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Category: Relation and Functions

let-S-n-p-k-0-n-k-p-prove-that-S-n-p-1-p-1-n-1-p-1-k-0-n-1-C-p-1-k-S-n-k-

Question Number 52669 by maxmathsup by imad last updated on 11/Jan/19 $${let}\:{S}_{{n}\:} \:\left({p}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}^{{p}} \\ $$$${prove}\:{that}\:{S}_{{n}} \left({p}\right)=\frac{\mathrm{1}}{{p}+\mathrm{1}}\left\{\:\left({n}+\mathrm{1}\right)^{{p}+\mathrm{1}} \:−\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{C}_{{p}+\mathrm{1}} ^{{k}} \:{S}_{{n}} \left({k}\right)\right\} \\…

Determine-all-function-f-R-0-1-R-satisfying-the-functional-relation-f-x-f-1-1-x-2-1-2x-x-1-x-for-x-0-and-x-1-

Question Number 118150 by bobhans last updated on 15/Oct/20 $$\mathrm{Determine}\:\mathrm{all}\:\mathrm{function}\:\mathrm{f}:\mathbb{R}\diagdown\left\{\mathrm{0},\mathrm{1}\right\}\:\rightarrow\mathbb{R} \\ $$$$\mathrm{satisfying}\:\mathrm{the}\:\mathrm{functional}\:\mathrm{relation} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\right)\:=\:\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2x}\right)}{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)};\:\mathrm{for}\:\mathrm{x}\neq\mathrm{0}\:\mathrm{and}\:\mathrm{x}\neq\mathrm{1} \\ $$ Commented by bemath last updated on 15/Oct/20 $${good}\:{question} \\…

Let-n-N-Find-the-number-of-polynomials-p-x-with-coefficients-in-0-1-2-3-such-that-p-2-n-

Question Number 117857 by bemath last updated on 14/Oct/20 $$\mathrm{Let}\:\mathrm{n}\in\mathbb{N}\:.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\: \\ $$$$\mathrm{polynomials}\:\mathrm{p}\left(\mathrm{x}\right)\:\mathrm{with}\:\mathrm{coefficients} \\ $$$$\mathrm{in}\:\left\{\:\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\:\right\}\:\mathrm{such}\:\mathrm{that}\:\mathrm{p}\left(\mathrm{2}\right)=\:\mathrm{n}\: \\ $$ Answered by mindispower last updated on 14/Oct/20 $${let}\:{p}\left({x}\right)=\underset{{k}\leqslant{n}} {\sum}{a}_{{k}}…

Given-f-x-1-3-x-2x-Ax-4-x-2-if-f-1-5-then-A-floor-function-

Question Number 183158 by cortano1 last updated on 21/Dec/22 $$\:{Given}\:{f}\left({x}\right)=\:\frac{\left[\frac{\mathrm{1}}{\mathrm{3}}{x}\right]\mid\mathrm{2}{x}\mid+{Ax}}{\mid\mathrm{4}−{x}^{\mathrm{2}} \mid} \\ $$$$\:{if}\:{f}\:'\left(−\mathrm{1}\right)=\:\mathrm{5}\:{then}\:{A}=? \\ $$$$\left[\:\:\:\:\right]\:=\:{floor}\:{function}\: \\ $$ Answered by TheSupreme last updated on 21/Dec/22 $${f}\left({x}\right)=\frac{{g}\left({x}\right){h}\left({x}\right)}{{u}\left({x}\right)}+\frac{{Ax}}{\mathrm{4}−{x}^{\mathrm{2}}…