Mixing
How does one evaluate the multiplication $f(2)cdot f(3)cdot f(4)cdots f(15)$ by formulating?
$begingroup$
Suppose
$$f : mathbb{Z}^+ rightarrow mathbb{R}$$
$$f(x) = 1-dfrac{1}{x^2}$$
How does one evaluate the multiplication $prod_{i=2}^{15} f(i)=f(2)cdot f(3)cdot f(4)cdots f(15)$?
Here I have to see the trick that directly yields the calculation.
How come that we write this using $Pi$ (product) notation? I'll be glad to hear your dear thoughts.
functions products
edited Jan 2 at 17:14
Henrik
5,99892030
asked Jan 2 at 17:08
EnzoEnzo
1537
$endgroup$
add a comment |
$begingroup$
Suppose
$$f : mathbb{Z}^+ rightarrow mathbb{R}$$
$$f(x) = 1-dfrac{1}{x^2}$$
How does one evaluate the multiplication $prod_{i=2}^{15} f(i)=f(2)cdot f(3)cdot f(4)cdots f(15)$?
Here I have to see the trick that directly yields the calculation.
How come that we write this using $Pi$ (product) notation? I'll be glad to hear your dear thoughts.
functions products
edited Jan 2 at 17:14
Henrik
5,99892030
asked Jan 2 at 17:08
EnzoEnzo
1537
$endgroup$
1
$begingroup$
Have you looked carefully at the product of the first few terms? I think if you use a bit of algebra, you'll see that there is some cancellation between numerators and denominators of adjacent terms.
$endgroup$
– hardmath
Jan 2 at 17:11
$begingroup$
@hardmath That's absolutely true. I also have looked carefully at the product of first few terms, no doubt.
$endgroup$
– Enzo
Jan 2 at 17:13
1
$begingroup$
It would also depend on how you write those terms. If you just write $left(1 - frac1{2^2}right)left(1 - frac1{3^2}right)left(1 - frac1{4^2}right)cdots$ then I don't think you will see a useful pattern.
$endgroup$
– David K
Jan 2 at 17:16
$begingroup$
I have got $$frac{8}{15}$$ with my TI calculator
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 17:18
add a comment |
$begingroup$
Suppose
$$f : mathbb{Z}^+ rightarrow mathbb{R}$$
$$f(x) = 1-dfrac{1}{x^2}$$
How does one evaluate the multiplication $prod_{i=2}^{15} f(i)=f(2)cdot f(3)cdot f(4)cdots f(15)$?
Here I have to see the trick that directly yields the calculation.
How come that we write this using $Pi$ (product) notation? I'll be glad to hear your dear thoughts.
functions products
edited Jan 2 at 17:14
Henrik
5,99892030
asked Jan 2 at 17:08
EnzoEnzo
1537
$endgroup$
Suppose
$$f : mathbb{Z}^+ rightarrow mathbb{R}$$
$$f(x) = 1-dfrac{1}{x^2}$$
How does one evaluate the multiplication $prod_{i=2}^{15} f(i)=f(2)cdot f(3)cdot f(4)cdots f(15)$?
Here I have to see the trick that directly yields the calculation.
How come that we write this using $Pi$ (product) notation? I'll be glad to hear your dear thoughts.
functions products
functions products
edited Jan 2 at 17:14
Henrik
5,99892030
asked Jan 2 at 17:08
EnzoEnzo
1537
edited Jan 2 at 17:14
Henrik
5,99892030
asked Jan 2 at 17:08
EnzoEnzo
1537
edited Jan 2 at 17:14
Henrik
5,99892030
edited Jan 2 at 17:14
Henrik
5,99892030
edited Jan 2 at 17:14
Henrik
5,99892030
5,99892030
asked Jan 2 at 17:08
EnzoEnzo
1537
asked Jan 2 at 17:08
EnzoEnzo
1537
asked Jan 2 at 17:08
EnzoEnzo
1537
1537
1
$begingroup$
Have you looked carefully at the product of the first few terms? I think if you use a bit of algebra, you'll see that there is some cancellation between numerators and denominators of adjacent terms.
$endgroup$
– hardmath
Jan 2 at 17:11
$begingroup$
@hardmath That's absolutely true. I also have looked carefully at the product of first few terms, no doubt.
$endgroup$
– Enzo
Jan 2 at 17:13
1
$begingroup$
It would also depend on how you write those terms. If you just write $left(1 - frac1{2^2}right)left(1 - frac1{3^2}right)left(1 - frac1{4^2}right)cdots$ then I don't think you will see a useful pattern.
$endgroup$
– David K
Jan 2 at 17:16
$begingroup$
I have got $$frac{8}{15}$$ with my TI calculator
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 17:18
add a comment |
1
$begingroup$
Have you looked carefully at the product of the first few terms? I think if you use a bit of algebra, you'll see that there is some cancellation between numerators and denominators of adjacent terms.
$endgroup$
– hardmath
Jan 2 at 17:11
$begingroup$
@hardmath That's absolutely true. I also have looked carefully at the product of first few terms, no doubt.
$endgroup$
– Enzo
Jan 2 at 17:13
1
$begingroup$
It would also depend on how you write those terms. If you just write $left(1 - frac1{2^2}right)left(1 - frac1{3^2}right)left(1 - frac1{4^2}right)cdots$ then I don't think you will see a useful pattern.
$endgroup$
– David K
Jan 2 at 17:16
$begingroup$
I have got $$frac{8}{15}$$ with my TI calculator
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 17:18
1
1
$begingroup$
Have you looked carefully at the product of the first few terms? I think if you use a bit of algebra, you'll see that there is some cancellation between numerators and denominators of adjacent terms.
$endgroup$
– hardmath
Jan 2 at 17:11
$begingroup$
Have you looked carefully at the product of the first few terms? I think if you use a bit of algebra, you'll see that there is some cancellation between numerators and denominators of adjacent terms.
$endgroup$
– hardmath
Jan 2 at 17:11
$begingroup$
@hardmath That's absolutely true. I also have looked carefully at the product of first few terms, no doubt.
$endgroup$
– Enzo
Jan 2 at 17:13
$begingroup$
@hardmath That's absolutely true. I also have looked carefully at the product of first few terms, no doubt.
$endgroup$
– Enzo
Jan 2 at 17:13
1
1
$begingroup$
It would also depend on how you write those terms. If you just write $left(1 - frac1{2^2}right)left(1 - frac1{3^2}right)left(1 - frac1{4^2}right)cdots$ then I don't think you will see a useful pattern.
$endgroup$
– David K
Jan 2 at 17:16
$begingroup$
It would also depend on how you write those terms. If you just write $left(1 - frac1{2^2}right)left(1 - frac1{3^2}right)left(1 - frac1{4^2}right)cdots$ then I don't think you will see a useful pattern.
$endgroup$
– David K
Jan 2 at 17:16
$begingroup$
I have got $$frac{8}{15}$$ with my TI calculator
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 17:18
$begingroup$
I have got $$frac{8}{15}$$ with my TI calculator
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 17:18
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: $f(x) = frac{x^2-1}{x^2} = frac{(x-1)(x+1)}{x^2}$
and take product of $f(2),f(3),...,f(n)$
Look:
$$
frac{(1)(3)}{2^2} cdot frac{(2)(4)}{3^2} cdot frac{(3)(5)}{4^2} cdot ... frac{(n-3)(n-1)}{(n-2)^2} cdot frac{(n-2)n}{(n-1)^2} cdot frac{(n-1)(n+1)}{n^2} = frac{1}{2} cdot frac{n+1}{n} $$
and take this for your task
answered Jan 2 at 17:15
VirtualUserVirtualUser
59412
$endgroup$
$begingroup$
That does not really help.
$endgroup$
– Enzo
Jan 2 at 17:16
1
$begingroup$
@Enzo Perhaps you should try harder to understand it. It is the same hint I would have given; I don't think there is an easier way.
$endgroup$
– David K
Jan 2 at 17:18
$begingroup$
Write as I said and look what will happen
$endgroup$
– VirtualUser
Jan 2 at 17:19
$begingroup$
@Enzo I'll go out on a limb and say that there is no easier way.
$endgroup$
– John
Jan 2 at 17:20
$begingroup$
$$biggr (frac{(2-1)(2+1)}{2^2} biggr ) biggr ( frac{(3-1)(3+1)}{3^2} biggr )$$ No such term cancels out here.
$endgroup$
– Enzo
Jan 2 at 17:22
|
show 2 more comments
$begingroup$
It is also convenient to use the factorial notation $n!=1cdot2cdots n$. We obtain
begin{align*}
color{blue}{prod_{i=2}^{15}f(i)}&=prod_{i=2}^{15}left(1-frac{1}{i^2}right)\
&=prod_{i=2}^{15}frac{i^2-1}{i^2}\
&=prod_{i=2}^{15}frac{(i-1)(i+1)}{i^2}\
&=frac{14!cdot 16!/2}{left(15!right)^2}tag{1}\
&,,color{blue}{=frac{8}{15}}tag{2}
end{align*}
Comment:
In (1) we use $prod_{i=2}^{15}(i-1)=1cdot 2cdots 14=14!$ and $prod_{i=2}^{15}(i+1)=3cdot4cdots 16=frac{1}{2}cdot 16!$.
In (2) we see that $14!/15!=1/15$ and $16!/15!=16$.
answered Jan 2 at 19:40
Markus ScheuerMarkus Scheuer
60.5k455144
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $f(x) = frac{x^2-1}{x^2} = frac{(x-1)(x+1)}{x^2}$
and take product of $f(2),f(3),...,f(n)$
Look:
$$
frac{(1)(3)}{2^2} cdot frac{(2)(4)}{3^2} cdot frac{(3)(5)}{4^2} cdot ... frac{(n-3)(n-1)}{(n-2)^2} cdot frac{(n-2)n}{(n-1)^2} cdot frac{(n-1)(n+1)}{n^2} = frac{1}{2} cdot frac{n+1}{n} $$
and take this for your task
answered Jan 2 at 17:15
VirtualUserVirtualUser
59412
$endgroup$
$begingroup$
That does not really help.
$endgroup$
– Enzo
Jan 2 at 17:16
1
$begingroup$
@Enzo Perhaps you should try harder to understand it. It is the same hint I would have given; I don't think there is an easier way.
$endgroup$
– David K
Jan 2 at 17:18
$begingroup$
Write as I said and look what will happen
$endgroup$
– VirtualUser
Jan 2 at 17:19
$begingroup$
@Enzo I'll go out on a limb and say that there is no easier way.
$endgroup$
– John
Jan 2 at 17:20
$begingroup$
$$biggr (frac{(2-1)(2+1)}{2^2} biggr ) biggr ( frac{(3-1)(3+1)}{3^2} biggr )$$ No such term cancels out here.
$endgroup$
– Enzo
Jan 2 at 17:22
|
show 2 more comments
$begingroup$
Hint: $f(x) = frac{x^2-1}{x^2} = frac{(x-1)(x+1)}{x^2}$
and take product of $f(2),f(3),...,f(n)$
Look:
$$
frac{(1)(3)}{2^2} cdot frac{(2)(4)}{3^2} cdot frac{(3)(5)}{4^2} cdot ... frac{(n-3)(n-1)}{(n-2)^2} cdot frac{(n-2)n}{(n-1)^2} cdot frac{(n-1)(n+1)}{n^2} = frac{1}{2} cdot frac{n+1}{n} $$
and take this for your task
answered Jan 2 at 17:15
VirtualUserVirtualUser
59412
$endgroup$
$begingroup$
That does not really help.
$endgroup$
– Enzo
Jan 2 at 17:16
1
$begingroup$
@Enzo Perhaps you should try harder to understand it. It is the same hint I would have given; I don't think there is an easier way.
$endgroup$
– David K
Jan 2 at 17:18
$begingroup$
Write as I said and look what will happen
$endgroup$
– VirtualUser
Jan 2 at 17:19
$begingroup$
@Enzo I'll go out on a limb and say that there is no easier way.
$endgroup$
– John
Jan 2 at 17:20
$begingroup$
$$biggr (frac{(2-1)(2+1)}{2^2} biggr ) biggr ( frac{(3-1)(3+1)}{3^2} biggr )$$ No such term cancels out here.
$endgroup$
– Enzo
Jan 2 at 17:22
|
show 2 more comments
$begingroup$
Hint: $f(x) = frac{x^2-1}{x^2} = frac{(x-1)(x+1)}{x^2}$
and take product of $f(2),f(3),...,f(n)$
Look:
$$
frac{(1)(3)}{2^2} cdot frac{(2)(4)}{3^2} cdot frac{(3)(5)}{4^2} cdot ... frac{(n-3)(n-1)}{(n-2)^2} cdot frac{(n-2)n}{(n-1)^2} cdot frac{(n-1)(n+1)}{n^2} = frac{1}{2} cdot frac{n+1}{n} $$
and take this for your task
answered Jan 2 at 17:15
VirtualUserVirtualUser
59412
$endgroup$
Hint: $f(x) = frac{x^2-1}{x^2} = frac{(x-1)(x+1)}{x^2}$
and take product of $f(2),f(3),...,f(n)$
Look:
$$
frac{(1)(3)}{2^2} cdot frac{(2)(4)}{3^2} cdot frac{(3)(5)}{4^2} cdot ... frac{(n-3)(n-1)}{(n-2)^2} cdot frac{(n-2)n}{(n-1)^2} cdot frac{(n-1)(n+1)}{n^2} = frac{1}{2} cdot frac{n+1}{n} $$
and take this for your task
answered Jan 2 at 17:15
VirtualUserVirtualUser
59412
edited Jan 2 at 17:23
answered Jan 2 at 17:15
VirtualUserVirtualUser
59412
answered Jan 2 at 17:15
VirtualUserVirtualUser
59412
answered Jan 2 at 17:15
VirtualUserVirtualUser
59412
59412
$begingroup$
That does not really help.
$endgroup$
– Enzo
Jan 2 at 17:16
1
$begingroup$
@Enzo Perhaps you should try harder to understand it. It is the same hint I would have given; I don't think there is an easier way.
$endgroup$
– David K
Jan 2 at 17:18
$begingroup$
Write as I said and look what will happen
$endgroup$
– VirtualUser
Jan 2 at 17:19
$begingroup$
@Enzo I'll go out on a limb and say that there is no easier way.
$endgroup$
– John
Jan 2 at 17:20
$begingroup$
$$biggr (frac{(2-1)(2+1)}{2^2} biggr ) biggr ( frac{(3-1)(3+1)}{3^2} biggr )$$ No such term cancels out here.
$endgroup$
– Enzo
Jan 2 at 17:22
|
show 2 more comments
$begingroup$
That does not really help.
$endgroup$
– Enzo
Jan 2 at 17:16
1
$begingroup$
@Enzo Perhaps you should try harder to understand it. It is the same hint I would have given; I don't think there is an easier way.
$endgroup$
– David K
Jan 2 at 17:18
$begingroup$
Write as I said and look what will happen
$endgroup$
– VirtualUser
Jan 2 at 17:19
$begingroup$
@Enzo I'll go out on a limb and say that there is no easier way.
$endgroup$
– John
Jan 2 at 17:20
$begingroup$
$$biggr (frac{(2-1)(2+1)}{2^2} biggr ) biggr ( frac{(3-1)(3+1)}{3^2} biggr )$$ No such term cancels out here.
$endgroup$
– Enzo
Jan 2 at 17:22
$begingroup$
That does not really help.
$endgroup$
– Enzo
Jan 2 at 17:16
$begingroup$
That does not really help.
$endgroup$
– Enzo
Jan 2 at 17:16
1
1
$begingroup$
@Enzo Perhaps you should try harder to understand it. It is the same hint I would have given; I don't think there is an easier way.
$endgroup$
– David K
Jan 2 at 17:18
$begingroup$
@Enzo Perhaps you should try harder to understand it. It is the same hint I would have given; I don't think there is an easier way.
$endgroup$
– David K
Jan 2 at 17:18
$begingroup$
Write as I said and look what will happen
$endgroup$
– VirtualUser
Jan 2 at 17:19
$begingroup$
Write as I said and look what will happen
$endgroup$
– VirtualUser
Jan 2 at 17:19
$begingroup$
@Enzo I'll go out on a limb and say that there is no easier way.
$endgroup$
– John
Jan 2 at 17:20
$begingroup$
@Enzo I'll go out on a limb and say that there is no easier way.
$endgroup$
– John
Jan 2 at 17:20
$begingroup$
$$biggr (frac{(2-1)(2+1)}{2^2} biggr ) biggr ( frac{(3-1)(3+1)}{3^2} biggr )$$ No such term cancels out here.
$endgroup$
– Enzo
Jan 2 at 17:22
$begingroup$
$$biggr (frac{(2-1)(2+1)}{2^2} biggr ) biggr ( frac{(3-1)(3+1)}{3^2} biggr )$$ No such term cancels out here.
$endgroup$
– Enzo
Jan 2 at 17:22
|
show 2 more comments
$begingroup$
It is also convenient to use the factorial notation $n!=1cdot2cdots n$. We obtain
begin{align*}
color{blue}{prod_{i=2}^{15}f(i)}&=prod_{i=2}^{15}left(1-frac{1}{i^2}right)\
&=prod_{i=2}^{15}frac{i^2-1}{i^2}\
&=prod_{i=2}^{15}frac{(i-1)(i+1)}{i^2}\
&=frac{14!cdot 16!/2}{left(15!right)^2}tag{1}\
&,,color{blue}{=frac{8}{15}}tag{2}
end{align*}
Comment:
In (1) we use $prod_{i=2}^{15}(i-1)=1cdot 2cdots 14=14!$ and $prod_{i=2}^{15}(i+1)=3cdot4cdots 16=frac{1}{2}cdot 16!$.
In (2) we see that $14!/15!=1/15$ and $16!/15!=16$.
answered Jan 2 at 19:40
Markus ScheuerMarkus Scheuer
60.5k455144
$endgroup$
add a comment |
$begingroup$
It is also convenient to use the factorial notation $n!=1cdot2cdots n$. We obtain
begin{align*}
color{blue}{prod_{i=2}^{15}f(i)}&=prod_{i=2}^{15}left(1-frac{1}{i^2}right)\
&=prod_{i=2}^{15}frac{i^2-1}{i^2}\
&=prod_{i=2}^{15}frac{(i-1)(i+1)}{i^2}\
&=frac{14!cdot 16!/2}{left(15!right)^2}tag{1}\
&,,color{blue}{=frac{8}{15}}tag{2}
end{align*}
Comment:
In (1) we use $prod_{i=2}^{15}(i-1)=1cdot 2cdots 14=14!$ and $prod_{i=2}^{15}(i+1)=3cdot4cdots 16=frac{1}{2}cdot 16!$.
In (2) we see that $14!/15!=1/15$ and $16!/15!=16$.
answered Jan 2 at 19:40
Markus ScheuerMarkus Scheuer
60.5k455144
$endgroup$
add a comment |
$begingroup$
It is also convenient to use the factorial notation $n!=1cdot2cdots n$. We obtain
begin{align*}
color{blue}{prod_{i=2}^{15}f(i)}&=prod_{i=2}^{15}left(1-frac{1}{i^2}right)\
&=prod_{i=2}^{15}frac{i^2-1}{i^2}\
&=prod_{i=2}^{15}frac{(i-1)(i+1)}{i^2}\
&=frac{14!cdot 16!/2}{left(15!right)^2}tag{1}\
&,,color{blue}{=frac{8}{15}}tag{2}
end{align*}
Comment:
In (1) we use $prod_{i=2}^{15}(i-1)=1cdot 2cdots 14=14!$ and $prod_{i=2}^{15}(i+1)=3cdot4cdots 16=frac{1}{2}cdot 16!$.
In (2) we see that $14!/15!=1/15$ and $16!/15!=16$.
answered Jan 2 at 19:40
Markus ScheuerMarkus Scheuer
60.5k455144
$endgroup$
It is also convenient to use the factorial notation $n!=1cdot2cdots n$. We obtain
begin{align*}
color{blue}{prod_{i=2}^{15}f(i)}&=prod_{i=2}^{15}left(1-frac{1}{i^2}right)\
&=prod_{i=2}^{15}frac{i^2-1}{i^2}\
&=prod_{i=2}^{15}frac{(i-1)(i+1)}{i^2}\
&=frac{14!cdot 16!/2}{left(15!right)^2}tag{1}\
&,,color{blue}{=frac{8}{15}}tag{2}
end{align*}
Comment:
In (1) we use $prod_{i=2}^{15}(i-1)=1cdot 2cdots 14=14!$ and $prod_{i=2}^{15}(i+1)=3cdot4cdots 16=frac{1}{2}cdot 16!$.
In (2) we see that $14!/15!=1/15$ and $16!/15!=16$.
answered Jan 2 at 19:40
Markus ScheuerMarkus Scheuer
60.5k455144
answered Jan 2 at 19:40
Markus ScheuerMarkus Scheuer
60.5k455144
answered Jan 2 at 19:40
Markus ScheuerMarkus Scheuer
60.5k455144
answered Jan 2 at 19:40
Markus ScheuerMarkus Scheuer
60.5k455144
60.5k455144
add a comment |
add a comment |
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Have you looked carefully at the product of the first few terms? I think if you use a bit of algebra, you'll see that there is some cancellation between numerators and denominators of adjacent terms.
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– hardmath
Jan 2 at 17:11
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@hardmath That's absolutely true. I also have looked carefully at the product of first few terms, no doubt.
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– Enzo
Jan 2 at 17:13
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It would also depend on how you write those terms. If you just write $left(1 - frac1{2^2}right)left(1 - frac1{3^2}right)left(1 - frac1{4^2}right)cdots$ then I don't think you will see a useful pattern.
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– David K
Jan 2 at 17:16
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I have got $$frac{8}{15}$$ with my TI calculator
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– Dr. Sonnhard Graubner
Jan 2 at 17:18