Geometric visualizations of b^{e}<e^{b} when e<b
aa r X i v : . [ m a t h . HO ] J a n GEOMETRIC VISUALIZATIONS OF b e < e b WHEN e < b
BIKASH CHAKRABORTY, SAGAR CHAKRABORTY AND REZA FARHADIAN
Abstract.
The aim of this short note is to provide three geometric visualizations of a fascinating inequality π e < e π . Geometric visualizations of π e < e π In connection to the two fascinating constants e and π , there are many beautiful equations as well as inequations.For example, the most beautiful equation is e iπ + 1 = 0; on the other hand, one of the most beautiful inequality is π e < e π .The inequality π e < e π has numerous geometric visualizations in the literature. In 1987, Nakhli ([6]) gave abeautiful geometric visualization to the inequality π e < e π using the fact that the global maximum of the curve y = ln xx occurs at the point x = e . The same inequality was visualized by Nelsen ([7]) using the fact that the curve y = e xe lies above the line y = x . Recently, the first Author([1]) visualized the inequality π e < e π from the Napier’sinequality. Moreover, the first Author together with Mukherjee([8]) gave another nice visualization of π e < e π usingthe fact that the line y = x − y = ln x at the point (1 , e b > b e when e < b and a b > b a when e ≤ a ≤ b . Also, Mukherjee and Chakraborty ([5]) provided another geometric proofto the inequality a b > b a when e ≤ a ≤ b , but their proof need some calculations to visualize the inequality.In this note, We provide three alternative geometric visualization to the inequality π e < e π in more generalsettings. Moreover, one of our visualization improves the visualization of Mukherjee and Chakraborty ([5]).1. Proof using the fact that y = e x lies above the line y = x + 1 Theorem 1.
For a real number b with e < b , b e < e b . ( be − y = e x y = x + 1 i.e., be < e be − ⇒ b e < e b . Proof using the Area Argument-I
Theorem 2.
For a real number b with e < b , b e < e b . ln( be ) y = e x y = 1 i.e., ln (cid:18) be (cid:19) < Z ln( be ) x =0 e x dx = be − ⇒ b e < e b Proof using the Area Argument-II
Theorem 3.
For two real numbers a and b with e ≤ a < b , b a < a b . e a b y = ln xy = 1 < " ( b − a ) · ln b − Z ba ln x dx + [( b − a ) · ln a − ( b − a ) ·
1] = b ln a − a ln b ⇒ b a < a b . Corollary 1. π e < e π . References [1] Bikash Chakraborty, A Visual Proof that π e < e π , Math. Intelligencer, 41(1) (2019), pp. 56.[2] Charles D. Gallant, A B > B A for e ≤ A < B , Mathematics Magazine, 64(1) (1991), pp. 31.
EOMETRIC VISUALIZATIONS OF b e < e b WHEN e < b [3] Nazrul Haque, A Visual proof that e < A ⇒ e A > A e , Math. Intelligencer, 42(3), (2020), pp. 74.[4] Nazrul Haque, A Visual proof: e ≤ A ≤ B ⇒ A B > B A , Resonance, 26(1), (2021), pp. 127-128.[5] Rajib Mukherjee and Manishita Chakraborty, Beyond π e < e π : Proof without words of b a < a b ( b > a ≥ e ), Resonance, 25(11),(2020), pp. 1631-1632.[6] Fouad Nakhli, e π > π e , Mathematics Magazine, 60(3) (1987), pp. 165.[7] Roger B. Nelsen, Proof Without Words: Steiner’s Problem on the Number e , Mathematics Magazine, 82(2) (2009), pp. 102.[8] Ananda Mukherjee, and Bikash Chakraborty, Yet Another Visual Proof that π e < e π , Math Intelligencer, 41 (2), (2019), pp 60-60. Department of Mathematics, Ramakrishna Mission Vivekananda Centenary College, Rahara, West Bengal 700 118,India.
Email address : [email protected], [email protected] Department of Mathematics, Jadavpur University, Kolkata - 700032.
Email address : [email protected] Department of Statistics, Lorestan University, Khorramabad, Iran
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