Mixing
Roots of $f$ and $f'$ for $1+sum_k=0^100frac(-1)^k+1(k+1)!x(x-1)(x-2)cdots (x-k)$
Clash Royale CLAN TAG#URR8PPP
up vote
6
down vote
favorite
I have this question from an admission exams. Given that $$f(x)=1+sum_k=0^100frac(-1)^k+1(k+1)!x(x-1)(x-2)cdots (x-k)$$ find $S(f(x))-S(f'(x))$ where $S$ denotes the sum of the real roots for $f(x)$ respectively $f'(x)$
Here is how I tried, since for $x=1,2,ldots,100, 101$ the sum part vanishes little by little and
$$f(1)=1-1=0$$
$$f(2)=1-2+frac12!2(2-1)=0$$
$$f(3)=1-3+frac12!3(3-1)-frac13!3(3-2)(3-1)=binom30-binom31+binom32-binom33$$ and we can see a pattern in the above equation so that we can rewrite $f(3)$ as $(1-1)^3=0,$ and indeed that $f(101)=(1-1)^101=0$
Since the polynomial is also of order $101$, the roots are $x=1,2, ldots, 101$ giving:$$S(f(x))=sum_j=1^101j =5151$$ Is this correct? And how can $S(f'(x))$ be evaluated ? This doesnt seem so obvious.. Also now we can rewrite $$f(x)=-frac1101!(x-1)(x-2)cdots(x-101)$$
polynomials roots
edited Jul 24 at 19:13
mechanodroid
22.2k52041
asked Jul 24 at 18:30
Zacky
2,1771326
 |Â
show 2 more comments
up vote
6
down vote
favorite
I have this question from an admission exams. Given that $$f(x)=1+sum_k=0^100frac(-1)^k+1(k+1)!x(x-1)(x-2)cdots (x-k)$$ find $S(f(x))-S(f'(x))$ where $S$ denotes the sum of the real roots for $f(x)$ respectively $f'(x)$
Here is how I tried, since for $x=1,2,ldots,100, 101$ the sum part vanishes little by little and
$$f(1)=1-1=0$$
$$f(2)=1-2+frac12!2(2-1)=0$$
$$f(3)=1-3+frac12!3(3-1)-frac13!3(3-2)(3-1)=binom30-binom31+binom32-binom33$$ and we can see a pattern in the above equation so that we can rewrite $f(3)$ as $(1-1)^3=0,$ and indeed that $f(101)=(1-1)^101=0$
Since the polynomial is also of order $101$, the roots are $x=1,2, ldots, 101$ giving:$$S(f(x))=sum_j=1^101j =5151$$ Is this correct? And how can $S(f'(x))$ be evaluated ? This doesnt seem so obvious.. Also now we can rewrite $$f(x)=-frac1101!(x-1)(x-2)cdots(x-101)$$
polynomials roots
edited Jul 24 at 19:13
mechanodroid
22.2k52041
asked Jul 24 at 18:30
Zacky
2,1771326
The sum of the roots of a polynomial of degree $n$ is the negative of the coefficient of the degree $n-1$ term, so you just have to calculate the derivative.
– saulspatz
Jul 24 at 18:50
You mean to use Vietta's relation?
– Zacky
Jul 24 at 18:55
Yes, exactly. Now that you've figured out a simple formula for $f$ it seems like it should be easy to finish off. (+1 for figuring out the formula for $f$ by the way.)
– saulspatz
Jul 24 at 18:58
Oh then it is just $$S(f'(x))=frac515100101 =5100 $$
– Zacky
Jul 24 at 19:06
I don't think that right. You might want to add your calculations to the body of the question.
– saulspatz
Jul 24 at 19:17
 |Â
show 2 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I have this question from an admission exams. Given that $$f(x)=1+sum_k=0^100frac(-1)^k+1(k+1)!x(x-1)(x-2)cdots (x-k)$$ find $S(f(x))-S(f'(x))$ where $S$ denotes the sum of the real roots for $f(x)$ respectively $f'(x)$
Here is how I tried, since for $x=1,2,ldots,100, 101$ the sum part vanishes little by little and
$$f(1)=1-1=0$$
$$f(2)=1-2+frac12!2(2-1)=0$$
$$f(3)=1-3+frac12!3(3-1)-frac13!3(3-2)(3-1)=binom30-binom31+binom32-binom33$$ and we can see a pattern in the above equation so that we can rewrite $f(3)$ as $(1-1)^3=0,$ and indeed that $f(101)=(1-1)^101=0$
Since the polynomial is also of order $101$, the roots are $x=1,2, ldots, 101$ giving:$$S(f(x))=sum_j=1^101j =5151$$ Is this correct? And how can $S(f'(x))$ be evaluated ? This doesnt seem so obvious.. Also now we can rewrite $$f(x)=-frac1101!(x-1)(x-2)cdots(x-101)$$
polynomials roots
edited Jul 24 at 19:13
mechanodroid
22.2k52041
asked Jul 24 at 18:30
Zacky
2,1771326
I have this question from an admission exams. Given that $$f(x)=1+sum_k=0^100frac(-1)^k+1(k+1)!x(x-1)(x-2)cdots (x-k)$$ find $S(f(x))-S(f'(x))$ where $S$ denotes the sum of the real roots for $f(x)$ respectively $f'(x)$
Here is how I tried, since for $x=1,2,ldots,100, 101$ the sum part vanishes little by little and
$$f(1)=1-1=0$$
$$f(2)=1-2+frac12!2(2-1)=0$$
$$f(3)=1-3+frac12!3(3-1)-frac13!3(3-2)(3-1)=binom30-binom31+binom32-binom33$$ and we can see a pattern in the above equation so that we can rewrite $f(3)$ as $(1-1)^3=0,$ and indeed that $f(101)=(1-1)^101=0$
Since the polynomial is also of order $101$, the roots are $x=1,2, ldots, 101$ giving:$$S(f(x))=sum_j=1^101j =5151$$ Is this correct? And how can $S(f'(x))$ be evaluated ? This doesnt seem so obvious.. Also now we can rewrite $$f(x)=-frac1101!(x-1)(x-2)cdots(x-101)$$
polynomials roots
edited Jul 24 at 19:13
mechanodroid
22.2k52041
asked Jul 24 at 18:30
Zacky
2,1771326
edited Jul 24 at 19:13
mechanodroid
22.2k52041
edited Jul 24 at 19:13
mechanodroid
22.2k52041
edited Jul 24 at 19:13
mechanodroid
22.2k52041
22.2k52041
asked Jul 24 at 18:30
Zacky
2,1771326
asked Jul 24 at 18:30
Zacky
2,1771326
asked Jul 24 at 18:30
Zacky
2,1771326
2,1771326
The sum of the roots of a polynomial of degree $n$ is the negative of the coefficient of the degree $n-1$ term, so you just have to calculate the derivative.
– saulspatz
Jul 24 at 18:50
You mean to use Vietta's relation?
– Zacky
Jul 24 at 18:55
Yes, exactly. Now that you've figured out a simple formula for $f$ it seems like it should be easy to finish off. (+1 for figuring out the formula for $f$ by the way.)
– saulspatz
Jul 24 at 18:58
Oh then it is just $$S(f'(x))=frac515100101 =5100 $$
– Zacky
Jul 24 at 19:06
I don't think that right. You might want to add your calculations to the body of the question.
– saulspatz
Jul 24 at 19:17
 |Â
show 2 more comments
The sum of the roots of a polynomial of degree $n$ is the negative of the coefficient of the degree $n-1$ term, so you just have to calculate the derivative.
– saulspatz
Jul 24 at 18:50
You mean to use Vietta's relation?
– Zacky
Jul 24 at 18:55
Yes, exactly. Now that you've figured out a simple formula for $f$ it seems like it should be easy to finish off. (+1 for figuring out the formula for $f$ by the way.)
– saulspatz
Jul 24 at 18:58
Oh then it is just $$S(f'(x))=frac515100101 =5100 $$
– Zacky
Jul 24 at 19:06
I don't think that right. You might want to add your calculations to the body of the question.
– saulspatz
Jul 24 at 19:17
The sum of the roots of a polynomial of degree $n$ is the negative of the coefficient of the degree $n-1$ term, so you just have to calculate the derivative.
– saulspatz
Jul 24 at 18:50
The sum of the roots of a polynomial of degree $n$ is the negative of the coefficient of the degree $n-1$ term, so you just have to calculate the derivative.
– saulspatz
Jul 24 at 18:50
You mean to use Vietta's relation?
– Zacky
Jul 24 at 18:55
You mean to use Vietta's relation?
– Zacky
Jul 24 at 18:55
Yes, exactly. Now that you've figured out a simple formula for $f$ it seems like it should be easy to finish off. (+1 for figuring out the formula for $f$ by the way.)
– saulspatz
Jul 24 at 18:58
Yes, exactly. Now that you've figured out a simple formula for $f$ it seems like it should be easy to finish off. (+1 for figuring out the formula for $f$ by the way.)
– saulspatz
Jul 24 at 18:58
Oh then it is just $$S(f'(x))=frac515100101 =5100 $$
– Zacky
Jul 24 at 19:06
Oh then it is just $$S(f'(x))=frac515100101 =5100 $$
– Zacky
Jul 24 at 19:06
I don't think that right. You might want to add your calculations to the body of the question.
– saulspatz
Jul 24 at 19:17
I don't think that right. You might want to add your calculations to the body of the question.
– saulspatz
Jul 24 at 19:17
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
We have
$$f(x) = a_101x^101 + a_100x^100 + cdots$$
and we know that $a_101 = -frac1101!$. Vieta's formulas give
$$5151 = text sum of roots of f = -fraca_100a_101$$
so $a_100 = frac5151101!$.
The derivative is
$$f'(x) = 101a_101x^100 + 100a_100x^99 + cdots$$
so Vieta's formulas give
$$text sum of roots of f' = -frac100a_100101a_101 = -frac100cdot frac5151101!101cdot frac-1101! = frac515100101= 5100$$
answered Jul 24 at 19:11
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
According to Vieta given the polynomial
$$
p_n(x) = a_n x^n+a_n-1x^n-1+cdots + a_0\
p'_n(x) = n a_n x^n-1 + (n-1)a_n-1x^n-2+cdots+a_1
$$
now
$$
S_100(x) = -fraca_100a_101\
S'_100(x) = -frac(101-1)a_100101 a_101
$$
Here
$$
a_101 = frac(-1)^101101!\
a_100 = -a_101frac100(100+1)2+frac(-1)^100100!
$$
then
$$
S_n(x) = 5151\
S'_n(x) = 5100
$$
and finally
$$
S_100(x)-S'_100(x) = 5151-5100 = 51
$$
answered Jul 24 at 21:01
Cesareo
5,6922412
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We have
$$f(x) = a_101x^101 + a_100x^100 + cdots$$
and we know that $a_101 = -frac1101!$. Vieta's formulas give
$$5151 = text sum of roots of f = -fraca_100a_101$$
so $a_100 = frac5151101!$.
The derivative is
$$f'(x) = 101a_101x^100 + 100a_100x^99 + cdots$$
so Vieta's formulas give
$$text sum of roots of f' = -frac100a_100101a_101 = -frac100cdot frac5151101!101cdot frac-1101! = frac515100101= 5100$$
answered Jul 24 at 19:11
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
accepted
We have
$$f(x) = a_101x^101 + a_100x^100 + cdots$$
and we know that $a_101 = -frac1101!$. Vieta's formulas give
$$5151 = text sum of roots of f = -fraca_100a_101$$
so $a_100 = frac5151101!$.
The derivative is
$$f'(x) = 101a_101x^100 + 100a_100x^99 + cdots$$
so Vieta's formulas give
$$text sum of roots of f' = -frac100a_100101a_101 = -frac100cdot frac5151101!101cdot frac-1101! = frac515100101= 5100$$
answered Jul 24 at 19:11
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We have
$$f(x) = a_101x^101 + a_100x^100 + cdots$$
and we know that $a_101 = -frac1101!$. Vieta's formulas give
$$5151 = text sum of roots of f = -fraca_100a_101$$
so $a_100 = frac5151101!$.
The derivative is
$$f'(x) = 101a_101x^100 + 100a_100x^99 + cdots$$
so Vieta's formulas give
$$text sum of roots of f' = -frac100a_100101a_101 = -frac100cdot frac5151101!101cdot frac-1101! = frac515100101= 5100$$
answered Jul 24 at 19:11
mechanodroid
22.2k52041
We have
$$f(x) = a_101x^101 + a_100x^100 + cdots$$
and we know that $a_101 = -frac1101!$. Vieta's formulas give
$$5151 = text sum of roots of f = -fraca_100a_101$$
so $a_100 = frac5151101!$.
The derivative is
$$f'(x) = 101a_101x^100 + 100a_100x^99 + cdots$$
so Vieta's formulas give
$$text sum of roots of f' = -frac100a_100101a_101 = -frac100cdot frac5151101!101cdot frac-1101! = frac515100101= 5100$$
answered Jul 24 at 19:11
mechanodroid
22.2k52041
answered Jul 24 at 19:11
mechanodroid
22.2k52041
answered Jul 24 at 19:11
mechanodroid
22.2k52041
answered Jul 24 at 19:11
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
up vote
0
down vote
According to Vieta given the polynomial
$$
p_n(x) = a_n x^n+a_n-1x^n-1+cdots + a_0\
p'_n(x) = n a_n x^n-1 + (n-1)a_n-1x^n-2+cdots+a_1
$$
now
$$
S_100(x) = -fraca_100a_101\
S'_100(x) = -frac(101-1)a_100101 a_101
$$
Here
$$
a_101 = frac(-1)^101101!\
a_100 = -a_101frac100(100+1)2+frac(-1)^100100!
$$
then
$$
S_n(x) = 5151\
S'_n(x) = 5100
$$
and finally
$$
S_100(x)-S'_100(x) = 5151-5100 = 51
$$
answered Jul 24 at 21:01
Cesareo
5,6922412
add a comment |Â
up vote
0
down vote
According to Vieta given the polynomial
$$
p_n(x) = a_n x^n+a_n-1x^n-1+cdots + a_0\
p'_n(x) = n a_n x^n-1 + (n-1)a_n-1x^n-2+cdots+a_1
$$
now
$$
S_100(x) = -fraca_100a_101\
S'_100(x) = -frac(101-1)a_100101 a_101
$$
Here
$$
a_101 = frac(-1)^101101!\
a_100 = -a_101frac100(100+1)2+frac(-1)^100100!
$$
then
$$
S_n(x) = 5151\
S'_n(x) = 5100
$$
and finally
$$
S_100(x)-S'_100(x) = 5151-5100 = 51
$$
answered Jul 24 at 21:01
Cesareo
5,6922412
add a comment |Â
up vote
0
down vote
up vote
0
down vote
According to Vieta given the polynomial
$$
p_n(x) = a_n x^n+a_n-1x^n-1+cdots + a_0\
p'_n(x) = n a_n x^n-1 + (n-1)a_n-1x^n-2+cdots+a_1
$$
now
$$
S_100(x) = -fraca_100a_101\
S'_100(x) = -frac(101-1)a_100101 a_101
$$
Here
$$
a_101 = frac(-1)^101101!\
a_100 = -a_101frac100(100+1)2+frac(-1)^100100!
$$
then
$$
S_n(x) = 5151\
S'_n(x) = 5100
$$
and finally
$$
S_100(x)-S'_100(x) = 5151-5100 = 51
$$
answered Jul 24 at 21:01
Cesareo
5,6922412
According to Vieta given the polynomial
$$
p_n(x) = a_n x^n+a_n-1x^n-1+cdots + a_0\
p'_n(x) = n a_n x^n-1 + (n-1)a_n-1x^n-2+cdots+a_1
$$
now
$$
S_100(x) = -fraca_100a_101\
S'_100(x) = -frac(101-1)a_100101 a_101
$$
Here
$$
a_101 = frac(-1)^101101!\
a_100 = -a_101frac100(100+1)2+frac(-1)^100100!
$$
then
$$
S_n(x) = 5151\
S'_n(x) = 5100
$$
and finally
$$
S_100(x)-S'_100(x) = 5151-5100 = 51
$$
answered Jul 24 at 21:01
Cesareo
5,6922412
answered Jul 24 at 21:01
Cesareo
5,6922412
answered Jul 24 at 21:01
Cesareo
5,6922412
answered Jul 24 at 21:01
Cesareo
5,6922412
5,6922412
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861632%2froots-of-f-and-f-for-1-sum-k-0100-frac-1k1k1xx-1x-2%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
The sum of the roots of a polynomial of degree $n$ is the negative of the coefficient of the degree $n-1$ term, so you just have to calculate the derivative.
– saulspatz
Jul 24 at 18:50
You mean to use Vietta's relation?
– Zacky
Jul 24 at 18:55
Yes, exactly. Now that you've figured out a simple formula for $f$ it seems like it should be easy to finish off. (+1 for figuring out the formula for $f$ by the way.)
– saulspatz
Jul 24 at 18:58
Oh then it is just $$S(f'(x))=frac515100101 =5100 $$
– Zacky
Jul 24 at 19:06
I don't think that right. You might want to add your calculations to the body of the question.
– saulspatz
Jul 24 at 19:17