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# summation

# Sigma notation

*Sigma notation* is often used to write complicated sums in a concise and compact way. One of the most common forms is

$\sum_{{n=a}}^{b}f(n).$ | (1) |

The $\sum$ symbol (capital sigma) means *sum* and the expression (1) is equivalent to the sum

$f(a)+f(a+1)+f(a+2)+\cdots+f(b)$ |

The starting and stopping values are written below and above the $\sum$ symbol respectively, and below we also specify which will be our *running variable* (or *summation index*) that will be changing values. So in the former expression, $n$ is the running variable, taking values starting at $a$ and stopping at $b$.
Usually it’s assumed that $a\leq b$ in (1) since otherwise there would be no summands. However it is also customary to take the value of the sum as zero when $a>b$.

For instance, if we wanted to represent the sum of all integers between $1$ and $100$ we would write:

$\sum_{{j=1}}^{{100}}j.$ |

here we are taking $j$ as summation index, and we are adding $f(j)=j$ where $j$ runs over all the integers starting at $1$ and ending at $100$.

Now suppose we wanted to represent the sum $4^{2}+5^{2}+6^{2}+\cdots+20^{2}$. The straightforward way to convert it to sigma notation is

$\sum_{{k=4}}^{{20}}k^{2}.$ |

However, it must be noted that such representation is not unique. For instance, here is another way to express the same summation:

$\sum_{{k=0}}^{{16}}(k+4)^{2}.$ |

If we wanted to sum all positive even numbers not greater than $50$ we would write

$\sum_{{k=1}}^{{25}}2k$ |

since if we substitute $k=1,2,3,\ldots,25$ into $f(k)=2k$ we get $2,4,6,8,\ldots,50$. Now, if we wanted to write all odd numbers not greater than $50$ we would write

$\sum_{{k=1}}^{{25}}(2k-1)\qquad\mbox{or}\qquad\sum_{{k=0}}^{{24}}(2k+1).$ |

It must be noted that, although the running variable usually takes integer values, the summation function needs not, and it can lie on any algebraic structure where a sum is defined. So, we can write $\sum_{{j=1}}^{{10}}\sqrt{j}$ for representing $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{10}$ even though the summing terms aren’t integers. If we wanted to sum all the fifth roots of unity (complex numbers) we could write $\sum_{{k=0}}^{4}e^{{2\pi ik/5}}$.

There are several variations to the notation. Often the starting and ending values for the running parameter are ommited if the set of values it can take is not relevant or is understood from context. So on some contexts one could see the sum

$\sum_{{k=1}}^{\infty}\frac{1}{2^{k}}$ |

written as $\sum_{k}\frac{1}{2^{k}}$ where the summation is understood from context to be done over all positive integers.

Also notice another variation in the preceding example: the upper limit is $\infty$, which means the sum doesn’t stop after some terms. In other words, the previous example represents the sum

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$ |

Another variation is to give further specifications for the allowed values of the running parameter. So, if we wanted to sum if we wanted to sum the reciprocals of all prime numbers we could simply write

$\sum_{{p\ \mathrm{prime}}}\frac{1}{p},$ |

where we also use the previous variation where omitting the omiting starting and stopping values indicates summing over all possible values the context allows. Now, if we wanted to sum all prime numbers between $1$ and $50$ we have choices:

$\sum_{{p=1\atop p\ \mathrm{prime}}}^{{50}}p,\qquad\sum_{{p\ \mathrm{prime}% \atop p\leq 50}}p.$ |

# Evaluating sums

Since we are working with sums, all the usual properties (commutativity, associativity, etc) hold. Two of the most useful formulas for dealing with summations are

$\displaystyle\sum_{k}(a_{k}+b_{k})$ | $\displaystyle=\Big(\sum a_{k}\Big)+\Big(\sum b_{k}\Big)$ | (2) | |

$\displaystyle\sum_{k}(c\cdot a_{k})$ | $\displaystyle=c\cdot\sum a_{k}.$ | (3) |

The first property allows us to split summations into simpler sums, and the second tells us we can factor out any constant term (actually, any expression not involving the summation index).

We have also a “changing limits” property, which we used to give two different expressions for $4^{2}+5^{2}+6^{2}+\cdots+20^{2}$:

$\sum_{{k=a}}^{b}f(k)=\sum_{{k=a+r}}^{{b+r}}f(k-r)$ | (4) |

where $r$ is any integer.

As example, supose we wanted to sum all the values for $x^{2}+2x+1$ when $x$ runs from $3$ to $7$. That is, we wish to evaluate $\sum_{{j=3}}^{7}j^{2}+2j+1$. By using the mentioneds properties we have

$\displaystyle\sum(j^{2}+2j+1)$ | $\displaystyle=\sum j^{2}+\sum 2j+\sum 1$ | |

$\displaystyle=\sum j^{2}+2\sum j+\sum 1$ |

where context indicates that in all previous sums, $j$ runs from $3$ to $7$. The problem now is reduced to find formulas for the sums in the last expression.

Notice also that, since $(x^{2}+2x+1)$ is the same as $(x+1)^{2}$ we could have also done the following changes:

$\displaystyle\sum_{{j=3}}^{7}(j^{2}+2j+1)$ | $\displaystyle=\sum_{{j=3}}^{7}(j+1)^{2}$ | |

$\displaystyle=\sum_{{j=4}}^{8}j^{2}.$ |

and again we are left with the task of evaluating a simpler summation.

# Useful formulas

We now give formulas for evaluating many common summations, which can be combined using the mentioned properties to evaluate a wide range of sums.

Constants. The simplest case is when the summation terms do not involve the running variable. Two examples are:

$\sum_{{k=1}}^{4}2,\qquad\sum_{{k=1}}^{4}x^{2},$ |

which respectively represent the sums $2+2+2+2$ and $x^{2}+x^{2}+x^{2}+x^{2}$. It’s obvious that if the summand does not depend on the running variable, all terms will be the same, and thus the sum will be the product of any summand by the nuber of summands. In other words,

$\sum_{{k=1}}^{n}c=nc.$ |

Remark: If a summand does not depend on the summation index, we say it is constant (with respect the summation). So in the previous example $x^{2}$ was “constant” since it didn’t depend on the running variable $k$, and thus

$\sum_{{k=1}}^{4}x^{2}=4x^{2}.$ |

Small powers. Suppose we wanted to calculate $1+2+3+\cdots+98+99+100$. In other words, we want to calculate $\sum_{{k=1}}^{{100}}k$. We can use the formula

$\sum_{{k=1}}^{n}k=\frac{n(n+1)}{2}$ |

which is credited to Gauss. Applying such formula we find that the sought answer is $100(101)/2=5050$. Notice that we can use the formula to evaluate sums of consecutive integers not necessarily at $1$. For instance, if we wanted $50+51+52+\cdots+77$ we could proceed as following:

$\displaystyle\sum_{{k=50}}^{{77}}k=50+51+52+\cdots+77$ | $\displaystyle=(1+2+\cdots+77)-(1+2+\cdots+49)$ | |

$\displaystyle=\sum_{{k=1}}^{{77}}k-\sum_{{k=1}}^{{49}}k$ | ||

$\displaystyle=\frac{77\cdot 78}{2}-\frac{49\cdot 50}{2}$ | ||

$\displaystyle=3003-225=1778$ |

Similar formulas exist for evaluating sums of small powers of consecutive integers:

$\displaystyle\sum_{{k=1}}^{n}k^{2}$ | $\displaystyle=\frac{n(n+1)(2n+1)}{6}$ | |

$\displaystyle\sum_{{k=1}}^{n}k^{3}$ | $\displaystyle=\frac{n^{2}(n+1)^{2}}{4}$ | |

$\displaystyle\sum_{{k=1}}^{n}k^{4}$ | $\displaystyle=\frac{n(n+1)(2n+1)(3n^{2}+3n-1)}{30}$ | |

$\displaystyle\sum_{{k=1}}^{n}k^{5}$ | $\displaystyle=\frac{n^{2}(n+1)^{2}(2n^{2}+2n-1)}{12}$ | |

$\displaystyle\sum_{{k=1}}^{n}k^{6}$ | $\displaystyle=\frac{n(n+1)(2n+1)(3n^{4}+6n^{3}-3n+1)}{42}$ | |

$\displaystyle\sum_{{k=1}}^{n}k^{7}$ | $\displaystyle=\frac{n^{2}(n+1)^{2}(3n^{4}+6n^{3}-n^{2}-4n+2)}{24}$ | |

$\displaystyle\sum_{{k=1}}^{n}k^{8}$ | $\displaystyle=\frac{n(n+1)(2n+1)(5n^{6}+15n^{5}+5n^{4}-15n^{3}-n^{2}+9n-3)}{90}$ | |

$\displaystyle\sum_{{k=1}}^{n}k^{9}$ | $\displaystyle=\frac{n^{2}(n+1)^{2}(n^{2}+n-1)(2n^{4}+4n^{3}-n^{2}-3n+3)}{20}$ | |

$\displaystyle\sum_{{k=1}}^{n}k^{{10}}$ | $\displaystyle=\frac{n(n+1)(2n+1)(n^{2}+n-1)(3n^{6}+9n^{5}+2n^{4}-11n^{3}+3n^{2% }+10n-5)}{66}$ |

Arithmetic progressions.
The *arithmetic progression* is a sequence where the difference of each term and the previous is always the same. In other words, is of the form

$a,a+d,a+2d,\ldots,a+nd$ |

(notice that the previous example has $n+1$ terms).

The corresponding sum $a+(a+d)+(a+2d)+\cdots+(a+nd)$ can be written with sigma notation as $\sum_{{k=0}}^{n}(a+kd)$. We can use all formulas we have so far to calculate it:

$\displaystyle\sum_{{k=0}}^{n}(a+kd)$ | $\displaystyle=\sum_{{k=0}}^{n}a+\sum_{{k=0}}^{n}kd$ | |

$\displaystyle=(n+1)a+d\sum_{{k=0}}^{n}k$ | ||

$\displaystyle=(n+1)a+d\frac{n(n+1)}{2}=\frac{(n+1)(dn+2a)}{2}$ |

If $a_{0},a_{1},a_{2},\ldots,a_{n}$ represent the terms of the progression, then we can also rewrite the last result as

$\sum_{{k=0}}^{n}(a+kd)=\frac{(n+1)(dn+2a)}{2}=\frac{(n+1)(a_{0}+a_{n})}{2}$ |

A particular case of arithmetic sequence is summing consecutive odd (or consecutive even ) numbers, since the common difference is always $2$.

$\displaystyle\sum_{{k=1}}^{n}2k=2\sum_{{k=1}}^{n}=n(n+1)$ | $\displaystyle\quad\mbox{even numbers}$ | ||

$\displaystyle\sum_{{k=1}}^{n}(2k-1)=2\sum_{{k=1}}^{n}k-\sum_{{k=1}}^{n}1=n(n+1% )-n=n^{2}$ | $\displaystyle\quad\mbox{odd numbers}$ |

Geometric progressions.The *geometric progression* is a sequence where the quotient between each term and the previous is always the same. in other words, has the form:

$a,ar,ar^{2},ar^{3},\ldots,ar^{n}.$ |

The corresponding sum $a+ar+ar^{2}+ar^{3}+\cdots+ar^{n}$ can be reduced to calculating $1+r+r^{2}+\cdots+r^{n}$ by factoring $a$ out of the summation. The corresponding formula is

$\sum_{{k=0}}^{n}r^{k}=1+r+r^{2}+\cdots+r^{n}=\frac{r^{{n+1}}-1}{r-1}$ |

A particularly nice formula is obtained when $r=2$:

$\sum_{{k=0}}^{n}2^{k}=2^{{n+1}}-1.$ |

So, in the old story about a chess board where the board is filled doubling the number of seeds on the previous position, the answer would be

$\sum_{{k=0}}^{{63}}2^{k}=2^{{64}}-1=18446744073709551615.$ |

When the common quotient $r$ has absolute value smaller than $1$, we can actually calculate the infinite series:

$\sum_{{k=0}}^{\infty}r^{k}=1+r+r^{2}+r^{3}+\cdots=\frac{1}{1-r}\qquad\mathrm{% for\ }|r|<1.$ |

Other formulas. Many other formulas can be found to evaluate sums. Here is a small miscellanea of remarkable formulas.

$\displaystyle\sum_{{k=1}}^{n}k(k+1)=\frac{n(n+1)(n+2)}{3}$ | ||

$\displaystyle\sum_{{k=1}}^{n}k(k+1)(k+2)=\frac{n(n+1)(n+2)(n+3)}{4}$ | ||

$\displaystyle\sum_{{k=1}}^{n}k(k+1)(k+2)(k+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$ |

$\displaystyle\sum_{{k=0}}^{n}{n\choose k}=2^{n}$ | |||

$\displaystyle\sum_{{k=0}}^{n}(-1)^{k}{n\choose k}=0$ | |||

$\displaystyle\sum_{{k=0}}^{n}k{n\choose k}=n2^{{n-1}}$ | |||

$\displaystyle\sum_{{k=0}}^{n}k^{2}{n\choose k}=n(n+1)2^{{n-2}}$ | |||

$\displaystyle\sum_{{k=0}}^{n}k^{3}{n\choose k}=n^{2}(n+3)2^{{n-3}}$ | |||

$\displaystyle\sum_{{k=0}}{n\choose k}^{2}={2n\choose n}$ |

If $f_{n}$ denotes the $n$-th Fibonacci number, then

$\displaystyle\sum_{{k=0}}^{n}f_{k}=f_{{n+2}}-1$ | ||

$\displaystyle\sum_{{k=0}}^{n}f_{k}^{2}=f_{n}f_{{n+1}}$ | ||

$\displaystyle\sum_{{k=0}}^{{\lfloor n/2\rfloor}}{n-k-1\choose k}=f_{n}$ |

# Notes

The sigma notation was introduced by the French mathematician Joseph Fourier in 1820 [1].

# References

- 1
N. Higham.
*Handbook of writing for the mathematical sciences*. Society for Industrial and Applied Mathematics, 1998. (pp. 25) - 2
Graham, Knuth, Patashnik.
*Concrete mathematics*. Addison-Wesley, 1994

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## Corrections

Small oversight by Poly ✓

Minor oversight by Poly ✓

Small oversight by Poly ✓

Small oversight by Poly ✓

## Comments

## Hi all old mathematicians

Which is the value of the sum

SIGMA_{i = 1}^4 of 2k+1,

is it 21 or 24? I have always (in 44 years) seen only such a practice, that the summing operator, similarly as the deriving and integral operators, affects only the first term (or addend) on its right side. This implies e.g. that SIGMA x + SIGMA y does not mean double summing

SIGMA (a + SIGMA b).

## Hi all old mathematicians (corrected)

Which is the value of the sum

SIGMA_{i = 1}^4 of 2i+1,

is it 21 or 24? I have always (in 44 years) seen only such a practice, that the summing operator, similarly as the deriving and integral operators, affects only the first term (or addend) on its right side. This implies e.g. that SIGMA x + SIGMA y does not mean double summing

SIGMA (a + SIGMA b).

## Re: Hi all old mathematicians (corrected)

Hi pahio,

If I'm reading your notation correctly you want to calculate:

SIGMA{i=1..4} (2*i+1)? (i.e. (2*1+1) + (2*2+1) + (2*3+1) + (2*4+1))?

Basically we have two properties of sigma we can use:

SIGMA(a+b) = SIGMA (a) + SIGMA (b) [where the range remains the same]

and SIGMA(c*a) = c * SIGMA (a) where c is a constant

so SIGMA{i=1..4} (2*i+1) = 2*SIGMA{i=1..4} (i) + SIGMA {i=1..4} (1)

= 2*SIGMA{i=1..4} (i) + 4

= 2 * 10 + 4 [using SIGMA{i=1..n} (i) = (n*(n+1)/2)]

= 24.

So no, there is no SIGMA double summing as you had presented:

SIGMA (a + SIGMA b).

Karli

## Re: Hi all old mathematicians (corrected)

and...

http://planetmath.org/?op=getobj&from=objects&id=6361

also gives these and other properties of SIGMA

## Re: Hi all old mathematicians (corrected)

I would have said it is

(2*1+1) + (2*2+1) + (2*3+1) + (2*4+1) = 3+5+7+9 = 24.

I agree that the SIGMA stops being active if it encounters

another SIGMA, but I think that it continues to be active after a +,

by default.

But of course if one has the option, one can and should use parentheses

to be completely clear.

I've recently had to endure some pain due to EXPECTATION X^2 -- no parens!

## Re: Hi all old mathematicians (corrected)

This reminds me of a saying

"write not to be understood, but so that you cannot be

misunderstood"

To make things precise, there should be parenthesis around

2i+1. Or if you really want to add one to the sum, then write

1 + \sum 2i

For simple sums, I think the below expressions are

unambigious:

1) b + \sum a_i

2) \sum a_i + b_i

(meaning \sum (a_i + b_i). This is implied by the

index in b_i)

3) \sum a_i + \sum b_i = (\sum a_i) + (\sum b_i)

(and not \sum (a_i + \sum b_i))

However, for any more comlpicated sums, I think that parentheses

should be used as much as necessary so that the sum is

readable. Clarity should be the aim.

## Re: Hi all old mathematicians (corrected)

I usually consider the "big operators" (\sum, \bigcup, etc) having lower precedence than "small operators" (+, \cup, etc)

so SIGMA 2k + 1 I take it as all parent post did:

(2*1+1)+(2*2+1) + (2*3+1) etc

and as matte points, that's specially true when the summnds all involve the summation index

\sum a_k + b_k

makes no sense expanding as (\sum ak) + bk

HOWEVER in case of doubt, and complex sums (well the example was a bit simple so it was understood out from context), it's much better to use parentheses

NOtice also that (on the top post), integral has the differential delimiting its scope, so it's not as good as analogy

f

G -----> H G

p \ /_ ----- ~ f(G)

\ / f ker f

G/ker f

## Re: Hi older and younger mathematicians

Matte is right, and I agree with him [her?] -- one should strive for precision and unambiguity in mathematical expressions. Therefore one should use parentheses after a sum sign when one means to sum an expression with more than one "terms"; dropping the parentheses is thus in principle worse than using them (so I hope that here in PM we should not set bad examples in this thing; cf. "summation").

Another motive: WHY should the summing operator behave in different way as other operators, e.g. deriving operators (d/dx, D etc.), nabla, integral operators? These all require parentheses for affecting sums and differences -- d/dx sin2x + cos3x is not same d/dx (sin2x + cos3x) in spite of the common variable x.

Jussi

## Re: Hi older and younger mathematicians

Don't get mad at me, but if I wanted to mean

(\sum 2k) + 1 I would write it in THAT particular way or as matte sugests

1+\sum 2k

and I think it's customary to understand that a summation "closes" revious summations

so \sum 2k + \sum 1 is not \sum (2k + \sum 1)

I ALSO said that when in doubt and/or complicated expressions, parenthesis should be used for the sake of clarity.

But let's not be too pedantic, and I don't think my entry sets a bad example or it's confusing (unless you WANT to be confused)

Notice all the first posters (stux and jac) that answered the original question said it was 24 so that means it's how a given mathematician would understand it out of context.

And I guess the particular place that bothers you was understood the proper way by everyone who looked at it.

HOWEVER, I'll grant you write access to ALL of my entries so you can fix whatever you take in consideration (so you don't think I'm stubborn or maliciously trying to confuse readers)

f

G -----> H G

p \ /_ ----- ~ f(G)

\ / f ker f

G/ker f

## Re: Hi older and younger mathematicians

We all have seen that the use of parentheses is a matter of taste -- e.g. drini and I evidently have the opposite tastes =o) I know that its always dangerous to dispute of such matters and hope that I have not made drini angry.

I have wanted to bring up my pedantic point of view, which is based on the similar behaving of operators. The cause was that I have in PM seen in two places the practice without parentheses (it shocked me since I have not before seen such, or do not remember such).

All in all, it is good that we discuss also the use of parentheses -- it belongs to mathematics.

Jussi

## Re: Hi older and younger mathematicians

An alternative to paretheses is a clearly stated convention.

If we agree that ^ has greater precidence than +, for example,

there is no need to parethesize 2+2^3. Otherwise, if we don't

agree, then there is potential for confusion.

Aaron has an entry called OrderOfOperations, I think this would

be the correct place to take up discussion of the conventions

used for SIGMA (and differentiation, etc.) on PlanetMath.

## Re: Hi older and younger mathematicians

Hi guys,

I wouldn't be surprised if there is already a clearly established convention for handling order of operations when the SIGMA or \sum operation is involved. However after a quick search I have not found reference to this topic.

However I do beleive, as drini stated, that the established convention is to give \sum a lower operator precedence than '+'. That means in the same way 2^3+4=(2^3)+4 differentiates 2^(3+4) (as you pointed out jac), \sum a + c should necessarily be understood as (\sum a) + c or c + (\sum a) as an clearer alternate notation drini suggested.

That said, it is my opinion, having \sum a_i + b_i (although understood to be part of \sum in context,) should be grouped by a parenthesis as \sum (a_i + b_i) for clarity.

An example of this can be seen in Wikipedia's entry for SERIES where they state an example as \sum (b_n-b_[n+1]) where the parenthesis is included for clarity.

The link to this is:

http://en.wikipedia.org/wiki/Series_%28mathematics%29#Examples

Also, I wanted to point out earlier that:

\sum (a + \sum b) would evaluate to: \sum a + \sum\sum b.

## Re: Hi older and younger mathematicians

yes, but as matter commented, if

\sum_j a+b

if both a,b involve "j" (the summation index) then it makes sense the sum sign applying to both of them

f

G -----> H G

p \ /_ ----- ~ f(G)

\ / f ker f

G/ker f

## Re: Hi older and younger mathematicians

Stux, I think you misspoke slightly...

"established convention is to give \sum a lower operator precedence than '+'"

yes, I agree.

But then

"\sum a + c should necessarily be understood as (\sum a) + c or c + (\sum a)"

does not make sense.

*Lower* precidence means sum last, and get \sum (a+c).

In the case of

\sum a + \sum b

(without parentheses), the above rule says we have

\sum (a + \sum b)

However, that is *not* always what is meant when the expression is

encountered in the literature! And I don't think one will frequently

see

(\sum a) + (\sum b)

even when that is what was meant. I'd have to say that the convention

that is actually employed in day-to-day math is actually more

complicated than simple operator precidence! But I think on PM, we

should only use one rule to determine where the parentheses: assume

everything in OrderOfOperations, and then add parentheses as needed.

## Re: Hi older and younger mathematicians

jac,

You know what? You're absolutely right, I definitely did 'misspeak' (If that's a word at all,) about the operator precedence for \sum. I had to think about it for a second! I'm sorry. So yeah it wouldn't be lower precedence. So pretty much the convention employed is a little more complicated than order of operations, but I don't think it's by much:

Generally, multiplications and powers are kept within the \sum, so I'd say (in a preliminary attempt to formalize the convention), it would have a higher precedence than addition but lower than multiplication/division, etc. -- I haven't even touched Integration, etc. The exception would be what drini mentioned: that it would make sense that \sum a_i + b_i is iterpreted as \sum (a_i + b_i); (granted I prefer the latter version cuz its prettier ;) ).

I am curious to know if there is already some kind of paper or documentation formalizing the convention. I wish I could find something like that -- it would be very helpful.

## Re: Hi older and younger mathematicians

Alright guys, your deep discussions on the scope of summations have borne fruit, although you might need a magnifying glass to see it. In the newly-added entry "proof of the multiplication formula for the gamma function" there is a summation in the exponent in the seventh equation. Thanks to you all, I was VERY, VERY, VERY CAREFUL to enclose the argument of that summation in parentheses lest there be confusion.

Pariuntur montes et nascitur ridicullisimus musculus.

## checking accuracy: first example with infinity

While I'm no mathematician, I remain an interested observer. It seems to me that the example given above for summation of 1/2^k for k=1 to infinity is incorrect.

It is given as 1/1+1/2+1/4+1/8+... but should be 1/2+1/4+1/8+... since the series starts at k=1 and 1/2^1 = 1/2.

If the series had started at k=0, then 1/2^0 would give us 1/1 and thus the series that was printed (1/1+1/2+1/4+...) would have been correct.

Am I correct or did I miss something in analytic geometry?

## Re: checking accuracy: first example with infinity

you're quite right. I fixed it.

## Negative index

dear friends.Whatshould we dowhen the index of a sum is negative?cananybody help?thanks

## Negative index

dear friends.What should we dowhen the index of a sum is negative?cananybody help?thanks