If $gleft(x-3right)=x^4$ what is $gleft(xright)$? The Next CEO of Stack OverflowHow many routes are there through from top left corner to top right in a 20x20 grid? Binomial Coefficent explanationIdentity for central binomial coefficientsBinomial Expansion Word Problem (Creating a Equation)Evaluate the sum $ sum_i=0^n (-1)^n-i binomni f(i)$Interesting functional equation: $f(x)=fracxx+fleft(fracxx+f(x)right)$Prove for $ forall n in mathbbN, exists x,y,z$ ( $0 leq x < y < z$ ) such that $ n = binomx1 + binomy2 + binomz3$What is $sum_i=0^n leftlfloor sqrtirightrfloor binomni$?Evaluating $lim_ntoinftysum_r=0^n(-1)^rbinomn2rleft(fracxnright)^2r$Calculate $(-1)^k left(beginmatrix -1/2 \ kendmatrixright)$. Is it equal to $frac (2k)!2^2k(k!)^2$?Binomial optimization
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If $gleft(x-3right)=x^4$ what is $gleft(xright)$?
The Next CEO of Stack OverflowHow many routes are there through from top left corner to top right in a 20x20 grid? Binomial Coefficent explanationIdentity for central binomial coefficientsBinomial Expansion Word Problem (Creating a Equation)Evaluate the sum $ sum_i=0^n (-1)^n-i binomni f(i)$Interesting functional equation: $f(x)=fracxx+fleft(fracxx+f(x)right)$Prove for $ forall n in mathbbN, exists x,y,z$ ( $0 leq x < y < z$ ) such that $ n = binomx1 + binomy2 + binomz3$What is $sum_i=0^n leftlfloor sqrtirightrfloor binomni$?Evaluating $lim_ntoinftysum_r=0^n(-1)^rbinomn2rleft(fracxnright)^2r$Calculate $(-1)^k left(beginmatrix -1/2 \ kendmatrixright)$. Is it equal to $frac (2k)!2^2k(k!)^2$?Binomial optimization
$begingroup$
Consider the following exercise:
Given that $gleft(x-3right)=x^4$ solve the equation $$gleft(xright)-81=gleft(x+3right)+108x$$
This appears in a textbook in a section about Combinatorics and Binomial Coefficients. Prior to this exercise there is no obvious reference to this kind of problem.
In the solution for this problem, the original equation is promptly substituted by $left(x+3right)^4-81=x^4+108x$. My question is: how did it go from $gleft(xright)$ to $left(x+3right)^4$?
functions binomial-coefficients
edited Mar 18 at 21:04
Maria Mazur
48.9k1260122
asked Mar 18 at 20:46
Daniel OscarDaniel Oscar
36711
$endgroup$
add a comment |
$begingroup$
Consider the following exercise:
Given that $gleft(x-3right)=x^4$ solve the equation $$gleft(xright)-81=gleft(x+3right)+108x$$
This appears in a textbook in a section about Combinatorics and Binomial Coefficients. Prior to this exercise there is no obvious reference to this kind of problem.
In the solution for this problem, the original equation is promptly substituted by $left(x+3right)^4-81=x^4+108x$. My question is: how did it go from $gleft(xright)$ to $left(x+3right)^4$?
functions binomial-coefficients
edited Mar 18 at 21:04
Maria Mazur
48.9k1260122
asked Mar 18 at 20:46
Daniel OscarDaniel Oscar
36711
$endgroup$
1
$begingroup$
Put y=x-3, then g(y)=x^4=(y+3)^4
$endgroup$
– Alexandros
Mar 18 at 20:52
add a comment |
$begingroup$
Consider the following exercise:
Given that $gleft(x-3right)=x^4$ solve the equation $$gleft(xright)-81=gleft(x+3right)+108x$$
This appears in a textbook in a section about Combinatorics and Binomial Coefficients. Prior to this exercise there is no obvious reference to this kind of problem.
In the solution for this problem, the original equation is promptly substituted by $left(x+3right)^4-81=x^4+108x$. My question is: how did it go from $gleft(xright)$ to $left(x+3right)^4$?
functions binomial-coefficients
edited Mar 18 at 21:04
Maria Mazur
48.9k1260122
asked Mar 18 at 20:46
Daniel OscarDaniel Oscar
36711
$endgroup$
Consider the following exercise:
Given that $gleft(x-3right)=x^4$ solve the equation $$gleft(xright)-81=gleft(x+3right)+108x$$
This appears in a textbook in a section about Combinatorics and Binomial Coefficients. Prior to this exercise there is no obvious reference to this kind of problem.
In the solution for this problem, the original equation is promptly substituted by $left(x+3right)^4-81=x^4+108x$. My question is: how did it go from $gleft(xright)$ to $left(x+3right)^4$?
functions binomial-coefficients
functions binomial-coefficients
edited Mar 18 at 21:04
Maria Mazur
48.9k1260122
asked Mar 18 at 20:46
Daniel OscarDaniel Oscar
36711
edited Mar 18 at 21:04
Maria Mazur
48.9k1260122
asked Mar 18 at 20:46
Daniel OscarDaniel Oscar
36711
edited Mar 18 at 21:04
Maria Mazur
48.9k1260122
edited Mar 18 at 21:04
Maria Mazur
48.9k1260122
edited Mar 18 at 21:04
Maria Mazur
48.9k1260122
48.9k1260122
asked Mar 18 at 20:46
Daniel OscarDaniel Oscar
36711
asked Mar 18 at 20:46
Daniel OscarDaniel Oscar
36711
asked Mar 18 at 20:46
Daniel OscarDaniel Oscar
36711
36711
1
$begingroup$
Put y=x-3, then g(y)=x^4=(y+3)^4
$endgroup$
– Alexandros
Mar 18 at 20:52
add a comment |
1
$begingroup$
Put y=x-3, then g(y)=x^4=(y+3)^4
$endgroup$
– Alexandros
Mar 18 at 20:52
1
1
$begingroup$
Put y=x-3, then g(y)=x^4=(y+3)^4
$endgroup$
– Alexandros
Mar 18 at 20:52
$begingroup$
Put y=x-3, then g(y)=x^4=(y+3)^4
$endgroup$
– Alexandros
Mar 18 at 20:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Write $t=x-3$ then $t=x+3$, so $$g(t)=g(x-3)= x^4= (t+3)^4$$
Since $xmapsto x-3$ is surjective you can write again it with $x$, so $g(x)=(x+3)^4$.
edited Mar 18 at 21:03
Henry Lee
2,181319
answered Mar 18 at 20:50
Maria MazurMaria Mazur
48.9k1260122
$endgroup$
add a comment |
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1 Answer
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votes
$begingroup$
Write $t=x-3$ then $t=x+3$, so $$g(t)=g(x-3)= x^4= (t+3)^4$$
Since $xmapsto x-3$ is surjective you can write again it with $x$, so $g(x)=(x+3)^4$.
edited Mar 18 at 21:03
Henry Lee
2,181319
answered Mar 18 at 20:50
Maria MazurMaria Mazur
48.9k1260122
$endgroup$
add a comment |
$begingroup$
Write $t=x-3$ then $t=x+3$, so $$g(t)=g(x-3)= x^4= (t+3)^4$$
Since $xmapsto x-3$ is surjective you can write again it with $x$, so $g(x)=(x+3)^4$.
edited Mar 18 at 21:03
Henry Lee
2,181319
answered Mar 18 at 20:50
Maria MazurMaria Mazur
48.9k1260122
$endgroup$
add a comment |
$begingroup$
Write $t=x-3$ then $t=x+3$, so $$g(t)=g(x-3)= x^4= (t+3)^4$$
Since $xmapsto x-3$ is surjective you can write again it with $x$, so $g(x)=(x+3)^4$.
edited Mar 18 at 21:03
Henry Lee
2,181319
answered Mar 18 at 20:50
Maria MazurMaria Mazur
48.9k1260122
$endgroup$
Write $t=x-3$ then $t=x+3$, so $$g(t)=g(x-3)= x^4= (t+3)^4$$
Since $xmapsto x-3$ is surjective you can write again it with $x$, so $g(x)=(x+3)^4$.
edited Mar 18 at 21:03
Henry Lee
2,181319
answered Mar 18 at 20:50
Maria MazurMaria Mazur
48.9k1260122
edited Mar 18 at 21:03
Henry Lee
2,181319
edited Mar 18 at 21:03
Henry Lee
2,181319
edited Mar 18 at 21:03
Henry Lee
2,181319
2,181319
answered Mar 18 at 20:50
Maria MazurMaria Mazur
48.9k1260122
answered Mar 18 at 20:50
Maria MazurMaria Mazur
48.9k1260122
answered Mar 18 at 20:50
Maria MazurMaria Mazur
48.9k1260122
48.9k1260122
add a comment |
add a comment |
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$begingroup$
Put y=x-3, then g(y)=x^4=(y+3)^4
$endgroup$
– Alexandros
Mar 18 at 20:52