# Thread: Proof needed that expression <=1 for all positive integers

1. ## Proof needed that expression <=1 for all positive integers

Consider the following function $f(k)$ where $k$ is a positive integer, and $\Omega$ is the Omega constant (0.5671432904097838729999686622) defined by $
ln({\frac{1}{\Omega}}) = \Omega
$

$
f(k) ={\frac{1}{\Omega}}\sum_{j=k}^{\lceil {\frac{k}{\Omega}} \rceil}{\frac{1}{j}}-{\frac{\lceil {\frac{k}{\Omega}} \rceil}{k}}+{\frac{1}{\Omega}}+{\frac{2}{k}}-{\frac{2}{k\Omega}}
$

Evaluating the funtion for small values of $k$ gives $f(1) = 1$, $f(2) \approx 0.97$, $f(3) \approx 0.97$, $f(4) \approx 0.97$ etc. Also $\lim{k \to \infty}, f(k)=1$

I am interested in finding a proof that for $k=1..\infty$, $f(k)\le1$. Any help, hints, or of course a really cool proof, would be much appreciated.

Thanks

Rob

2. Originally Posted by Evo Rob
Consider the following function $f(k)$ where $k$ is a positive integer, and $\Omega$ is the Omega constant (0.5671432904097838729999686622) defined by $
ln({\frac{1}{\Omega}}) = \Omega
$

$
f(k) ={\frac{1}{\Omega}}\sum_{j=k}^{\lceil {\frac{k}{\Omega}} \rceil}{\frac{1}{j}}-{\frac{\lceil {\frac{k}{\Omega}} \rceil}{k}}+{\frac{1}{\Omega}}+{\frac{2}{k}}-{\frac{2}{k\Omega}}
$

Evaluating the funtion for small values of $k$ gives $f(1) = 1$, $f(2) \approx 0.97$, $f(3) \approx 0.97$, $f(4) \approx 0.97$ etc. Also $\lim{k \to \infty}, f(k)=1$

I am interested in finding a proof that for $k=1..\infty$, $f(k)\le1$. Any help, hints, or of course a really cool proof, would be much appreciated.

Thanks

Rob
I believe that I can do this, but the proof is very fussy so I would rather not have to type it out if anyone else has a neater proof.

I will start typing it off line and I should be able to post it later other things being equal.

Outline:

relpace the sum by an integral which is an upper bound for the sum.

ignore the first two terms outside the summation as they are always negative

show that for k>13 the integral and the last two terms outside the summation are strictly decreasing as a function of k, and that at k=14 this is less that 1.

This shows that the function is bounded above by 1 for k>13, and so we need only check the first 13 terms to prove that f(k) is less than 1 for all positive k.

We also seem to have a discrepancy with our evaluation of the first few valuea of f(k), maybe these need checking

RonL

3. Originally Posted by CaptainBlack
I believe that I can do this, but the proof is very fussy so I would rather not have to type it out if anyone else has a neater proof.

I will start typing it off line and I should be able to post it later other things being equal.

Outline:

relpace the sum by an integral which is an upper bound for the sum.

ignore the first two terms outside the summation as they are always negative

show that for k>13 the integral and the last two terms outside the summation are strictly decreasing as a function of k, and that at k=14 this is less that 1.

This shows that the function is bounded above by 1 for k>13, and so we need only check the first 13 terms to prove that f(k) is less than 1 for all positive k.

We also seem to have a discrepancy with our evaluation of the first few valuea of f(k), maybe these need checking

RonL
Because $1/x$ is a positive decreasing function for $x>0$, for $k>1$ we have:

$\sum_{j=k}^{\lceil(k/\Omega\rceil} \frac{1}{j}\le \int_{x=k}^{\frac{k}{\Omega}+1} \frac{1}{x-1}\ dx=\ln(k/\Omega) - \ln(k-1)=\ln\left( \frac{k}{\Omega(k-1)}\right)$

Hence:

$
f(k) \le {\frac{1}{\Omega}}\ln\left( \frac{k}{\Omega(k-1)}\right)
-{\frac{\lceil {\frac{k}{\Omega}} \rceil}{k}}+{\frac{1}{\Omega}}+{\frac{2}{k}}-{\frac{2}{k\Omega}}
$

Now $-{\frac{\lceil {\frac{k}{\Omega}} \rceil}{k}}+{\frac{1}{\Omega}}<0$, so

$
f(k) \le {\frac{1}{\Omega}}\ln\left( \frac{k}{\Omega(k-1)}\right)
+{\frac{2}{k}}-{\frac{2}{k\Omega}}
$

Now let:

$
h(k)= {\frac{1}{\Omega}}\ln\left( \frac{k}{\Omega(k-1)}\right)
+{\frac{2}{k}}-{\frac{2}{k\Omega}}
$

Then:

$\frac{dh}{dk}=\frac{1}{\Omega k(k-1)}-\frac{2}{k^2}+\frac{2}{k^2\Omega}$

But $\lim_{k\to \infty}h(k)=1$ , and we can see that $\frac{dh}{dk}>0\ \forall k>1$ , so $h(k)$ is an increasing function bounded above by $1$, so we have:

$f(k)\le h(k) <1$

as required.

(Note this worked out simpler than originaly thought due to correction of some minor errors in my original working)

RonL

4. Firstly, please accept my humble apologies, there was a mistake in my typing of the problem
formulation. For some inexplicable reason, I typed in ceiling functions when they should have been
floors. This is possibly what caused the discrepancy in the first few values of f(k)

However, the suggested approach works, with a few minor alterations, for the correct problem
formulation:

$
f(k) ={\frac{1}{\Omega}}\sum_{j=k}^{\lfloor {k/\Omega}
\rfloor}{\frac{1}{j}}-{\frac{\lfloor{\frac{k}{\Omega}}
\rfloor}{k}}+{\frac{1}{\Omega}}+{\frac{2}{k}}-{\frac{2}{k\Omega}}
$

As $1/x$ is a positive decreasing function for $k>1$, we have:
$
\sum_{j=k}^{\lfloor(k/\Omega\rfloor} \frac{1}{j}\le \int_{x=k}^{\frac{k}{\Omega}} \frac{1}{x-1}\
dx=\ln(k/\Omega -1) - \ln(k-1)=\ln\left( \frac{k/\Omega -1}{(k-1)}\right)
$

Also, as
$
\lfloor{k/\Omega}\rfloor\ge{k/\Omega}-1
$

we have:
$
f(k)\le {\frac{1}{\Omega}}\ln\left( \frac{k/\Omega -1}{(k-1)}\right) + {\frac{3}{k}}-{\frac{2}{k\Omega}}
$

Now let:
$
h(k)={\frac{1}{\Omega}}\ln\left( \frac{k/\Omega -1}{(k-1)}\right) + {\frac{3}{k}}-{\frac{2}{k\Omega}}
$

$
{\frac{dh}{dk}}=-{\frac{(1-\Omega)}{(k-1)(k-\Omega)}}- {\frac{3}{k^2}}+{\frac{2}{k^2\Omega}}
$

Now for ${\frac{dh}{dk}}$ to be positive, requires that:
$
{\frac{(k-1)(k-\Omega)}{k^2}}>{\frac{\Omega(1-\Omega)}{(2-3\Omega)}} = 0.8222248$

This is true $\forall k\ge9$.

As $\lim_{k\to \infty}h(k)=1$ and $
\frac{dh}{dk}>0\ \forall k>8
$

$
f(k)\le h(k) \le1\ \forall k>8
$

Finally, $f(k)$ can be evaluted for $k=1$ to $8$, completing the proof.

Once again, many thanks for the excellent and timely help with this, it is much appreciated.

Rob

5. Originally Posted by CaptainBlack
I believe that I can do this, but the proof is very fussy so I would rather not have to type it out if anyone else has a neater proof.

I will start typing it off line and I should be able to post it later other things being equal.

Outline:

relpace the sum by an integral which is an upper bound for the sum.

ignore the first two terms outside the summation as they are always negative

show that for k>13 the integral and the last two terms outside the summation are strictly decreasing as a function of k, and that at k=14 this is less that 1.

This shows that the function is bounded above by 1 for k>13, and so we need only check the first 13 terms to prove that f(k) is less than 1 for all positive k.

We also seem to have a discrepancy with our evaluation of the first few valuea of f(k), maybe these need checking

RonL
A big improvement on the sad old line "I have no time now but will reply later if no-one else does ....."

6. Originally Posted by mr fantastic
A big improvement on the sad old line "I have no time now but will reply later if no-one else does ....."
I had thought of writting exactly that but thought I would only get stick for doing so. Also I had the solution (complete with some errors) in my note pad so could give an outline, it was the prospect of typing it that put me off giving the solution right away (that and having to walk the dog).

RonL

7. Originally Posted by Evo Rob
Firstly, please accept my humble apologies, there was a mistake in my typing of the problem
formulation. For some inexplicable reason, I typed in ceiling functions when they should have been
floors. This is possibly what caused the discrepancy in the first few values of f(k)

However, the suggested approach works, with a few minor alterations, for the correct problem
formulation:

$
f(k) ={\frac{1}{\Omega}}\sum_{j=k}^{\lfloor {k/\Omega}
\rfloor}{\frac{1}{j}}-{\frac{\lfloor{\frac{k}{\Omega}}
\rfloor}{k}}+{\frac{1}{\Omega}}+{\frac{2}{k}}-{\frac{2}{k\Omega}}
$

As $1/x$ is a positive decreasing function for $k>1$, we have:
$
\sum_{j=k}^{\lfloor(k/\Omega\rfloor} \frac{1}{j}\le \int_{x=k}^{\frac{k}{\Omega}} \frac{1}{x-1}\
dx=\ln(k/\Omega -1) - \ln(k-1)=\ln\left( \frac{k/\Omega -1}{(k-1)}\right)
$

Also, as
$
\lfloor{k/\Omega}\rfloor\ge{k/\Omega}-1
$

we have:
$
f(k)\le {\frac{1}{\Omega}}\ln\left( \frac{k/\Omega -1}{(k-1)}\right) + {\frac{3}{k}}-{\frac{2}{k\Omega}}
$

Now let:
$
h(k)={\frac{1}{\Omega}}\ln\left( \frac{k/\Omega -1}{(k-1)}\right) + {\frac{3}{k}}-{\frac{2}{k\Omega}}
$

$
{\frac{dh}{dk}}=-{\frac{(1-\Omega)}{(k-1)(k-\Omega)}}- {\frac{3}{k^2}}+{\frac{2}{k^2\Omega}}
$

Now for ${\frac{dh}{dk}}$ to be positive, requires that:
$
{\frac{(k-1)(k-\Omega)}{k^2}}>{\frac{\Omega(1-\Omega)}{(2-3\Omega)}} = 0.8222248$

This is true $\forall k\ge9$.

As $\lim_{k\to \infty}h(k)=1$ and $
\frac{dh}{dk}>0\ \forall k>8
$

$
f(k)\le h(k) \le1\ \forall k>8
$

Finally, $f(k)$ can be evaluted for $k=1$ to $8$, completing the proof.

Once again, many thanks for the excellent and timely help with this, it is much appreciated.

Rob
if $f_1(k)$ is the function defined in your original postm and $f_2(k)$ the corrected function, is it not the case that:

$f_2(k)\le f_1(k)$

and so the original proof still proves the required result!

RonL

8. As the second term in each function is negative, I think:
$
f_2(k)\ge f_1(k)
$

Either we have
$
{\lceil{\frac{k}{\Omega}} \rceil}= {\lfloor{\frac{k}{\Omega}} \rfloor}
$

in which case the two functions are the same, or we have
$
{\lceil{\frac{k}{\Omega}} \rceil}= {\lfloor{\frac{k}{\Omega}} \rfloor}+1
$

in which case
$
f_2(k)-f_1(k)={\frac{1}{k}}-{\frac{1}{\Omega({\lfloor{k/\Omega}\rfloor + 1})}} \ge0
$

Rob

9. Unfortunately, both of the proofs given in this thread appear to be faulty.

In the original proof (given by Captain Black), there is a missing minus sign, which prevents the proof from working.

$
\frac{dh}{dk}={\color{red}-}\frac{1}{\Omega k(k-1)}-\frac{2}{k^2}+\frac{2}{k^2\Omega}
$

And in my attempted proof for the correct version of the problem (with floor functions rather than ceilings), there is a missing $\Omega$ which also foils the proof.

$
{\frac{dh}{dk}}=-{\frac{(1-\Omega)}{{\color{red}\Omega}(k-1)(k-\Omega)}}- {\frac{3}{k^2}}+{\frac{2}{k^2\Omega}}
$

So, if anyone would like to try and solve the problem, its still open...

Given:

$
f(k) ={\frac{1}{\Omega}}\sum_{j=k}^{\lfloor {k/\Omega}
\rfloor}{\frac{1}{j}}-{\frac{\lfloor{\frac{k}{\Omega}}
\rfloor}{k}}+{\frac{1}{\Omega}}+{\frac{2}{k}}-{\frac{2}{k\Omega}}
$

Where k is a positive integer and $\Omega$ is the Omega constant, prove that $\forall k>0\ \ f(k)\le1$

Any help, or ideally a really cool proof would be much appreciated.

Thanks

Rob

10. Originally Posted by Evo Rob
Unfortunately, both of the proofs given in this thread appear to be faulty.

In the original proof (given by Captain Black), there is a missing minus sign, which prevents the proof from working.

$
\frac{dh}{dk}={\color{red}-}\frac{1}{\Omega k(k-1)}-\frac{2}{k^2}+\frac{2}{k^2\Omega}
$

And in my attempted proof for the correct version of the problem (with floor functions rather than ceilings), there is a missing $\Omega$ which also foils the proof.

$
{\frac{dh}{dk}}=-{\frac{(1-\Omega)}{{\color{red}\Omega}(k-1)(k-\Omega)}}- {\frac{3}{k^2}}+{\frac{2}{k^2\Omega}}
$

So, if anyone would like to try and solve the problem, its still open...

Given:

$
f(k) ={\frac{1}{\Omega}}\sum_{j=k}^{\lfloor {k/\Omega}
\rfloor}{\frac{1}{j}}-{\frac{\lfloor{\frac{k}{\Omega}}
\rfloor}{k}}+{\frac{1}{\Omega}}+{\frac{2}{k}}-{\frac{2}{k\Omega}}
$

Where k is a positive integer and $\Omega$ is the Omega constant, prove that $\forall k>0\ \ f(k)\le1$

Any help, or ideally a really cool proof would be much appreciated.

Thanks

Rob
Well I can make mine work by putting back the two terms that I took out, replacing the $\lceil k/\Omega\rceil$ by $k/\Omega+1$ and then we have a bound which is again increasing after the first term and tending to 1, so it will then work.

(I think - I haven't worked through the algebra in detail yet and I have to go out in a moment so won't be able to fix it untill the tommorow)

RonL

11. Originally Posted by CaptainBlack
Well I can make mine work by putting back the two terms that I took out, replacing the $\lceil k/\Omega\rceil$ by $k/\Omega+1$ and then we have a bound which is again increasing after the first term and tending to 1, so it will then work.

(I think - I haven't worked through the algebra in detail yet and I have to go out in a moment so won't be able to fix it untill the tommorow)

RonL
Having worked through this in detail I find that this approach without some modification cannot be made to work.

RonL

12. Problem statement:

The function $f(k)$ is defined as follows:

$
f(k) ={\frac{1}{\Omega}}\sum_{j=k}^{\lfloor {k/\Omega}
\rfloor}{\frac{1}{j}}-{\frac{\lfloor{\frac{k}{\Omega}}
\rfloor}{k}}+{\frac{1}{\Omega}}+{\frac{2}{k}}-{\frac{2}{k\Omega}}
$

where k is a positive integer and $\Omega$ is the Omega constant, defined by $\Omega=ln({\frac{1}{\Omega}})$

Prove that $\forall k>0 \ \ f(k)\le1$

Solution:

Inductive proof using discrete maths.

Proof sketch:
1. Show that $kf(k)\le{k}$ for $k=1,2$
2. Show that $\forall k\ge2 \ \ (k+1)f(k+1)-kf(k)\le1$
3. By induction $\forall k\ge1 \ \ f(k)\le1$

Step 1. Show that $kf(k)\le{k}$ for $k=1,2$
$f(1) = 1$ , $f(2) = {\frac{5}{6\Omega}}-{\frac{1}{2}}\approx0.97$

Step 2. Show that $\forall k\ge2 \ \ (k+1)f(k+1)-kf(k)\le1$
$
kf(k) =\sum_{j=k}^{\lfloor {k/\Omega}
\rfloor}{\frac{k/\Omega}{j}}-{\lfloor{k/\Omega}
\rfloor}+{\frac{k}{\Omega}}+2-{\frac{2}{\Omega}}
$

$
=\sum_{j=k}^{\lfloor {k/\Omega}
\rfloor}{\frac{(k/\Omega)-j}{j}}+k({\frac{1}{\Omega}}-1)+3-{\frac{2}{\Omega}}
$

$
(k+1)f(k+1)-kf(k) =\sum_{j=k+1}^{\lfloor {(k+1)/\Omega}
\rfloor}{\frac{((k+1)/\Omega)-j}{j}} - \sum_{j=k}^{\lfloor {k/\Omega}
\rfloor}{\frac{(k/\Omega)-j}{j}}+({\frac{1}{\Omega}}-1)
$

$
={\frac{1}{\Omega}}\sum_{j=k+1}^{\lfloor {k/\Omega}
\rfloor}{\frac{1}{j}} + \sum_{j={\lfloor {k/\Omega}
\rfloor + 1}}^{\lfloor {(k+1)/\Omega}
\rfloor}{\frac{((k+1)/\Omega)-j}{j}}
$

(Note this holds for $k\ge2$ but not for $k=1$ as in that case $k+1>\lfloor(k/\Omega)\rfloor$)

Now there are two cases to consider:

Case 1: ${\lfloor {(k+1)/\Omega}
\rfloor}={\lfloor {k/\Omega}
\rfloor}+1
$

The second summation term above can be expressed as:
${\frac{(k/\Omega)-{\lfloor {(k/\Omega}
\rfloor}}{
{\lfloor {k/\Omega}
\rfloor}+1
}}
\le
{\frac{1}{{\lfloor {k/\Omega}
\rfloor}+1
}}
$

Hence:
$
(k+1)f(k+1)-kf(k) \le {\frac{1}{\Omega}}\sum_{j=k+1}^{\lfloor {(k+1)/\Omega}
\rfloor}{\frac{1}{j}}$

$
\le {\frac{1}{\Omega}}\int_{x=k+1}^{((k+1)/\Omega)
}{\frac{1}{x}}={\frac{1}{\Omega}}\ln({\frac{(k+1)/\Omega
}{k+1}})=1$

So we have:

$
(k+1)f(k+1)-kf(k) \le 1
$

Case 2: ${\lfloor {(k+1)/\Omega}
\rfloor}={\lfloor {k/\Omega}
\rfloor}+2
$

The second summation term can be expressed as:
${\frac{(k/\Omega)-{\lfloor {(k/\Omega}
\rfloor}}{
{\lfloor {k/\Omega}
\rfloor}+1
}}
+
{\frac{(k/\Omega)-{\lfloor {(k/\Omega}
\rfloor}-1}{
{\lfloor {k/\Omega}
\rfloor}+1
}}
$

The first of the above terms is the same as case 1 and can be treated in the same way. The second term is negative and can therefore be safely disgarded.

Again we have:

$
(k+1)f(k+1)-kf(k) \le 1
$

Step 3: By induction $\forall k\ge1 \ \ f(k)\le1$

By step 1, we have $f(2)\le1$ (Basis).

By step 2, we have $\forall k>2 \ \ (k+1)f(k+1)-kf(k)\le1$

Inductive step: assuming that $f(k)\le1$, we have
${(k+1)f(k+1)} \le {kf(k)+1} \le {k+1}$ and hence
$f(k+1)\le1$

As $f(1)=1$ (by step 1), we have:

$\forall k>0 \ \ f(k)\le1$

Theorem proved!

Rob

13. Originally Posted by Evo Rob
Problem statement:

The function $f(k)$ is defined as follows:

$
f(k) ={\frac{1}{\Omega}}\sum_{j=k}^{\lfloor {k/\Omega}
\rfloor}{\frac{1}{j}}-{\frac{\lfloor{\frac{k}{\Omega}}
\rfloor}{k}}+{\frac{1}{\Omega}}+{\frac{2}{k}}-{\frac{2}{k\Omega}}
$

where k is a positive integer and $\Omega$ is the Omega constant, defined by $\Omega=ln({\frac{1}{\Omega}})$

Prove that $\forall k>0 \ \ f(k)\le1$

Solution:

Inductive proof using discrete maths.

Proof sketch:
1. Show that $kf(k)\le{k}$ for $k=1,2$
2. Show that $\forall k\ge2 \ \ (k+1)f(k+1)-kf(k)\le1$
3. By induction $\forall k\ge1 \ \ f(k)\le1$

Step 1. Show that $kf(k)\le{k}$ for $k=1,2$
$f(1) = 1$ , $f(2) = {\frac{5}{6\Omega}}-{\frac{1}{2}}\approx0.97$

Step 2. Show that $\forall k\ge2 \ \ (k+1)f(k+1)-kf(k)\le1$
$
kf(k) =\sum_{j=k}^{\lfloor {k/\Omega}
\rfloor}{\frac{k/\Omega}{j}}-{\lfloor{k/\Omega}
\rfloor}+{\frac{k}{\Omega}}+2-{\frac{2}{\Omega}}
$

$
=\sum_{j=k}^{\lfloor {k/\Omega}
\rfloor}{\frac{(k/\Omega)-j}{j}}+k({\frac{1}{\Omega}}-1)+3-{\frac{2}{\Omega}}
$

$
(k+1)f(k+1)-kf(k) =\sum_{j=k+1}^{\lfloor {(k+1)/\Omega}
\rfloor}{\frac{((k+1)/\Omega)-j}{j}} - \sum_{j=k}^{\lfloor {k/\Omega}
\rfloor}{\frac{(k/\Omega)-j}{j}}+({\frac{1}{\Omega}}-1)
$

$
={\frac{1}{\Omega}}\sum_{j=k+1}^{\lfloor {k/\Omega}
\rfloor}{\frac{1}{j}} + \sum_{j={\lfloor {k/\Omega}
\rfloor + 1}}^{\lfloor {(k+1)/\Omega}
\rfloor}{\frac{((k+1)/\Omega)-j}{j}}
$

(Note this holds for $k\ge2$ but not for $k=1$ as in that case $k+1>\lfloor(k/\Omega)\rfloor$)

Now there are two cases to consider:

Case 1: ${\lfloor {(k+1)/\Omega}
\rfloor}={\lfloor {k/\Omega}
\rfloor}+1
$

The second summation term above can be expressed as:
${\frac{(k/\Omega)-{\lfloor {(k/\Omega}
\rfloor}}{
{\lfloor {k/\Omega}
\rfloor}+1
}}
\le
{\frac{1}{{\lfloor {k/\Omega}
\rfloor}+1
}}
$

Hence:
$
(k+1)f(k+1)-kf(k) \le {\frac{1}{\Omega}}\sum_{j=k+1}^{\lfloor {(k+1)/\Omega}
\rfloor}{\frac{1}{j}}$

$
\le {\frac{1}{\Omega}}\int_{x=k+1}^{((k+1)/\Omega)
}{\frac{1}{x}}={\frac{1}{\Omega}}\ln({\frac{(k+1)/\Omega
}{k+1}})=1$

So we have:

$
(k+1)f(k+1)-kf(k) \le 1
$

Case 2: ${\lfloor {(k+1)/\Omega}
\rfloor}={\lfloor {k/\Omega}
\rfloor}+2
$

The second summation term can be expressed as:
${\frac{(k/\Omega)-{\lfloor {(k/\Omega}
\rfloor}}{
{\lfloor {k/\Omega}
\rfloor}+1
}}
+
{\frac{(k/\Omega)-{\lfloor {(k/\Omega}
\rfloor}-1}{
{\lfloor {k/\Omega}
\rfloor}+1
}}
$

The first of the above terms is the same as case 1 and can be treated in the same way. The second term is negative and can therefore be safely disgarded.

Again we have:

$
(k+1)f(k+1)-kf(k) \le 1
$

Step 3: By induction $\forall k\ge1 \ \ f(k)\le1$

By step 1, we have $f(2)\le1$ (Basis).

By step 2, we have $\forall k>2 \ \ (k+1)f(k+1)-kf(k)\le1$

Inductive step: assuming that $f(k)\le1$, we have
${(k+1)f(k+1)} \le {kf(k)+1} \le {k+1}$ and hence
$f(k+1)\le1$

As $f(1)=1$ (by step 1), we have:

$\forall k>0 \ \ f(k)\le1$

Theorem proved!

Rob
Well that will save me having to worry about it anymore, my investigations were leading me in a vaguely similar direction, but unfortunaly work has been getting in the way of progressing the task.

RonL