hcju/proofwikiqa huggingface dataset:
The proof proceeds by induction on $n$, where $n$ is the degree of the polynomial. === Basis for the Induction === When $n = 0$, the mapping is a constant mapping. By Constant Function is of Exponential Order Zero and Raising Exponential Order, the mapping is of exponential order $\epsilon$. This is the basis for the induction. === Induction Hypothesis === Fix $k \in \N$ with $k \ge 0$. Assume: :$t^k \in \mathcal E_\epsilon$ That is,  :$\forall t \ge M : \size {t^k} < K e^{\epsilon t}$ For some $M > 0$, $K > 0$, and for any $\epsilon > 0$ arbitrarily small. This is our induction hypothesis. === Induction Step === We have: : $x^k$ is of exponential order epsilon by the induction hypothesis : Identity is of Exponential Order Epsilon : Product of Functions of Exponential Order Therefore, $x^{k + 1}$ is also of exponential order epsilon. The result follows by the Principle of Mathematical Induction. {{qed}} Category:Exponential Order k2fx965jtou6vvzxjm2d5a7hzvw8m14
From the definition of exponential order, there exist strictly positive real numbers $M$ and $K$ such that: :$\forall t \ge M: \size {\map f t} < K e^{a t}$ From Exponential is Strictly Increasing, we have: :$K e^{a t} < K e^{b t}$ Therefore: :$\forall t \ge M: \size {\map f t} < K e^{b t}$ The result follows from the definition of exponential order. {{qed}} Category:Exponential Order a3cg99grb6mk3m8fwjljixt9ee4q53j
From Gamma Function is Continuous on Positive Reals, $f$ is continuous for $t \ge 0$. Set $t > 0$. From Stirling's Formula: :$\displaystyle t! \sim \sqrt {2 \pi t} \left({\frac t e}\right)^t$ where $\sim$ denotes asymptotic equality. That is,  {{begin-eqn}} {{eqn | l = t!       | o = \sim       | r = \sqrt {2 \pi t} \left({\frac t e}\right)^t }} {{eqn | o = >       | r = t^t \, e^{-t} }} {{end-eqn}} Seeking a contradiction, assume $t^t \, e^{-t}$ is of exponential order $a$.  Then, for $t$ sufficiently large, there exists a $K>0$ such that: {{begin-eqn}} {{eqn | l =  t^t \, e^{-t}       | o = <        | r = K e^{at} }} {{eqn | ll = \impliedby       | l = t^t       | o = <       | r = K e^{\left({a+1}\right)t} }} {{end-eqn}} This implies that $t^t$ is of exponential order, which is false. From this contradiction it follows that $t!$ is also not of exponential order. {{qed}} Category:Gamma Function Category:Exponential Order 86kidv62y1m6yyivarqkxpg05e7y4ls
=== Lemma === {{:X to the x is not of Exponential Order/Lemma}}{{qed|lemma}} By the definition of power: :$\map f t = \map \exp {t \ln t}$  The theorem is equivalent to that there do not exist strictly positive real constants $M$, $K$, $a$ such that: :$\forall t \ge M: \size {\map f t} < K e^{a t}$ {{AimForCont}} such constants $M$, $K$ and $a$ exist. From the lemma, there exists a constant $C$ such that: :$\forall t > C: \size {\map f t} > K e^{a t}$ For any $M$ chosen, $M + C > C$, thus from the lemma: :$\size {\map f {M + C} } > K e^{a t}$ However, $M + C > M$, thus from the assumption: :$\size {\map f {M + C} } < K e^{a t}$ Thus a contradiction has arisen. It follows by Proof by Contradiction that our assumption was false. Hence the result. {{qed}}
By Empty Set is Subset of All Sets: :$\forall S \in \powerset X: \O \subseteq S$ By Empty Set is Element of Power Set: :$\O \in \powerset X$ Thus by definition of the smallest element: :$\O = \bot$ {{qed}}
{{begin-eqn}} {{eqn | l = \lim_{x \mathop \to 0^+} x^x       | r = \lim_{x \mathop \to 0^+} \exp \left({x \ln x}\right)        | c = definition of power }} {{eqn | r = \exp \left({\lim_{x \mathop \to 0^+}x \ln x}\right)        | c = Exponential Function is Continuous }} {{eqn | r = \exp \left({\lim_{x \mathop \to 0^+}\frac{\ln x} {\frac 1 x} } \right)  }} {{eqn | r = \exp \left({\lim_{x \mathop \to 0^+}\frac{\frac 1 x} {\frac {-1} {x^2} } } \right)        | c = L'Hôpital's Rule }} {{eqn | r = \exp \left({\lim_{x \mathop \to 0^+}\frac {-x^2} x } \right)  }} {{eqn | r = \exp 0 }} {{eqn | r = 1       | c = Exponential of Zero }} {{end-eqn}} {{qed}} Category:Limits of Functions d3ezyfqkokyp48hiq97gx2m67hxzv27
This follows trivially from Hermite-Lindemann-Weierstrass Theorem by taking $n = 1$. {{qed}} {{Namedfor|Charles Hermite|name2 = Carl Louis Ferdinand von Lindemann|name3 = Karl Theodor Wilhelm Weierstrass|cat = Hermite|cat2 = Lindemann|cat3 = Weierstrass}} Category:Transcendental Number Theory oe1v9bdjojaoblepzd3sz6role0rg5y
Assume the truth of Schanuel's Conjecture. From Schanuel's Conjecture Implies Algebraic Independence of Pi and Log of Pi over the Rationals, $\ln \pi$ and $\pi$ are algebraically independent over the rational numbers $\Q$. Therefore, if Schanuel's Conjecture holds, $\ln \pi$ must be transcendental. {{qed}} Category:Transcendental Numbers Category:Pi Category:Logarithms Category:Schanuel's Conjecture mrrbpogcvdpyxo7t1dygcbazwaircyy
Assume the truth of Schanuel's Conjecture. By Schanuel's Conjecture Implies Algebraic Independence of Pi and Euler's Number over the Rationals, $\pi$ and $e$ are algebraically independent over the rational numbers $\Q$. That is, no non-trivial polynomials $\map f {x, y}$ with rational coefficients satisfy: :$\map f {\pi, e} = 0$ {{AimForCont}} $\pi \times e$ is algebraic. Then there would be a non-trivial polynomial $\map g z$ with rational coefficients satisfying: :$\map g {\pi \times e} = 0$ However, $\map f {x, y} := \map g {x \times y}$ would be a non-trivial polynomial with rational coefficients satisfying: :$\map f {\pi, e} = 0$ which contradicts the earlier statement that no such polynomials exist. Therefore, if Schanuel's Conjecture holds, $\pi \times e$ is transcendental. {{qed}} Category:Transcendental Numbers Category:Pi Category:Euler's Number Category:Schanuel's Conjecture mvuajogd9vg72l5c5ibnbbdt67oty2s
Assume the truth of Schanuel's Conjecture. By Schanuel's Conjecture Implies Algebraic Independence of Pi and Euler's Number over the Rationals, $\pi$ and $e$ are algebraically independent over the rational numbers $\Q$. That is, no non-trivial polynomials $\map f {x, y}$ with rational coefficients satisfy: :$\map f {\pi, e} = 0$ {{AimForCont}} $\pi + e$ is algebraic. Then there would be a non-trivial polynomial $\map g z$ with rational coefficients satisfying: :$\map g {\pi + e} = 0$ However, $\map f {x, y} := \map g {x + y}$ would be a non-trivial polynomial with rational coefficients satisfying: :$\map f {\pi, e} = 0$ which contradicts the earlier statement that no such polynomials exist. Therefore, if Schanuel's Conjecture holds, $\pi + e$ is transcendental. {{qed}} Category:Transcendental Numbers Category:Pi Category:Euler's Number Category:Schanuel's Conjecture aowdcvwj0okwfdcove95p30qj98qlre
Proof by Contradiction: {{AimForCont}} $a + b$ and $a \times b$ are both not transcendental. Hence by definition, they are both algebraic. Hence, $\left({z - a}\right) \left({z - b}\right)$ is a polynomial with algebraic coefficients. Therefore, $a$ and $b$ must both be algebraic. However, this contradicts with the assumption that $a$ and $b$ are both transcendental. Hence by Proof by Contradiction it must follow that at least one of $a + b$ and $a \times b$ is transcendental. {{qed}} Category:Transcendental Numbers q034nv8hp89erw2p8k2vyzxhde00y19
Assume the truth of Schanuel's Conjecture. Let $z_1 = 1$ and $z_2 = i \pi$. Note that $z_1$ is wholly real and $z_2$ is wholly imaginary. Hence, by Wholly Real Number and Wholly Imaginary Number are Linearly Independent over the Rationals, they are linearly independent over $\Q$. By Schanuel's Conjecture, the extension field $\map \Q {z_1, z_2, e^{z_1}, e^{z_2} }$ has transcendence degree at least $2$ over $\Q$. That is, the extension field $\map \Q {1, i \pi, e, -1}$ has transcendence degree at least $2$ over $\Q$. However, $1$ and $-1$ are algebraic. Therefore $i \pi$ and $e$ must be algebraically independent over $\Q$. {{AimForCont}} $\pi$ and $e$ are not algebraically independent over $\Q$. Then, then there would be a non-trivial polynomial $\map g {x, y}$ with rational coefficients satisfying: :$\map g {\pi, e} = 0$ Then, one can construct a non-trivial polynomial $\map f {x, y} = \map g {i x, y} \map g {-i x, y}$ with rational coefficients satisfying: :$\map f {i \pi, e} = 0$ which is contradictory to the previous statement that no such polynomials exist. Therefore, if Schanuel's Conjecture is true, then $\pi$ and $e$ are algebraically independent over $\Q$. {{qed}} Category:Transcendental Numbers Category:Pi Category:Euler's Number Category:Schanuel's Conjecture bca5qabvbylpy9qfnt7qhdjk2dw6izw
By definition of surjection: :$f: X \to f \sqbrk X$ is a surjection. The case when $X \ne \O$: By Surjection iff Cardinal Inequality: :$\card {f \sqbrk X} \le \card X$ Thus by Set of Cardinality not Greater than Cardinality of Finite Set is Finite: :$f \sqbrk X$ is finite. {{qed|lemma}} The case when $X = \O$: By the corollary to Image of Empty Set is Empty Set: :$f \sqbrk X = \O$ Thus the result holds. {{qed}}
Let $M_3$ denote the order $3$ magic square By Sum of Terms of Magic Square, the total of all the entries in $M_3$ is given by: :$T_3 = \dfrac {3^2 \left({3^2 + 1}\right)} 2 = \dfrac {9 \times 10} 2 = 45$ As there are $3$ rows of $M_3$, the magic constant of $M_3$ is given by: :$S_3 = \dfrac {45} 3 = 15$ {{qed}}
Let $M_n$ denote the magic square of order $n$. By Magic Constant of Magic Square, the magic constant of $M_n$ is given by: :$S_n = \dfrac {n \left({n^2 + 1}\right)} 2$ Setting $n = 3$: :$S_3 = \dfrac {3 \times 10} 2 = 15$ {{qed}}
Let $M_n$ denote a magic square of order $n$. $M_n$ is by definition a square matrix of order $n$ containing the positive integers from $1$ upwards. Thus there are $n^2$ entries in $M_n$, going from $1$ to $n^2$. Thus: {{begin-eqn}} {{eqn | l = T_n       | r = \sum_{k \mathop = 1}^{n^2} k       | c =  }} {{eqn | r = \frac {n^2 \paren {n^2 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{qed}} === Sequence === {{:Sum of Terms of Magic Square/Sequence}} Category:Magic Squares qhd8dfr9zzpr71zt75jgci7eitv0ipv
Assume the truth of Schanuel's Conjecture. Let $z_1 = \ln \ln 2$, $z_2 = 1 + \ln \ln 2$, $z_3 = \ln 2$, and $z_4 = e \ln 2$. By Lemma, they are linearly independent over the rational numbers $\Q$. Observe that $z_3 = e^{z_1}$ and $z_4 = e^{z_2}$. By Schanuel's Conjecture, the extension field $\map \Q {z_1, z_2, z_3, z_4, e^{z_1}, e^{z_2}, e^{z_3}, e^{z_4} }$ has transcendence degree at least $4$ over the rational numbers $\Q$. That is, the extension field $\map \Q {\ln \ln 2, 1 + \ln \ln 2, \ln 2, e \ln 2, \ln 2, e \ln 2, 2, 2^e}$ has transcendence degree at least $4$ over $\Q$. However, $2$ is algebraic. Also, $\ln \ln 2$ and $1 + \ln \ln 2$ are not algebraically independent over $\Q$. Therefore, $\ln \ln 2$, $\ln 2$, $e \ln 2$, and $2^e$ must all be transcendental. Therefore, if Schanuel's Conjecture holds, then $2^e$ is transcendental. {{qed}} Category:Transcendental Numbers Category:2 Category:Euler's Number Category:Schanuel's Conjecture p4yeslr5n0d1ccazm9ubdxffvot025o
Assume the truth of Schanuel's Conjecture. Let $z_1 = \ln \ln \pi$, $z_2 = 1 + \ln \ln \pi$, $z_3 = \ln \pi$, $z_4 = e \ln \pi$, and $z_5 = i \pi$. By Lemma, they are linearly independent over the rational numbers $\Q$. Observe that $z_3 = e^{z_1}$ and $z_4 = e^{z_2}$. By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_2, z_3, z_4, z_5, e^{z_1}, e^{z_2}, e^{z_3}, e^{z_4}, e^{z_5}}\right)$ has transcendence degree at least $5$ over the rational numbers $\Q$. That is, the extension field $\Q \left({\ln \ln \pi, 1 + \ln \ln \pi, \ln \pi, e \ln \pi, i \pi, \ln \pi, e \ln \pi, \pi, \pi^e, e^{i \pi}}\right)$ has transcendence degree at least $5$ over $\Q$. However, by Euler's Identity, $e^{i \pi} = -1$ is algebraic. Also, $\ln \ln \pi$ and $1 + \ln \ln \pi$ are not algebraically independent over $\Q$. Also, $\pi$ and $i \pi$ are not algebraically independent over $\Q$. Therefore, $\ln \ln \pi$, $\ln \pi$, $e \ln \pi$, $\pi$, and $\pi^e$ must all be transcendental. Therefore, if Schanuel's Conjecture holds, then $\pi^e$ is transcendental. {{qed}} Category:Transcendental Numbers Category:Pi Category:Euler's Number Category:Schanuel's Conjecture kn6giozv3jtiyvbe6njft4bf8fpiznk
Assume the truth of Schanuel's Conjecture. Let $z_1 = 1$, $z_2 = e$. By Euler's Number is Irrational, $z_1$ and $z_2$ are linearly independent over $\Q$. By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_2, e^{z_1}, e^{z_2}}\right)$ has transcendence degree at least $2$ over $\Q$. That is, the extension field $\Q \left({1, e, e, e^e}\right)$ has transcendence degree at least $2$ over $\Q$. However, $1$ is algebraic. Therefore, if Schanuel's Conjecture holds, $e^e$ must be transcendental. {{qed}} Category:Transcendental Numbers Category:Euler's Number Category:Schanuel's Conjecture o4jst2u7k117zspwek0wwym8z4aolfw
Let $M_n$ denote a magic square of order $n$. By Sum of Terms of Magic Square, the total of all the entries in a magic square of order $n$ is given by: :$T_n = \dfrac {n^2 \paren {n^2 + 1} } 2$ There are $n$ rows in $M_n$, each one with the same magic constant. Thus the magic constant $S_n$ of the magic square $M_n$ is given by: {{begin-eqn}} {{eqn | l = S_n       | r = \dfrac {T_n} n       | c =  }} {{eqn | r = \dfrac {n^2 \paren {n^2 + 1} } {2 n}       | c = Sum of Terms of Magic Square }} {{eqn | r = \dfrac {n \paren {n^2 + 1} } 2       | c =  }} {{end-eqn}} {{qed}} === Sequence === {{:Magic Constant of Magic Square/Sequence}} Category:Magic Squares ogo3904guqwq2mjl8olycsiyk6l4ykq
Let $M_3$ denote the magic square of order $3$. Each row, column and diagonal of $M_3$ must be a different set of $3$ elements of $\N_9$, where $\N_9$ denotes the set $\set {1, 2, 3, 4, 5, 6, 7, 8, 9}$. The sets of $3$ elements of $\N_9$ adding to $15$ can be stated: :$\set {1, 5, 9}, \set {1, 6, 8}$ :$\set {2, 4, 9}, \set {2, 5, 8}, \set {2, 6, 7}$ :$\set {3, 4, 8}, \set {3, 5, 7}$ :$\set {4, 5, 6}$ The number of rows, columns and diagonals of $M_3$ passing through each cell of $M_3$ depends upon where in $M_3$ that cell is positioned. :The corner cells are each on one row, one column and one diagonal :The edge cells are each on one row and one column :The center cells is on one row, one column and both diagonals. Thus the center cell has to contain an integer which is in at least $4$ of the $3$-element subsets of $\N^*_9$ above. There is only one such element, which is $5$. Thus $5$ goes in the center square of $M_3$. By a similar analysis, it is seen that: :the corner cells contain $2, 4, 6, 8$ :the edge cells contain $1, 3, 7, 9$. {{WLOG}}, let $2$ be placed in the upper left corner of $M_3$. There is no $3$-element subset of $\N^*_9$ adding to $15$ containing both $1$ and $2$. It follows that $1$ cannot be placed on either the upper edge or left hand edge. Thus $1$ must go into the right hand edge or the bottom edge. {{WLOG}}, let $1$ be placed on the right hand edge. Up to rotations and reflections, the placements of these three elements is unique. Hence $M_3$ so far looks like this: :$\begin {array} {|c|c|c|} \hline 2 &    &   \\ \hline   &  5 & 1 \\ \hline   &    &   \\ \hline \end {array}$ The population of the remaining cells is forced: :$\begin {array} {|c|c|c|} \hline 2 &   &   \\ \hline 9 & 5 & 1 \\ \hline   &   & 8 \\ \hline \end {array}$ :$\begin {array} {|c|c|c|} \hline 2 &   & 6 \\ \hline 9 & 5 & 1 \\ \hline 4 &   & 8 \\ \hline \end {array}$ :$\begin {array} {|c|c|c|} \hline 2 & 7 & 6 \\ \hline 9 & 5 & 1 \\ \hline 4 & 3 & 8 \\ \hline \end {array}$ {{qed}}
From Rational Numbers form Subfield of Complex Numbers, the unitary module $\left({\C, +, \times}\right)_\Q$ over $\Q$ satisfies the unitary module axioms: * $\Q$-Action: $\C$ is closed under multiplication, so $\Q \times \C \subset \C$. * Distributive: $\times$ distributes over $+$. * Associativity: $\times$ is associative. * Multiplicative Identity: $1$ is the multiplicative identity in $\C$. Let $a, b \in \Q$ such that: :$a z_1 + b z_2 = 0$ By the definition of wholly imaginary, there is a real number $c$ such that: :$c i = z_2$ where $i$ is the imaginary unit. Therefore: :$a z_1 + b c i = 0$ Equating real parts and imaginary parts: :$a z_1 = 0$ :$b c = 0$ Since $z_1$ and $c$ are both non-zero, $a$ and $b$ must be zero. The result follows from the definition of linear independence. {{qed}} Category:Complex Analysis hg3uhv1nrtiqrbzqa9ukeygksck8qbv
Assume the truth of Schanuel's Conjecture. Let $z_1 = \ln \pi$, $z_2 = i \pi$. Note that $z_1$ is wholly real and $z_2$ is wholly imaginary. Hence, by Wholly Real Number and Wholly Imaginary Number are Linearly Independent over the Rationals, they are linearly independent over $\Q$. By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_2, e^{z_1}, e^{z_2}}\right)$ has transcendence degree at least $2$ over $\Q$. That is, the extension field $\Q \left({\ln \pi, i \pi, \pi, e^{i \pi}}\right)$ has transcendence degree at least $2$ over $\Q$. However, by Euler's Identity, $e^{i \pi} = -1$ is algebraic. Also, $i \pi$ and $\pi$ are not algebraically independent, as they satisfy $x^2 + y^2 = 0$, where $x = i \pi$ and $y = \pi$. Therefore, if Schanuel's Conjecture holds, then $\ln \pi$ and $\pi$ are algebraically independent. {{qed}} Category:Transcendental Numbers Category:Pi Category:Logarithms Category:Schanuel's Conjecture sibv8slqarxunibd8j24sjrfl6cl59o
{{AimForCont}} $\ln a$ is not transcendental. Hence, by definition, it is algebraic. Since $a$ is not $1$, $\ln a$ cannot be $0$. Hence, by the weaker Hermite-Lindemann-Weierstrass theorem, $e^{\ln a} = a$ is transcendental. This contradicts with the assumption that $a$ is algebraic. Hence, $\ln a$ must be transcendental. {{qed}} Category:Transcendental Number Theory 5jdtvg2bmxt9g67744j9k9o41r3lq2m
By observation: :$\tuple {1, 5, 9}$: common difference $4$ :$\tuple {2, 5, 8}$: common difference $3$ :$\tuple {3, 5, 7}$: common difference $2$ :$\tuple {4, 5, 6}$: common difference $1$ {{qed}}
From the definition of omega constant, it is the real number $\Omega$ such that: :$\Omega \, e^\Omega = 1$ where $e$ denotes Euler's number. {{AimForCont}} $\Omega$ is not transcendental. Hence, by definition, $\Omega$ is algebraic. Then $e^\Omega$ is also algebraic, because: :$e^\Omega = \dfrac 1 \Omega$ However, by the weaker Hermite-Lindemann-Weierstrass theorem, $e^\Omega$ is transcendental. From this contradiction it follows that $\Omega$ is transcendental. {{qed}} Category:Omega Constant Category:Lambert W Function Category:Transcendental Number Theory pm1dchvhwhp8u9q4xsxay4jhnq3w6tx
From the definition of the principal branch of the Lambert W function: :$y = W_0 \left({x}\right) \iff x = y e^y$ where $x \in \left[{-\dfrac 1 e \,.\,.\, \to}\right)$ and $y \in \left[{-1 \,.\,.\, \to}\right)$. The result follows from substituting $x = 0$ and $y = 0$. {{qed}} Category:Lambert W Function eqh9slzcxvkfb6vlmfhrna606w7srbu
The proof proceeds by induction. Let $n, b \in \Z_{>0}$, where $b \ge 3$. For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^n j b^{n - j} + n = \sum_{j \mathop = 1}^n \left({b - j}\right) b^{n - j}$ === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \left({b - 2}\right) \sum_{j \mathop = 1}^1 j b^{1 - j} + 1       | r = \left({b - 2}\right) b^0 + 1       | c =  }} {{eqn | r = b - 1       | c =  }} {{eqn | r = \sum_{j \mathop = 1}^1 \left({b - j}\right) b^{1 - j}       | c =  }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^k j b^{k - j} + k = \sum_{j \mathop = 1}^k \left({b - j}\right) b^{k - j}$ from which it is to be shown that: :$\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + \left({k + 1}\right) = \sum_{j \mathop = 1}^{k + 1} \left({b - j}\right) b^{k + 1 - j}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \left({b - 2}\right) \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + \left({k + 1}\right)       | r = \left({b - 2}\right) \sum_{j \mathop = 1}^k j b^{k + 1 - j} + k + 1 + \left({b - 2}\right) \left({k + 1}\right)       | c =  }} {{eqn | r = \left({b - 2}\right) b \sum_{j \mathop = 1}^k j b^{k - j} + \left({b k - b k}\right) + k + 1 + b k - 2 k + b - 2       | c =  }} {{eqn | r = b \left({\left({b - 2}\right) \sum_{j \mathop = 1}^k j b^{k - j} + k}\right) + b - k - 1       | c =  }} {{eqn | r = b \left({\sum_{j \mathop = 1}^k \left({b - j}\right) b^{k - j} }\right) + b - k - 1       | c = Induction Hypothesis }} {{eqn | r = \sum_{j \mathop = 1}^k \left({b - j}\right) b^{k + 1 - j} + b - \left({k + 1}\right)       | c =  }} {{eqn | r = \sum_{j \mathop = 1}^{k + 1} \left({b - j}\right) b^{k + 1 - j}       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^n j b^{n - j} + n = \sum_{j \mathop = 1}^n \left({b - j}\right) b^{n - j}$ {{qed}} Category:Recreational Mathematics oe909ismpfuwtghb0q4rabt8zru2xcq
The proof proceeds by induction. Let $n, b \in \Z_{>0}$, where $b \ge 3$. For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \left({b - 1}\right) \sum_{j \mathop = 1}^n j b^{n - j} + n + 1 = \sum_{j \mathop = 0}^n b^j$ === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \left({b - 1}\right) \sum_{j \mathop = 1}^1 j b^{1 - j} + 2       | r = \left({b - 1}\right) b^0 + 2       | c =  }} {{eqn | r = b + 1       | c =  }} {{eqn | r = \sum_{j \mathop = 0}^1 b^j       | c =  }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \left({b - 1}\right) \sum_{j \mathop = 1}^k j b^{k - j} + k + 1 = \sum_{j \mathop = 0}^k b^j$ from which it is to be shown that: :$\displaystyle \left({b - 1}\right) \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + \left({k + 2}\right) = \sum_{j \mathop = 0}^{k + 1} b^j$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | r = \left({b - 1}\right) \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + \left({k + 2}\right)       | o =        | c =  }} {{eqn | r = \left({b - 1}\right) \sum_{j \mathop = 1}^k j b^{k + 1 - j} + \left({k + 2}\right) + \left({b - 1}\right) \left({k + 1}\right)       | c =  }} {{eqn | r = \left({b - 1}\right) b \sum_{j \mathop = 1}^k j b^{k - j} + \left({b \left({k + 1}\right) - b \left({k + 1}\right)}\right) + \left({k + 2}\right) + \left({b - 1}\right) \left({k + 1}\right)       | c =  }} {{eqn | r = b \left({\left({b - 1}\right) \sum_{j \mathop = 1}^k j b^{k - j} + k + 1}\right) + 1       | c =  }} {{eqn | r = b \left({\sum_{j \mathop = 0}^k b^j}\right) + 1       | c = Induction Hypothesis }} {{eqn | r = \sum_{j \mathop = 0}^k b^{j + 1} + 1       | c =  }} {{eqn | r = \sum_{j \mathop = 1}^{k + 1} b^j + 1       | c = Translation of Index Variable of Summation: Corollary }} {{eqn | r = \sum_{j \mathop = 0}^{k + 1} b^j       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \left({b - 1}\right) \sum_{j \mathop = 1}^n j b^{n - j} + n + 1 = \sum_{j \mathop = 0}^n b^j$ {{qed}} Category:Recreational Mathematics 54w5ccp5anqbfvjlmd530mhqddb2am1