Question

Solve this summation

Answer 1

\[\Huge{\sum\limits_{n_0=0}^4 \sum\limits_{n_1=0}^{n_0} \sum\limits_{n_2=0}^{n_1} = 1}\]

Answer 2

@mathstudent55?

Answer 3

Help, @perl @rational

Answer 4

hi :)

Answer 5

DO you know how to do it, @perl?

Answer 6

we can try to expand the left side

Answer 7

Ok

Answer 8

$$\Large{\sum\limits_{n_0=0}^4 \sum\limits_{n_1=0}^{n_0} \sum\limits_{n_2=0}^{n_1} =\\ \sum\limits_{n_1=0}^{0} \sum\limits_{n_2=0}^{n_1} + \sum\limits_{n_1=0}^{1} \sum\limits_{n_2=0}^{n_1}+ \sum\limits_{n_1=0}^{2} \sum\limits_{n_2=0}^{n_1}
+\sum\limits_{n_1=0}^{3} \sum\limits_{n_2=0}^{n_1}
\\ +\sum\limits_{n_1=0}^{4} \sum\limits_{n_2=0}^{n_1}+ \sum\limits_{n_1=0}^{5} \sum\limits_{n_2=0}^{n_1}
}

$$

Answer 9

that removes one sum, now try to remove another

Answer 10

are you allowed to use negative integers, thats the only way i can see how to solve this

Answer 11

Sure, go for it @perl

Answer 12

im having a little trouble understanding the question.
is the question to evaluate the sum on the left?

Answer 13

Actually, wait. Forget I put the = sign. They want us to solve that

Answer 14

ok that makes more sense

Answer 15

$$
\Large{\sum\limits_{n_0=0}^4 \sum\limits_{n_1=0}^{n_0} \sum\limits_{n_2=0}^{n_1} = 35}

$$

Answer 16

Exactly. How?

Answer 17

you can manually expand the double sum expression above

Answer 18

Woah. How do we solve that monstrosity?

Answer 19

$$
\large{\sum\limits_{n_0=0}^4 \sum\limits_{n_1=0}^{n_0} \sum\limits_{n_2=0}^{n_1} =

\\ \sum\limits_{n_1=0}^{0} \sum\limits_{n_2=0}^{n_1} + \sum\limits_{n_1=0}^{1} \sum\limits_{n_2=0}^{n_1}+ \sum\limits_{n_1=0}^{2} \sum\limits_{n_2=0}^{n_1}
+\sum\limits_{n_1=0}^{3} \sum\limits_{n_2=0}^{n_1}
\\ +\sum\limits_{n_1=0}^{4} \sum\limits_{n_2=0}^{n_1}+ \sum\limits_{n_1=0}^{5} \sum\limits_{n_2=0}^{n_1}
\\
=\sum\limits_{n_2=0}^{0} \\+
\sum\limits_{n_2=0}^{0}+ \sum\limits_{n_2=0}^{1} \\+
\sum\limits_{n_2=0}^{0} +\sum\limits_{n_2=0}^{1}
+\sum\limits_{n_2=0}^{2} \\
+ \sum\limits_{n_2=0}^{0} +\sum\limits_{n_2=0}^{1}
+\sum\limits_{n_2=0}^{2}+ \sum\limits_{n_2=0}^{3}\\
+ \sum\limits_{n_2=0}^{0} +\sum\limits_{n_2=0}^{1}
+\sum\limits_{n_2=0}^{2}+ \sum\limits_{n_2=0}^{3} +\sum\limits_{n_2=0}^{4}\\
= 1 \\
+ 1 + 2\\
+ 1 + 2 + 3\\
+ 1 + 2 + 3 + 4\\
+ 1 + 2 + 3 + 4 + 5

}


$$

Answer 20

Ahh, I get it. Thanks alot @perl :)

Answer 21

what you can do is , imagine there is a 1 inside the sum, since there is nothing there

Answer 22

Ok

Answer 23

example:
$$
\Large \sum_{0}^{2} = \sum_{0}^{2}(1) =1 + 1 + 1

$$
note that we count from zero. 0,1,2 is three numbers

Answer 24

the logic might be clearer here. we can pull (factor out) constants inside or outside the sum operator.
$$
\Large \sum_{0}^{2}=1\cdot \sum_{0}^{2} = \sum_{0}^{2}(1) =1 + 1 + 1

$$
note that
$$

\Large \sum_{0}^{0}=1\cdot \sum_{0}^{0} = \sum_{0}^{0}(1)= 1
$$
since you counted the index 0 only once, zero to zero

Answer 25

I get it. So it adds one to how many there is on the top - 1?

Answer 26

More generally
$$
\Large{
\sum_{m}^{n}a = (n-m + 1)\cdot a\\
\sum_{m}^{n}1 = (n-m + 1)\cdot 1
}
$$

Answer 27

where a is a constant

Answer 28

and that is because you are counting starting from m:
\( \bf m, m+1, m+2, ... ,n-1, n \)
How many indices did you count?

Answer 29

What do you mean, indices?

Answer 30

i call these indices , here 2 and 5
$$\Large \sum_{2}^{5} $$

Answer 31

i guess you can call them limits too :)

Answer 32

Ahh, that makes sense

Answer 33

but the limit need not start at zero.
this is logical as well

$$

\Large \sum_{2}^{5} (1) = 1 + 1 + 1 + 1

$$

Answer 34

the number of times to add 1 you count by starting with 2, then 3, then 4, then 5. So a total of 4 times you are summing 1

Answer 35

this problem is great practice on double sums , and i've never seen a triple sum used (outside of calculus)

Answer 36

I see. So, for :\[\Huge{\sum_{5}^{10}(2)}=2+2+2+2+2+2\]

Answer 37

note that to actually solve the problem above i entered the outermost sum index first , so i am working from the outside to the inside

Answer 38

and yes thats correct :)

Answer 39

$$

\Large {\sum_{5}^{5}(2)}=2
$$

Answer 40

Yay. But how do you solve when both indices are \(n_1 \text{ & } n_2\)

Answer 41

you mean like in the problem above ?

Answer 42

Yeah

Answer 43

if both indices are unknown, (and you have only one sum) , then you can't solve it.
but note that in the above problem, you have sums that are outside .
and you do the outermost sum first

Answer 44

Wow. That makes way more sense. You litrally added the summation 5 times, because the difference in position from4 to 0 is 5?

Answer 45

start with the outermost Sum, n=0 to 4.
now count starting with zero, then 1, then 2,
$$
\Large {\sum\limits_{\color{red}{n_0}= \color{red}0}^\color{red}4 \sum\limits_{n_1=0}^{\color{red}{n_0}} \sum\limits_{n_2=0}^{n_1}
\\
= \sum\limits_{n_1=0}^{\color{red}0} \sum\limits_{n_2=0}^{n_1} + \sum\limits_{n_1=0}^{\color{red}1} \sum\limits_{n_2=0}^{n_1}+ \sum\limits_{n_1=0}^{\color{red}2} \sum\limits_{n_2=0}^{n_1}
+\sum\limits_{n_1=0}^{\color{red}3} \sum\limits_{n_2=0}^{n_1}
\\ +\sum\limits_{n_1=0}^{\color{red}4} \sum\limits_{n_2=0}^{n_1}


}

$$

Answer 46

exactly :)

Answer 47

Would it get more complicated if n_0 was something other than 0?

Answer 48

no, it can be anything. we can even solve this problem more generally

Answer 49

I actually made a typo above.
but let me fix that before solving the general problem

$$
\large{\sum\limits_{n_0=0}^4 \sum\limits_{n_1=0}^{n_0} \sum\limits_{n_2=0}^{n_1} =

\\ \sum\limits_{n_1=0}^{0} \sum\limits_{n_2=0}^{n_1} + \sum\limits_{n_1=0}^{1} \sum\limits_{n_2=0}^{n_1}+ \sum\limits_{n_1=0}^{2} \sum\limits_{n_2=0}^{n_1}
+\sum\limits_{n_1=0}^{3} \sum\limits_{n_2=0}^{n_1}
+\sum\limits_{n_1=0}^{4} \sum\limits_{n_2=0}^{n_1}\\
=\sum\limits_{n_2=0}^{0} \\+
\sum\limits_{n_2=0}^{0}+ \sum\limits_{n_2=0}^{1} \\+
\sum\limits_{n_2=0}^{0} +\sum\limits_{n_2=0}^{1}
+\sum\limits_{n_2=0}^{2} \\
+ \sum\limits_{n_2=0}^{0} +\sum\limits_{n_2=0}^{1}
+\sum\limits_{n_2=0}^{2}+ \sum\limits_{n_2=0}^{3}\\
+ \sum\limits_{n_2=0}^{0} +\sum\limits_{n_2=0}^{1}
+\sum\limits_{n_2=0}^{2}+ \sum\limits_{n_2=0}^{3} +\sum\limits_{n_2=0}^{4}\\
= 1 \\
+ 1 + 2\\
+ 1 + 2 + 3\\
+ 1 + 2 + 3 + 4\\
+ 1 + 2 + 3 + 4 + 5

}

$$

Answer 50

I get it now. It makes it easier if we split it up, then solve individually, and then add the values together. Awesome :D

Answer 51

now if we want to solve it more generally, we can find a general formula.
and to make it even more fun we can add more sums. but lets keep it to three sums for now

Answer 52

$$
\Large{\sum\limits_{n_0=a}^b \sum\limits_{n_1=c}^{n_0} \sum\limits_{n_2=d}^{n_1} = 1} $$

there are some caveats or warnings here.
the sum is not defined if

Answer 53

If?

Answer 54

we want the bottom limit to be less than the top limit

Answer 55

ok i guess it could still be defined, but it can get tricky, for example what is

$$
\Large \sum_{5}^{2}(2) ~?

$$

Answer 56

What happens then?

Answer 57

$$
\Large {

\Large \sum_{5}^{5}(2) = 2\\

\Large \sum_{5}^{4}(2) = 2-2=0\\

\Large \sum_{5}^{3}(2) = 2-2-2=-2\\
}

$$

Answer 58

Ahh, Ok

Answer 59

so we avoid this problem by stipulating that a <= b
$$
\Large{\sum\limits_{n_0=a}^b \sum\limits_{n_1=c}^{n_0} \sum\limits_{n_2=d}^{n_1} = ?
\\ \text{ where }a \leq b
}$$

Answer 60

Ok

Answer 61

so the general solution will be solve similar to the above.
you can first look at the case when a = b , and that clearly will equal to just 1.

now the really cool result, can you find a closed form for k sums:

$$
\LARGE \sum\limits_{n_0=a_0}^b \sum\limits_{n_1=a_1}^{n_0} \sum\limits_{n_2=a_2}^{n_1} ...\sum\limits_{n_2=a_k}^{n_k}
\\


\Large \text{ where } a_n \leq a_{n-1} \leq a_{n-2} \leq ... \leq a_0 \leq b


$$

Answer 62

and this ties in with pascals triangle and binomial coefficients (you might have noticed a nice pattern above, the sum of triangular numbers)

Answer 63

typo
see if you find a closed form expression for k sums
$$
\LARGE \sum\limits_{n_0=a_0}^b \sum\limits_{n_1=a_1}^{n_0} \sum\limits_{n_2=a_2}^{n_1} ...\sum\limits_{n_2=a_k}^{n_k}
\\


\Large \text{ where } a_k \leq a_{k-1} \leq... \leq a_1 \leq a_0 \leq b


$$

Answer 64

and this is how math mushrooms from a small problem into a general problem.

Answer 65

I see. I was really confused when I was showed this by my teacher, since I never worked with Summations before

Answer 66

yes , its quite a remarkable notation

Answer 67

Yes it is @perl :). Can you help me with one more question.

Answer 68

ok

Answer 69

If \(f(x)=2x-1\)
Find:
\(f(f(f(f(f(x))))=1\)

Answer 70

let me start you out and you can 'see' the pattern

Answer 71

also there is a clever notation to condense this expression as well. mathematicians always seem to find the most condense way to express ideas

Answer 72

Cool

Answer 73

$$
\Large { f(x) = 2x - 1 \\
\\ \therefore \\
f(f(f(f(f(x)))) \\
= f(f(f(f(2x-1))) \\
= f(f(f(2\cdot (2x-1)-1 )) \\


}
$$

Answer 74

now you can simplify a bit before applying the next composition

Answer 75

4x-3

Answer 76

$$
\Large { f(x) = 2x - 1 \\
\\ \therefore \\
f(f(f(f(f(x))))) \\
= f(f(f(f(2x-1)))) \\
= f(f(f(2\cdot (2x-1)-1 ))) \\
= f(f(f(4x-2 - 1)))\\
= f(f(f(4x-3))) \\
= f(f(2(4x-3)-1)) \\
= f(f(8x - 6 -1 )) \\
= f(f(8x -7 )) \\
= f(2(8x -7 )-1) \\
= f(16x -14-1) \\
= f(16x -15) \\
= 2(16x - 15) - 1 \\
= 32x - 30 -1 \\
= 32x - 31

\\
\\


}



$$

Answer 77

I see

Answer 78

the cool notation that condenses the expression:
$$
\Large f(f(f(f(f(x))))) = f^{(n)}(x)

$$

Answer 79

http://en.wikipedia.org/wiki/Iterated_function

Answer 80

I see. That makes it much more readable. Is there a simpler way of solving this without simply substitutiing over and over?

Answer 81

i dont know of any way, but if you were asked to find f^100 (x) , then you might start looking for patterns . or use a computer to iterate it

Answer 82

actually the notation i used isnt quite accurate, with parentheses that can refer to the n'th derivative

Answer 83

if you look here
http://en.wikipedia.org/wiki/Taylor_series

Answer 84

I see.

Answer 85

$$

\Large{ f^{(n)}(a)\\ \text{ denotes the nth derivative of f evaluated}\\ \text{ at the point a, } \text{but}\\
f^{n} \\
\text{ denotes the nth iteration of f : }\\ f(f(f(...f(x)...)))



}




$$

Answer 86

\[\sum_{10}^{5}2 = 12 ????\]

Answer 87

reverse the 5 and 10

Answer 88

is that correct??