Series with x^y?
- Thread starter Damoo
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Hello,
Can anybody please be so kind to show me how to calculate this series?
thanks
I think you mis-stated the problem the problem should be:
\(\displaystyle \sum_{x=1}^{10}{x^3} \ = \ ? \)
It is not a geometric series
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Hello, Damoo!
\(\displaystyle \text{[1] Direct addition: }\:1^3 + 2^3+ 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3 + 10^3\)
. . . \(\displaystyle =\; 1 + 8 + 27 + 64 + 125 + 216 + 343 + 512 + 729 + 1000 \;=\;3025\)
\(\displaystyle \displaystyle\text{[2] Formula: }\:\sum^n_{k=1}k^3 \:=\:\frac{n^2(n+1)^2}{4}\)
\(\displaystyle \text{For }n =10\!:\;S \;=\;\dfrac{10^2\cdot11^2}{4} \;=\;3025\)
\(\displaystyle \displaystyle\text{[3] It happens that: }\:\sum^{10}_{x=1}x^3 \;=\;\left(\sum^{10}_{x=1} x\right)^2\)
\(\displaystyle \text{That is: }\:1^2 + 2^3 + 3^3 + \cdots + 10^3 \;=\;(1+2+3+\cdots + 10)^2\)
. . . . . . . . \(\displaystyle =\;55^2 \;=\;3025\)
I know of three ways to calculate this sum.
\(\displaystyle \text{[1] Direct addition: }\:1^3 + 2^3+ 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3 + 10^3\)
. . . \(\displaystyle =\; 1 + 8 + 27 + 64 + 125 + 216 + 343 + 512 + 729 + 1000 \;=\;3025\)
\(\displaystyle \displaystyle\text{[2] Formula: }\:\sum^n_{k=1}k^3 \:=\:\frac{n^2(n+1)^2}{4}\)
\(\displaystyle \text{For }n =10\!:\;S \;=\;\dfrac{10^2\cdot11^2}{4} \;=\;3025\)
\(\displaystyle \displaystyle\text{[3] It happens that: }\:\sum^{10}_{x=1}x^3 \;=\;\left(\sum^{10}_{x=1} x\right)^2\)
\(\displaystyle \text{That is: }\:1^2 + 2^3 + 3^3 + \cdots + 10^3 \;=\;(1+2+3+\cdots + 10)^2\)
. . . . . . . . \(\displaystyle =\;55^2 \;=\;3025\)
There is a fourth way to express the sum, namely
\(\displaystyle \dbinom{n + 1}{n - 1}^2.\)
Of course, computationally \(\displaystyle \dbinom{n + 1}{n - 1}^2 = \left(\dfrac{(n + 1)!}{(n - 1)! * \{(n + 1) - (n - 1)\}}\right)^2 = \left(\dfrac{n(n + 1)}{2}\right)^2.\)
For this portion, it could also be done as:
\(\displaystyle \displaystyle\sum^{10}_{x=1} \ x \ =\)
\(\displaystyle \int_{0.5}^{10.5} x \ dx \ =\)
\(\displaystyle \dfrac{1}{2}\bigg\{x^2 \ \bigg|_{0.5}^{10.5}\bigg\} \ =\)
\(\displaystyle \frac{1}{2}(110.25 - .25) \ = \)
\(\displaystyle \frac{1}{2}(110) \ = \)
\(\displaystyle \text{55}\)
\(\displaystyle \displaystyle\sum^{10}_{x=1} \ x \ =\)
\(\displaystyle \int_{0.5}^{10.5} x \ dx \ =\)
\(\displaystyle \dfrac{1}{2}\bigg\{x^2 \ \bigg|_{0.5}^{10.5}\bigg\} \ =\)
\(\displaystyle \frac{1}{2}(110.25 - .25) \ = \)
\(\displaystyle \frac{1}{2}(110) \ = \)
\(\displaystyle \text{55}\)
Since we're all having fun with different answers,
\(\displaystyle \displaystyle \sum_{x=1}^{10} x^3 = \left(\dfrac{\int_0^{\infty} \left(\alpha/\sqrt{2}\right)\cdot \alpha^{10}e^{-\alpha}d\alpha}{\int_0^{\infty} \left(\alpha/\sqrt{2}\right)^{-1}\cdot \alpha^{10}e^{-\alpha}d\alpha}\right)^2\)
\(\displaystyle \displaystyle \sum_{x=1}^{10} x^3 = \left(\dfrac{\int_0^{\infty} \left(\alpha/\sqrt{2}\right)\cdot \alpha^{10}e^{-\alpha}d\alpha}{\int_0^{\infty} \left(\alpha/\sqrt{2}\right)^{-1}\cdot \alpha^{10}e^{-\alpha}d\alpha}\right)^2\)
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Wow... never even knew there were so many ways. Math is hard....
I got it now though, thanks everybody!
For anybody curious
The specific way my prof was looking for was using factorial powers and finite differences.
Something like...
Where
=
=
(as a sum of factorial powers)
Then using
= Δ
and
g(x)=g(h+1)-g(x).
Dunno what the formula is exactly called but it seemed long and needlessly long.
I got it now though, thanks everybody!
For anybody curious
The specific way my prof was looking for was using factorial powers and finite differences.
Something like...
Where
=
Then using
Dunno what the formula is exactly called but it seemed long and needlessly long.
HallsofIvy
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One method of dealing with problems of this sort, in general, is "Newton's divided difference" formula. First evaluate \(\displaystyle \sum_{x=1}^n x^3\) for a number of different values of n: n= 0; sum= 0; n= 1, sum= 1; n= 2, sum= 1+ 8= 9; n= 3, sum= 1+ 8+ 27= 36; n= 4, sum= 1+ 8+ 27+ 64= 100; n= 5, sum= 1+ 8+ 27+ 64+ 125= 225; n= 6, sum= 1+ 8+ 27+ 64+ 125+ 216= 441
(Actually, had you done this initially, you should have realized you were getting squares! n= 1, the sum is 12, n= 2, the sum is 32, n= 3, the sum is 62, etc. Can you find the formula for the base?)
The "first difference", \(\displaystyle \Delta f\), of those is is \(\displaystyle \Delta_n= f(n+1)- f(n)\) which gives: n= 0, 1-0= 1, n= 1, 9- 1= 8; n= 2, 36- 9= 27; n= 3, 100- 36= 64; n= 4, 225- 100= 125; n= 5, 441- 225= 216.
The "second difference", \(\displaystyle \Delta^2 f\), is the difference between those first differences: n= 0, 8- 1= 7; n= 1, 27- 8= 19; n= 2, 64- 27= 37;n= 3, 125- 64= 61;n= 4, 216- 125= 91.
The "third difference", \(\displaystyle \Delta^3 f\), is the difference between those second differences: n= 0, 19- 7= 12; n= 1, 37- 19= 18; n= 2, 61- 37= 24; n= 3, 91= 61= 30.
The "fourth difference", \(\displaystyle \Delta^4 f\), is the difference between those third differences: n= 0, 18- 12= 6; n= 1, 24- 18= 6; n= 2, 30- 24= 6.
Here, we can stop: those are all equal to 6 so the "fourth difference", and any higher differences, are all 0 (and it should be easy to see that if we continued beyond n= 6 in the original list, we would get the same result).
"Newton's divided difference" formula says that if f(x) has nth divided difference at 0 of \(\displaystyle \Delta^n f(0)\), then, for integer x, \(\displaystyle f(x)= f(0)+ (\Delta f)x+ \left(\frac{\Delta^2 f(0)}{2}\right)x(x-1)+ (\frac{\Delta^3f(0)}{3!}x(x-1)(x-2)+ \cdot\cdot\cdot\). Because all differences beyond \(\displaystyle \Delta^4\) are 0, this will give a fourth degree polynomial for f.
Here, f(0)= 0, \(\displaystyle \Delta f(0)= 1\), \(\displaystyle \Delta^2 f(0)= 7\), \(\displaystyle \Delta^3 f(0)= 12\), \(\displaystyle \Delta^4 f(0)= 6\), and all higher diffences are 0 so we have
\(\displaystyle f(x)= 0+ 1(x)+ (7/2)x(x-1)+ (12/6)x(x-1)(x-2)+ (6/24)x(x-1)(x-2)(x-3)\)
a fourth degree polynomial. You can multiply that out to find the polynomial in "standard form".
Yet another method: The logic used to find the Newton divided difference polynomial indicates that if we have a sum of terms \(\displaystyle x^n\), then the sum itself will be a polynomial of degree n+1. Since, here, we have a sum of \(\displaystyle x^3\), the sum must be a polynomial of degree 4. That is, it can be written as \(\displaystyle f(x)= Ax^4+ Bx^3+ Cx^2+ Dx+ E\) and we only need to find the 5 numbers A, B, C, D, and E.
We can get five equations to do that by using the values of the sum:
x= 0 gave \(\displaystyle f(0)= A(0)^4+ B(0)^3+ C(0)^2+ D(0)+ E= 0\) so E= 0[/tex]
x= 1 gave \(\displaystyle f(1)= A(1)^4+ B(1)^3+ C(1)^2+ D(1)+ E= A+ B+ C+ D= 1\)
x= 2 gave \(\displaystyle f(2)= A(2)^4+ B(2)^3+ C(2)^2+ D(2)+ E= 16A+ 8B+ 4C+ 2D= 9\)
x= 3 gave \(\displaystyle f(3)= A(3)^4+ B(3)^3+ C(3)^2+ D(3)+ E= 64A+ 27B+ 9C+ 3D= 36\)
x= 4 gave \(\displaystyle f(4)= A(4)^4+ B(4)^3+ C(4)^3+ D(4)+ E= 256A+ 64B+ 16C+ 4D= 125\)
Solve those four equations for A, B, C, and D.
(Actually, had you done this initially, you should have realized you were getting squares! n= 1, the sum is 12, n= 2, the sum is 32, n= 3, the sum is 62, etc. Can you find the formula for the base?)
The "first difference", \(\displaystyle \Delta f\), of those is is \(\displaystyle \Delta_n= f(n+1)- f(n)\) which gives: n= 0, 1-0= 1, n= 1, 9- 1= 8; n= 2, 36- 9= 27; n= 3, 100- 36= 64; n= 4, 225- 100= 125; n= 5, 441- 225= 216.
The "second difference", \(\displaystyle \Delta^2 f\), is the difference between those first differences: n= 0, 8- 1= 7; n= 1, 27- 8= 19; n= 2, 64- 27= 37;n= 3, 125- 64= 61;n= 4, 216- 125= 91.
The "third difference", \(\displaystyle \Delta^3 f\), is the difference between those second differences: n= 0, 19- 7= 12; n= 1, 37- 19= 18; n= 2, 61- 37= 24; n= 3, 91= 61= 30.
The "fourth difference", \(\displaystyle \Delta^4 f\), is the difference between those third differences: n= 0, 18- 12= 6; n= 1, 24- 18= 6; n= 2, 30- 24= 6.
Here, we can stop: those are all equal to 6 so the "fourth difference", and any higher differences, are all 0 (and it should be easy to see that if we continued beyond n= 6 in the original list, we would get the same result).
"Newton's divided difference" formula says that if f(x) has nth divided difference at 0 of \(\displaystyle \Delta^n f(0)\), then, for integer x, \(\displaystyle f(x)= f(0)+ (\Delta f)x+ \left(\frac{\Delta^2 f(0)}{2}\right)x(x-1)+ (\frac{\Delta^3f(0)}{3!}x(x-1)(x-2)+ \cdot\cdot\cdot\). Because all differences beyond \(\displaystyle \Delta^4\) are 0, this will give a fourth degree polynomial for f.
Here, f(0)= 0, \(\displaystyle \Delta f(0)= 1\), \(\displaystyle \Delta^2 f(0)= 7\), \(\displaystyle \Delta^3 f(0)= 12\), \(\displaystyle \Delta^4 f(0)= 6\), and all higher diffences are 0 so we have
\(\displaystyle f(x)= 0+ 1(x)+ (7/2)x(x-1)+ (12/6)x(x-1)(x-2)+ (6/24)x(x-1)(x-2)(x-3)\)
a fourth degree polynomial. You can multiply that out to find the polynomial in "standard form".
Yet another method: The logic used to find the Newton divided difference polynomial indicates that if we have a sum of terms \(\displaystyle x^n\), then the sum itself will be a polynomial of degree n+1. Since, here, we have a sum of \(\displaystyle x^3\), the sum must be a polynomial of degree 4. That is, it can be written as \(\displaystyle f(x)= Ax^4+ Bx^3+ Cx^2+ Dx+ E\) and we only need to find the 5 numbers A, B, C, D, and E.
We can get five equations to do that by using the values of the sum:
x= 0 gave \(\displaystyle f(0)= A(0)^4+ B(0)^3+ C(0)^2+ D(0)+ E= 0\) so E= 0[/tex]
x= 1 gave \(\displaystyle f(1)= A(1)^4+ B(1)^3+ C(1)^2+ D(1)+ E= A+ B+ C+ D= 1\)
x= 2 gave \(\displaystyle f(2)= A(2)^4+ B(2)^3+ C(2)^2+ D(2)+ E= 16A+ 8B+ 4C+ 2D= 9\)
x= 3 gave \(\displaystyle f(3)= A(3)^4+ B(3)^3+ C(3)^2+ D(3)+ E= 64A+ 27B+ 9C+ 3D= 36\)
x= 4 gave \(\displaystyle f(4)= A(4)^4+ B(4)^3+ C(4)^3+ D(4)+ E= 256A+ 64B+ 16C+ 4D= 125\)
Solve those four equations for A, B, C, and D.
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Nature and its laws
Lay hid in night.
God said, "Let Newton be,"
And all was light.
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