https://suriyaprakash.blogspot.com

Wednesday, November 28, 2007. E pi or pi e is greater. Lets consider e x and x e. When x is 0, e x = 1 and x e is 0. Now we know e x x e when x is 0. Lets find the mininum difference between the two. Now consider the equation. When the value of y is minimum, dy/dx = 0. D/dx (e x - x e) = 0. E x - e x (e-1) = 0. E x - e(x e)/x = 0. X(e x) - e(x e) = 0. Hence the minimum value of y is at x = e. The value of y at x=e is 0. Now that we proved e x - x e is positive at x = 0 and the minimum is 0 at x = e,.

INTERNAL PAGES

suriyaprakash.blogspot.com

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Suriya: Calculating EMI or Mortgage

http://suriyaprakash.blogspot.com/2007/10/calculating-emi.html

Thursday, October 11, 2007. Calculating EMI or Mortgage. Lets use the following symbols. R - Rate of interest (For monthly installment with 6% APR, r is .06/12 = .005). Y - Number of installments (For 30 yrs mortgage y = 30 * 12 = 600). E - EMI or Mortgage or Monthly payment. P - Loan/Mortgage amount. In this blog I am going to show the mathematical derivation of each of the parameters (r, y, E, P) based on the known rest. Calculation of EMI E. Based on P, r and y. B1 = P - (E - (P * r). P (1 r) - E.

2

Suriya: November 2007

http://suriyaprakash.blogspot.com/2007_11_01_archive.html

Wednesday, November 28, 2007. E pi or pi e is greater. Lets consider e x and x e. When x is 0, e x = 1 and x e is 0. Now we know e x x e when x is 0. Lets find the mininum difference between the two. Now consider the equation. When the value of y is minimum, dy/dx = 0. D/dx (e x - x e) = 0. E x - e x (e-1) = 0. E x - e(x e)/x = 0. X(e x) - e(x e) = 0. Hence the minimum value of y is at x = e. The value of y at x=e is 0. Now that we proved e x - x e is positive at x = 0 and the minimum is 0 at x = e,.

3

Suriya: October 2007

http://suriyaprakash.blogspot.com/2007_10_01_archive.html

Thursday, October 11, 2007. Calculating EMI or Mortgage. Lets use the following symbols. R - Rate of interest (For monthly installment with 6% APR, r is .06/12 = .005). Y - Number of installments (For 30 yrs mortgage y = 30 * 12 = 600). E - EMI or Mortgage or Monthly payment. P - Loan/Mortgage amount. In this blog I am going to show the mathematical derivation of each of the parameters (r, y, E, P) based on the known rest. Calculation of EMI E. Based on P, r and y. B1 = P - (E - (P * r). P (1 r) - E.

4

Suriya: e^pi or pi^e is greater.

http://suriyaprakash.blogspot.com/2007/11/epi-or-pie-is-greater.html

Wednesday, November 28, 2007. E pi or pi e is greater. Lets consider e x and x e. When x is 0, e x = 1 and x e is 0. Now we know e x x e when x is 0. Lets find the mininum difference between the two. Now consider the equation. When the value of y is minimum, dy/dx = 0. D/dx (e x - x e) = 0. E x - e x (e-1) = 0. E x - e(x e)/x = 0. X(e x) - e(x e) = 0. Hence the minimum value of y is at x = e. The value of y at x=e is 0. Now that we proved e x - x e is positive at x = 0 and the minimum is 0 at x = e,.

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Wednesday, November 28, 2007. E pi or pi e is greater. Lets consider e x and x e. When x is 0, e x = 1 and x e is 0. Now we know e x x e when x is 0. Lets find the mininum difference between the two. Now consider the equation. When the value of y is minimum, dy/dx = 0. D/dx (e x - x e) = 0. E x - e x (e-1) = 0. E x - e(x e)/x = 0. X(e x) - e(x e) = 0. Hence the minimum value of y is at x = e. The value of y at x=e is 0. Now that we proved e x - x e is positive at x = 0 and the minimum is 0 at x = e,.

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