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# Driving an LED from a uC pin with the Anode above uC's Vcc rail

I am using a 3.3V uC (ATMega 328p in Arduino Pro Mini) running from 3 alkalines (nominal 4.5V). So Vin=4.5V, Vcc=3.3V

I'd like to drive LEDs from some pins, say at 20ma each. And I was thinking to connect the LED anodes to Vin (4.5v) with the cathode going to the uC pin, rather than putting all the LED currents through the 3.3v onboard regulator. (Also, while I'll use Red LEDs in the example here, I may want to use Blue as well, which would be more comfortable with a bit more driving voltage).

I figure a red LED at about 2v Vfwd, so the resistor would need to drop 2.5v @ 20ma, say 1K2 ohms.

The issue: this connect a uC pin to a source 1.2V about its Vcc rail (through an LED).

I think we are fine when the LED is on (pin low). But what about turning the LED off?

simulate this circuit – Schematic created using CircuitLab

Approach 1: Pull uC pin high (3.3v) to turn off the LED. The LED is between 4.5V and 3.3V, and the 1.2v difference will not be enough to light it - and I THINK it also won't bring the pin too high above its rail. But LEDs are not ideal devices, hence this question.

Approach 2: tristate uC pin (make it an input) to turn off the LED. The LED is between 4.5V and the high impedance input (along with it's static protection diode to Vcc). Again, I hope the LED's Vfwd would be enough protection for the uC pin.

Is either or both of these approaches workable and safe for the uC?

EDIT: substitute 120 ohm resistor above, oops.

• I -think- that going high impedance or active low should be safe enough. But at these current levels why not just source current from the microcontroller's VCC rail as an output rather than a current sink/active low how it is now? Sep 24, 2014 at 18:21
• Remember that every pin/port of the ATMEGA has CMOS style clamping diodes to VCC and GND to deal with VERY low power over-voltage/negative voltage issues, but if you have a constant over-voltage condition with only a 1.2k resistor I think it will bust the CMOS diode clamps over time and break your entire port Sep 24, 2014 at 18:25
• @KyranF - If the pin is 'tristated', i.e. an input, then the LED should be below it's threshold, and be a very large resistance; it will be a "VERY low power over-voltage", so it won't be "constant over-voltage condition with only a 1.2k resistor". Having said that, I wouldn't do this unless I needed to fix something already broken, and this was my 'best option'. Sep 24, 2014 at 19:15
• Agreed. unless the regulator is some ooddly weak thing, the only thing you are doing is shifting the power\heat loss from the regulator to the led resistors. While still having the 40/100/240 pin/port/total mA current limits. Sinking current to a low pin, has the same issues as sourcing current out of the pin. In this case, positive Voltage Droop. (er, lift?) Sep 24, 2014 at 19:24
• @Kyranf I used a red LED as an example, but a blue or green LED with Vfwd closer to 3V gets a bit tighter in terms of margins when powered from 3.3v out of the Arduino Pro Mini's (only about 0.3 V for the current limiting resistor). That's one reason I'd like to explore powering the LEDs directly from the 4.5v Vin.
– Zeph
Sep 24, 2014 at 21:12

What you're talking about is not huge risk for a hobby project.. a few things to think about.

It's not really the $V_F$ at the operating current that matters, it's the $V_F$ at the maximum 'off' brightness that is acceptable. That voltage might be as low as 1.4-1.5V volt for a low-current red LED- in a dark room they can be quite visible with microamperes of current. Driving the output to the 3.3V (nominal level) gets us 3.3V on the output. A fresh alkaline cell with minimal load might be 1.63V/cell at room temperature (just measured one), so 3 would be 4.89V. That leaves you with 1.59V across the LED + resistor (nominal, not allowing that the 3.3V might be a few percent low).

That's way too much to ensure it's not emitting a whole bunch of light.

So, we tri-state it- that allows the output to go maybe a bit above Vcc without much current flowing. 300mV is safe, the datasheet says 500mV absolute maximum. At 500mV, we'd have 1.09V across the LED, probably enough to ensure it's off, at least under nominal conditions. The 'absolute maximum' figure is never a good thing to design to, but usually there's a caveat on this particular figure that allows that voltage or a bit more if the current is limited.

So, I do think this will work (with tri-state, not push-pull), and I also think it's acceptable enough for a hobby project assuming nobody is going to be using a battery eliminator on the circuit in the future*. Keep in mind the margin on the red LEDs is minimal and consider eschewing red in favor of yellow or orange. Or, simply add a silicon diode in series with the red LEDs (one diode can be used for several LEDs).

• If the ESD protection network in the ATMEGA328P does begin to conduct it will tend to raise up the 3.3V supply, out of control of the regulator. This is not a good thing for stability and could conceivably damage something, though the ATMEGA328P-M itself is rated for 5V operation.

I once did something like this in a commercial product (to drive a series string of LEDs with high Vf using a 5V constant-current output), but I designed a power supply with just the correct oddball voltage and appropriate temperature coefficient to match the LEDs and thus optimize the situation. I think the supply was around 8-9V. Worked a treat, easily from -20°C to 80°C (spec was 0-50°C).

• Your reasoning is good and I would be convinced, but @Olin Lathrop comes to the opposite conclusion and alas, his reasoning sounds good to me too. Could you look over his answer and take his points into account? Thanks
– Zeph
Sep 25, 2014 at 18:06
• I agree with Olin's reasoning, he's just got valid concerns about different things happening (that could affect, say, an analog input). If you want to be really safe, you can use a series diode as I suggested above with a bleeder resistor to ground for the red LEDs only, and drive the outputs push-pull as Olin suggests. Something like a 1N4148 with 10K to ground could supply many red LEDs. Sep 25, 2014 at 18:52

Subjecting a I/O pin to voltages above the processor's power level is not a good idea.

Use a transistor:

The correct solution that doesn't violate any parameters in the datasheet is above. However, if you are trying to get away with it, then it will most likely work if you make sure the output pin is always driven.

We are assuming a modern processor with a CMOS output. CMOS transistor can conduct current in both directions when on. The little bit of current you will get thru the LED when the output is high will be conducted thru the top driver transistor to Vdd. The value of that current at 1.2 V forward bias will be hard to predict, will vary significantly with temperature, and will vary part to part. However, it will be "very small" compared to the overall current draw of the processor.

If you switch the output between driven low and high impedance instead of driven low and high, then the protection circuitry for that pin will take the little LED current instead. For some designs, this is just a diode to Vdd and therefore no worse than having the top FET conduct. However, other designs have more complicated circuitry which would be worse to dump the current thru. Guessing what causes the least trouble is one of the problems you run into when ignoring specs.

There are two more ways to be fully in spec:

1. Add a small bleeder resistor to ground. Size this to take whatever you determine will be the maximum LED current under all conditions. The LED may be very dimly lit when it's supposed to be off.

2. Use a pin that has "5 V tolerant" input, and switch the pin between driven low and floating.

• doing this would need an extra transistor and resistor than before, could be a bulky solution for driving lots of LEDs as seems to be their intent.. Sep 24, 2014 at 19:01
• As op has explained, the pins will never see a raw voltage higher than vcc. you Don't even attempt to answer the question. Sep 24, 2014 at 19:26
• Thanks for taking the time to draw up the conventional transistor approach. I am aware of that option, but as @KyranF understood, I'm checking whether I can trust Vfwd of the LED to protect the protection diodes, and feedback about which of the two approaches (pin = VCC or pin high impedance) would be better, in a context where adding the transistor and both resistors would be less convenient.
– Zeph
Sep 25, 2014 at 1:33
• @Passe: No, the pin can see a voltage above Vdd. The LED will allow some current at 1.2 V forward bias. That could easily be, and probably is, enough to raise a high impedance pin above its supply voltage. Sep 25, 2014 at 13:17
• This answer is much improved, and has good reasoning for why active high for LED off would be safer than high impedance. Alas, another good answer from @Spehro Pefhany also has reasoning for why high impedance would be better. Could you take his/her reasoning into account and help me sort out which approach is better? Thanks
– Zeph
Sep 25, 2014 at 18:04

Your approach 2 (tristate the D3 pin) is okay except as other posters pointed out the ESD protection diode from D3 to VCC is in series with the LED (and resistor) when the output is tristated. There will be some small current flow there at (4.5V-3.3V)=1.2V across both diodes from the 4.5V supply to VCC. The LED threshold is much higher than the ESD diode's threshold of about 0.5V to 0.6V for appreciable current (microamps or more; they're big diodes), so at 1.2V across both the current is likely to be in the microamp range. However, the D3 pin voltage will go above VCC by about 0.5V to 0.6V and this probably exceeds the recommended voltage at the pad (for stress on the gate oxide of the input circuit transistors).

You're essentially using the D3 pin as an open drain NMOS switch. You could always add a discrete NMOS transistor off chip that can withstand > 4.5V. While it adds a component and inverts the sense of the port, it prevents overvoltage at the IC.

Note, none of the approaches so far provide current regulation, so the brightness will vary with resistor tolerance, temperature and supply variation. There are some simple schemes for that if you're interested, but they add a couple more components.

I figure a red LED at about 2v Vfwd, so the resistor would need to drop 2.5v @ 20ma, say 1K2 ohms.

You have to check your math on this one for your resistor value.

$$R = \frac{V}{I}$$

R = 2.5V dropped by the resistor / 20mA through the resistor = 125 ohms.

If you are not interested in adding a transistor, why not source the LED from the microcontroller pin?

• Oops, did the divide quickly and wrongly in my head, you are correct. What advice would you give for sourcing a blue or green LED from a microcontroller pin with Vcc at 3.3v? I figure that the 0.3 V drop across the resistor is going to be highly variable, as the nominal Vfwd of the LED (around 3V) can vary by batch, temperature, etc. Vfwd varying by 0.1v makes a small difference when driving from 4.5v (1.5 vs 1.6 or 1.4 v drop across resister) but much more when driving from 3.3v. Or so I've heard.
– Zeph
Sep 25, 2014 at 17:54


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Wetting in Color

 Title: Wetting in Color Author: Burgess, Ian Bruce Full Text & Related Files: Burgess_gsas.harvard_0084L_10524.pdf (5.690Mb; PDF) Abstract: Colorimetric litmus tests such as pH paper have enjoyed wide commercial success due to their inexpensive production and exceptional ease of use. However, expansion of colorimetry to new sensing paradigms is challenging because macroscopic color changes are seldom coupled to arbitrary differences in the physical/chemical properties of a system. In this thesis I present in detail the development of Wetting in Color Technology, focusing primarily on its application as an inexpensive and highly selective colorimetric indicator for organic liquids. The technology exploits chemically-encoded inverse-opal photonic crystals to control the infiltration of fluids to liquid-specific spatial patterns, projecting minute differences in liquids’ wettability to macroscopically distinct, easy-to-visualize structural color patterns. It is shown experimentally and corroborated with theoretical modeling using percolation theory that the high selectivity of wetting, upon-which the sensitivity of the indicator relies, is caused by the highly symmetric structure of our large-area, defect-free $$SiO_2$$ inverse-opals. The regular structure also produces a bright iridescent color, which disappears when infiltrated with liquid naturally coupling the optical and fluidic responses. Surface modification protocols are developed, requiring only silanization and selective oxidation, to facilitate the deterministic design of an indicator that differentiates a broad range of liquids. The resulting tunable, built-in horizontal and vertical chemistry gradients allow the wettability threshold to be tailored to specific liquids across a continuous range, and make the readout rely only on countable color differences. As wetting is a generic fluidic phenomenon, Wetting in Color technology could be suitable for applications in authentication or identification of unknown liquids across a broad range of industries. However, the generic nature of the response also ensures chemical non-specificity. It is shown that combinatorial measurements from an array of indicators add a degree of chemical specificity to the platform, which can be further improved by monitoring the drying of the inverse-opal films. While colorimetry is the central focus of this thesis, applications of this platform in encryption, fluidics and nanofabrication are also briefly explored. Terms of Use: This article is made available under the terms and conditions applicable to Other Posted Material, as set forth at http://nrs.harvard.edu/urn-3:HUL.InstRepos:dash.current.terms-of-use#LAA Citable link to this page: http://nrs.harvard.edu/urn-3:HUL.InstRepos:9876056


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# Thread: evaluatng others from extreme value of one function

1. ## evaluatng others from extreme value of one function

Suppose f is differentiable on -infinity and +infinity(everywhere) and assume that it has a local extreme value at the point (2,0). Let g(x)=xf(x)+1 and h(x)=xf(x)+x+1 for all values of x.
Evaluate g(2), h(2), g'(2) and h'(2).

2. ## Re: evaluatng others from extreme value of one function

Hello, asifrahman1988!

$\text{Suppose }f(x)\text{ is differentiable everywhere.}$
$\text{Assume that }f(x)\text{ has a local extreme value at the point }(2,0).$
$\text{Let }g(x)\:=\:x\!\cdot\!f(x)+1\,\text{ and }\,h(x)\:=\:x\!\cdot\!f(x)+x+1\,\text{ for all values of }x.$

$\text{Evaluate: }g(2),\;h(2),\;g'(2),\;h'(2)$

We are given some information . . .
. . $(2,0)\text{ is on the graph: }\:f(2) = 0$
. . $\text{The first derivative is 0 when }x=2\!:\;f'(2) \,=\,0$

$g(x) \;=\;x\!\cdot\!f(x) + 1$
$g(2) \;=\;2\!\cdot\!f(2) + 1 \;=\;2\!\cdot\!0 + 1 \;=\;1$

$h(x) \;=\;x\!\cdot\!f(x) + x + 1$
$h(2) \;=\;2\!\cdot\!f(2) + 2 + 1 \;=\;2\!\cdot\!0 + 2 + 1 \;=\;3$

$g'(x) \;=\;x\!\cdot\!f'(x) + f(x)$
$g'(2) \;=\;2\!\cdot\!f'(2) + f(2) \;=\;2\!\cdot\!0 + 0 \;=\;0$

$h'(x) \;=\;x\!\cdot\!f'(x) + f(x) + 1$
$h'(2) \;=\;2\!\cdot\!f'(2) + f(2) + 1 \;=\;2\!\cdot\!0+ 0 + 1 ;=\;1$


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# Problem 6-32 Gerald/Brooke, Ltd. manufactures shirts, which it sells to customers for embroiderin...

Problem 6-32

Gerald/Brooke, Ltd. manufactures shirts, which it sells to customers for embroidering with various slogans and emblems. The standard cost card for the shirts is as follows.

 Standard Price Standard Quantity Standard Cost Direct materials $1.6 per yard 1.25 yards$2 Direct labor $12 per DLH 0.25 DLH 3 Variable overhead$4 per DLH 0.25 DLH 1 Fixed overhead $6 per DLH 0.25 DLH 1.5$7.50

Bobby Brickley, operations manager, was reviewing the results for November when he became upset by the unfavorable variances he was seeing. In an attempt to understand what had happened, Bobby asked CFO Lila Davis for more information. She provided the following overhead budgets, along with the actual results for November.

The company purchased and used 119,500 yards of fabric during the month. Fabric purchases during the month were made at $1.45 per yard. The direct labor payroll ran$257,730, with an actual hourly rate of $12.1 per direct labor hour. The annual budgets were based on the production of 1,003,870 shirts, using 255,100 direct labor hours. Though the budget for November was based on 84,900 shirts, the company actually produced 85,200 shirts during the month.  Variable Overhead Budget Annual Budget Per Shirt November—Actual Indirect material$453,300 $0.45$36,800 Indirect labor 301,600 0.3 33,850 Equipment repair 202,500 0.2 18,000 Equipment power 52,000 0.05 13,100 Total $1,009,400$1.00 $101,750  Fixed Overhead Budget Annual Budget November—Actual Supervisory salaries$263,500 $22,400 Insurance 351,000 33,100 Property taxes 80,200 8,700 Depreciation 323,100 34,700 Utilities 212,500 24,300 Quality inspection 282,000 32,500 Total$1,512,300 $155,700 (a) Calculate the direct materials price and quantity variances for November. (If variance is zero, select "Not Applicable" and enter 0 for the amounts.)  Direct material price variance$ FavorableUnfavorableNot Applicable Direct material quantity variance $UnfavorableFavorableNot Applicable (b) Calculate the direct labor rate and efficiency variances for November. (Round answers to 0 decimal places, e.g. 125. If variance is zero, select "Not Applicable" and enter 0 for the amounts.)  Direct labor rate variance$ UnfavorableFavorableNot Applicable Direct labor efficiency variance $FavorableUnfavorableNot Applicable (c) Calculate the variable overhead spending and efficiency variances for November. (Round answers to 0 decimal places, e.g. 125. If variance is zero, select "Not Applicable" and enter 0 for the amounts.)  Variable overhead spending variance$ FavorableNot ApplicableUnfavorable Variable overhead efficiency variance $UnfavourableNot ApplicableFavourable (d) Calculate the fixed overhead spending variance for November. (Round answer to 0 decimal places, e.g. 125. If variance is zero, select "Not Applicable" and enter 0 for the amounts.)  Fixed overhead spending variance$ FavorableUnfavorableNot Applicable

Manufactures shirts, which it sells to customers for with the standard cost card for the shirts is as follows. Bobby when he became upset by the The company and used 119. shirts, the company actually produced 85, 200 shirts during the month. Calculate the direct materials price and quantity for November. Direct material price variance Direct material quantity variance Calculate the direct labor rate and variance for November. Calculate the variable overhead for November. Variable overhead spending variance Calculate the fixed for November. Fixed overhead spending variance


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# Continuing a Charge after enemy shifts

Can a player doing a Charge action, continue the movement after the target uses an Immediate Interrupt to do something, e.g. Shift or Teleport, to move away ?

And what direction do they have to go? ... *I'm thinking here of the target moving back along the line of the charge, and the player having to reverse the direction of the Charge ... but what about the pseudo "momentum" effect*

-

It depends on what triggered the shift.

As @ioanwigmore pointed out, PHB1 p.287 describes a Charge as (emphasis mine):

Move and Attack: Move your speed as part of a charge and make a melee basic attack or a bull rush at the end of your move.

As @Ravn pointed out, PHB1 pg 268 describes an Immediate Interrupt as (emphasis mine):

An immediate interrupt lets you jump in when a certain trigger condition arises, acting before the trigger resolves. If an interrupt invalidates a triggering action, that action is lost. For example, an enemy makes a melee attack against you, but you use a power that lets you shift away as an immediate interrupt. If your enemy can no longer reach you, the enemy’s attack action is lost.

Case 1: If the attack was the trigger (for example, Trigger: You are the target of a melee attack.), then movement is already done. The charger has already transitioned from the "movement" phase of the charge to the "attack" phase, and can't move any more.

Case 2: If the movement was the trigger (for example, Trigger: A foe enters a square adjacent to you.) then the shift happens before the "movement" phase of the charge resolves (i.e. finishes). In that case the charger can continue as long as they have movement left and each square of movement brings them closer to the target's new location.

The pseudo-momentum doesn't say you can't change directions, only that you must always be getting closer to your target. For example, because of how 4e counts diagonals, the following 90 degree turn in mid-charge is perfectly valid:

oTooooo
ooFoooo
oooPooo
ooooPoo
XXXooPo
ooooPoo
oooPooo
ooSoooo

o = empty space, X = obstacles, T = target, S = charger's starting position, F = charger's final position, P = charger's path

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No.

As the rules in PH1 (page 287) for charge read

Move and Attack: Move your speed as part of a charge and make a melee basic attack or a bull rush at the end of your move.

I'd argue that if the interrupt occurs during the melee basic attack (or bull rush or substituted power), and not during the move, then by RAW "end of your move" means no you cannot keep on charging, as it is the attack roll that has been interrupted, not the movement.

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This is true. I guess it depends on what the interrupt is in response to. If it's the MBA, then ioanwigmore's answer seems right. If it's the charger moving adjacent to the interrupter, see my reply below. – Ravn Jun 3 '12 at 17:03

Assuming the interrupt trigger was the charger moving adjacent to the interrupter:

In Player's Handbook 1, page 268, it says about immediate interrupts;

lets you jump in when a certain trigger condition arises, acting before the trigger resolves. If an interrupt invalidates a triggering action, that action is lost. For example, an enemy makes a melee attack against you, but you use a power that lets you shift away as an immediate interrupt. If your enemy can no longer reach you, the enemy’s attack action is lost.

Since the interrupt acts before the trigger (the charge) is fully resolved, I'd say the charging player could finish his move, provided he still has squares of movement left for it.

If the interrupter moves out of range of the charge, it's invalidated and lost (the charge stops).

If the interrupt trigger is the MBA, the charger cannot again. As ioanwigmore noted in his answer, the MBA part of a charge comes after the movement is resolved.

As for direction, I'm not completely sure. As far as I can tell from the RAW on charging (PH1, page 287 and errata), there's no reason why a player couldn't continue charging and just change direction, as long as the movement gets him closer to the target.

To preserve momentum, one could argue that the charging player picks a specific square as his charge target. This way, if the interrupter simply shifts, the charger should still be able to hit him with a MBA, but if the interrupter teleports behind the charger, he's potentially out of reach.

-


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# Math Help - Question about triangles and their exterior angle

1. ## Question about triangles and their exterior angle

I got this question wrong in my GRE math prep, and I have no idea why.

Triangle ABC is an isosceles triangle with AB = BC. The side AC is extended to point D, so that BCD forms an exterior angle. (See attached picture)

Which of the following can be concluded?

Among a few other answers, I circled: Measure of angle BCD is greater than measure of angle BCA.

This was not one of the answers in the answer key, and the explanation said that the relative measures of angle BCD and BCA could not be determined.

I feel that that is somehow wrong because triangle ABC is an isosceles.

I played with it for a long time, trying to prove it, and this is what I came up with, but I'm not certain at all. (Also I don't know how to do the angle sign, so I'm just going to use an < for now, and I've used a picture with the angles labeled with numbers to make it easier to identify).

<1 = <3 because triangle ABC is isosceles, with AB=BC
<1 + <2 +<3 = 180
<3 + <4 = 180

If <1=<3, then 2 times <3 + <2 = 180.

2 times <3 + <2 = 180 = <3 + <4
<3 + <2 = <4

So, <4 must be greater than <3 unless <2 =0

I don't know if this is correct or not, can anyone explain it to me in another way?

2. ## Re: Question about triangles and their exterior angle

$\angle 1 + \angle 2 + \angle 3 = 180^\circ$. Since $\angle 1 = \angle 3$ it can easily be shown that

$\angle 3 < 90^{\circ}$. It follows that $\angle 4 > 90^{\circ}$, and $\angle 4 > \angle 3$.

3. ## Re: Question about triangles and their exterior angle

Oh, I hadn't even thought of that. Of course, because otherwise <3 + <1 would be 180.

So, that is something I can conclude after all? Silly prep book

4. ## Re: Question about triangles and their exterior angle

Yep. A lot of prep books are badly flawed.

5. ## Re: Question about triangles and their exterior angle

Originally Posted by Blueberry
I got this question wrong in my GRE math prep, and I have no idea why.

Triangle ABC is an isosceles triangle with AB = BC. The side AC is extended to point D, so that BCD forms an exterior angle. (See attached picture)

Which of the following can be concluded?

Among a few other answers, I circled: Measure of angle BCD is greater than measure of angle BCA.

This was not one of the answers in the answer key, and the explanation said that the relative measures of angle BCD and BCA could not be determined.

I feel that that is somehow wrong because triangle ABC is an isosceles.

I played with it for a long time, trying to prove it, and this is what I came up with, but I'm not certain at all. (Also I don't know how to do the angle sign, so I'm just going to use an < for now, and I've used a picture with the angles labeled with numbers to make it easier to identify).

<1 = <3 because triangle ABC is isosceles, with AB=BC
<1 + <2 +<3 = 180
<3 + <4 = 180

If <1=<3, then 2 times <3 + <2 = 180.

2 times <3 + <2 = 180 = <3 + <4
<3 + <2 = <4

So, <4 must be greater than <3 unless <2 =0

I don't know if this is correct or not, can anyone explain it to me in another way?
Since

$\angle1 + \angle2 + \angle3 = 180^\circ$ ....... and

$\angle3 + \angle4 = 180^\circ$

follows: $\angle4 = \angle1 + \angle2$

6. ## Re: Question about triangles and their exterior angle

Hello, Blueberry!

I got this question wrong in my GRE math prep, and I have no idea why.

Triangle ABC is an isosceles triangle with AB = BC.
The side AC is extended to point D, so that BCD forms an exterior angle. (See attached picture)

Code:
            B
o
* *
* 2 *
*     *
*       *
* 1     3 * 4
A o * * * * * o * * * o D
C
Which of the following can be concluded?

Among a few other answers, I circled: Measure of / BCD is greater than measure of / BCA.
. . You are correct!

The explanation said that the relative measures of / BCD and BCA could not be determined.
. . This is wrong!

I feel that that is somehow wrong because triangle ABC is an isosceles. . Right!

Putting it simply: $\angle 3$ must be acute; therefore, $\angle 4$ is obtuse.

Here is another proof . . .

We know that: . $\angle 1 + \angle 2 + \angle 3 \:=\:180^o$

Since $AB = BC,\:\angle 1 = \angle 3.$

So we have: . $2\angle 3 + \angle 2 \:=\:180^o$

Since $\angle 2 > 0^o,\,\text{ then: }\,2\angle 3 \:<\:180^o$

Therefore: . $\angle 3 \:<\:90^o\:\text{ . . . }\:\angle 3\text{ is acute.}$

Even more obvious: $\text{If }\angle 3 \ge 90^o,\:there\;is\;no\;triangle.$


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# How to calculate the mass of a component in battery acid based on the density and mass fraction?

Battery acid has a density of $\pu{1.285 g cm^-3}$ and contains $38\%$ by weight sulfuric acid. How many grams of pure sulfuric acid are contained in a liter of battery acid?

I just took $1.285 \times 0.38$ and got how many grams of sulfuric acid are in $\pu{1 cm^3}$ of battery acid. So now I just converted it into grams per liter and got $\pu{488.3 gL^-1}$ of sulfuric acid in battery acid. I'm not sure if I did it right though.

• Please have a look at this tutorial to acquaint yourself with the way math and chemical formulae can be nicely formatted on this site. – Philipp Sep 14 '14 at 5:06
• duplicate of math.stackexchange.com/questions/930540/… – RE60K Sep 14 '14 at 6:36
• Please do not cross-post your questions. It results in a duplication of efforts and wastes the time of the people that have answered it. – jonsca Sep 14 '14 at 10:36

$\textrm{density} = \frac{\textrm{mass}}{\textrm{volume}}$
$\textrm{mass}=\textrm{density} \times \textrm{volume} = 1.285 \times 1000=1285\ \textrm{g}$


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# Chromatic polynomial of a $8$-vertex graph

I'm supposed to determine chromatic polynomial of a $8$-vertex graph, which is the result of sticking

There are two complete $5$-vertex graphs, each with one edge removed, in the first graph we remove the edge $uv$ and in the second graph we remove $u'v'$. Then we stick the graphs together by putting $u=u'$ and $v=v'$ . I hope the description is quite clear, because I don't know how to insert the image of this graph.

It seems pointless to try using the recurrence formula for chromatic polynomial of a graph (by trying to get complete graphs by contracting together two unconnected vertices and drawing an additional edge linking them).

Could you help me with that?

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If we take two copies of $K_5$ and merge two vertices in one copy with two vertices in the other, the resulting graph has only eight vertices. –  Chris Godsil Jan 29 '13 at 19:29
Oh, right, it is an 8-vertice graph. –  Hagrid Jan 29 '13 at 20:59

It's easy to see this from first principles: remember that $P(x)$ is the number of colourings of the graph using $x$ colours.

If you fix the colours of the two vertices in the middle (which used to be $u$ and $v$), then the remaining graph splits into two separate complete graphs which are very simple to count colourings in.

(At this point you should ask yourself: if there are $3$ ways to colour each copy, how many different ways are there to colour both together? Hint: it's not $6$.)

The exact number of colourings of each complete graph (as a function of $x$) depends on whether you colour $u$ and $v$ the same colour or distinct colours, so consider these two cases separately, and remember to account for how many ways there are to choose the two initial colours.

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Well, if we consider a graph with 5 vertices where $u$ and $v$ are not connected then for such a graph if $u$ and $v$ have different colours we get x(x-1)(x-2)(x-3)(x-4) possible colourings and if $u$ and $v$ have the same colours we get x(x-1)(x-2)(x-3) possible colourings. Now, the answer in my textbook is $\frac{(x(x-1)(x-2)(x-3)(x-4))^2}{x(x-1)} + \frac{(x(x-1)(x-2)(x-3))^2}{x}$. Could you tell me what to do next? Or did I just understand you incorrectly? –  Hagrid Jan 29 '13 at 21:23
@Hagrid You're on the right track but you need to take into account that your graph contains two such graphs of 5 vertices. There are $x(x-1)$ ways to choose $u$ and $v$ with different colours, and $x$ ways to choose them with the same colour. How many choices for the remaining 6 vertices? It helps if you have given some thought about where formulas like $x(x-1)(x-2)(x-3)$ come from. –  Erick Wong Jan 29 '13 at 21:41
From the picture that @Robert Israel presented below we can see that if $u, \ v$ have the same colour, then we choose colour of the base and then of the vertices of three triangles in x(x-1)(x-2)(x-3) ways and if their colours are different we do it in x(x-1)(x-2)(x-3)(x-4) ways. Is that correct? –  Hagrid Jan 29 '13 at 22:09
@Hagrid Let's make this more concrete by picking, say $x=6$. There are $6\cdot 5 = 30$ ways to colour $u$ and $v$ differently. Now there are $4$ colours remaining, since we cannot reuse $u$ and $v$'s colours: how many ways can you colour all six remaining vertices? (Certainly there is more freedom in colouring two triangles than just one of them.) –  Erick Wong Jan 29 '13 at 22:41
Having coloured the two middle vertices (1 and 2), we can the colour the "3" in 4 ways, "4" in 3 ways and "5" in 2 ways. Since We can repeat it for "3'", "4'", "5'" because they are not directly linked to "3", "4", "5". –  Hagrid Jan 29 '13 at 23:10

I assume this is your graph. Shown is a 4-colouring. It's easy to see there are $4$-cliques, so the chromatic number is $4$.

EDIT: I suggest you look at two separate cases, where $1$ and $2$ have the same colour and where they have different colours. Note that given the colours of vertices $1$ and $2$, vertices $3$, $4$ and $5$ can have any three different colours that are not those of $1$ and $2$, and similarly for $3'$, $4'$ and $5'$.

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But the question asks for the chromatic polynomial, not the chromatic number. –  Chris Godsil Jan 29 '13 at 21:33
Yes, but see the last paragraph. –  Robert Israel Jan 29 '13 at 21:47

So, here's the graph:

and another method for finding the chromatic polynomial.

We can use the addition/identification law on the vertices $x$ and $y$ to compute the chromatic polynomial. The law states that if $x$ and $y$ are not adjacent, then the $$P(G,\lambda)=P(G+xy,\lambda)+P(G/xy,\lambda).$$

In this case, the chromatic polynomial of the above graph is the sum of the chromatic polynomials of the graphs:

and

These graphs are clique separable copies of complete graphs. Thus, they have the chromatic polynomials $$\frac{P(K_5;\lambda)^2}{P(K_2;\lambda)}$$ and $$\frac{P(K_4;\lambda)^2}{P(K_1;\lambda)},$$ respectively.

The chromatic polynomial of the complete graph $K_n$ is $$P(K_n;\lambda)=\lambda(\lambda-1)\cdots(\lambda-n+1).$$

Hence the chromatic polynomial of the original graph is: $$\frac{\big(\lambda(\lambda-1)(\lambda-2)(\lambda-3)(\lambda-4)\big)^2}{\lambda(\lambda-1)}+\frac{\big(\lambda(\lambda-1)(\lambda-2)(\lambda-3)\big)^2}{\lambda}$$ which simplifies to $$\lambda(\lambda-1)(\lambda-2)^2(\lambda-3)^2(\lambda-7\lambda+15).$$

-


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TROPICAL GEOGRAPHY ›› 2015, Vol. 35 ›› Issue (4): 601-606.

### Applications of EEMD in the Trends Analysis of the Thunderstorm Days

CHEN Zehuang1，ZHANG Yufeng1，XIE Fei2，HUO Guang3，CAO Hongliang1

1. （1．Nanjing University of Information Science & Technology，Collaborative Innovation Center on Forecast and Evaluation of Meteorological Disasters// Key Laboratory for Aerosol-Cloud-Precipitation of China Meteorological Administration，Nanjing 210044，China；2．College of Meteorology and Oceanography，PLA University of Science and Technology，Nanjing 211101，China；3．Qinggang County Meteorological Bureau，Suihua 152002，China）
• Online:2015-07-03 Published:2015-07-03

Abstract:

The EEMD (Ensemble empirical mode decomposition) was used to analyze the thunderstorm days of Hong Kong to clarify the applicability in the trends analysis of lightning day. At first, the IMF (intrinsic mode function) components were decomposed based on EEMD, and then the Hilbert transform was also used to extract the features of each IMF component of the thunderstorm days. Also both the Hilbert spectrum and the marginal spectrum were showed in this paper to illustrate the variation features of the days. Lastly, the significance test of the thunderstorm days of IMF component was also made to illustrate the reliability of the IMFs in analyzing the trends. According to the research the conclusions can be drawn as follows: the thunderstorm days in Hong Kong could be decomposed into a trend term and five IMF components with different center frequencies. And the energies of the IMFs were mainly concentrated in 0.35~0.5 Hz and 0~0.05 Hz. Based on the analysis of the energy spectrum density and period distribution of the IMFs, it proved that 2.8 a interannual variation and 25 a generational variation of lightning day were the main cycles, and 4.5 and 7.1 a were the secondary cycle. Besides, from the trend term, the thunderstorm days in Hong Kong grown in wave-like form. From the results, the EEMD algorithm can be applied to the analysis of the characteristics of thunderstorm day’s trend better.


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# How do you find the derivative of [e^x / (1 - e^x)]?

Sep 7, 2016

$y ' = {e}^{x} / {\left(1 - {e}^{x}\right)}^{2}$.

#### Explanation:

Let $y = {e}^{x} / \left(1 - {e}^{x}\right)$.

To find $y '$, we will use the Quotient Rule , which states, :

$\left(\frac{u}{v}\right) ' = \frac{v u ' - u v '}{v} ^ 2$.

$y = {e}^{x} / \left(1 - {e}^{x}\right)$

$\Rightarrow y ' = \frac{\left(1 - {e}^{x}\right) \left({e}^{x}\right) ' - {e}^{x} \left(1 - {e}^{x}\right) '}{1 - {e}^{x}} ^ 2$.

$= \frac{\left(1 - {e}^{x}\right) \left({e}^{x}\right) - {e}^{x} \left(1 ' - \left({e}^{x}\right) '\right)}{1 - {e}^{x}} ^ 2$.

$= \frac{\left(1 - {e}^{x}\right) {e}^{x} - {e}^{x} \left(0 - {e}^{x}\right)}{1 - {e}^{x}} ^ 2$.

$= \frac{{e}^{x} - {e}^{2 x} + {e}^{2 x}}{1 - {e}^{x}} ^ 2$

$\therefore y ' = {e}^{x} / {\left(1 - {e}^{x}\right)}^{2}$.

Enjoy Maths.!


text:
### Home > CCG > Chapter 1 > Lesson 1.2.5 > Problem1-98

1-98.

At right is a diagram of a regular hexagon with center $C$. A polygon is regular if all sides are equal and all angles are equal. Copy this figure on your paper, then answer the questions below.

1. Draw the result of rotating the hexagon about its center $180º$. Explain what happened. When this happens, the shape has rotation symmetry.

The orientation of the hexagon does not change.

2. What is the result when the original hexagon is reflected across line $n$, as shown at right? A shape with this quality is said to have reflection symmetry and line $n$ is a line of symmetry of the hexagon (not of the reflection).

The orientation of the hexagon does not change.

3. Does a regular hexagon have any other lines of symmetry? That is, are there any other lines you could fold across so that both halves of the hexagon will match up? Find as many as you can.


text:
# Oblem 1: An air conditioner based on the theoretical single-stage cycle is using Refrigerant- 134a as...

###### Question:

oblem 1: An air conditioner based on the theoretical single-stage cycle is using Refrigerant- 134a as the refrigerant operates with a condensing temperature of 100°F and evaporating temperature of 0°F. The air conditioner produces 5 tons of refrigeration. Determine: (a) The coefficient of performance of the cycle and the rate of refrigerant flow. (b) The condenser of the air conditioner is designed to reject the heat from the refrigerant. Air (Cp 0.24 Btu/lb. °F, p 0.077 lb/ft3) flows across the finned condenser coils, entering at 77°F and leaving at 95oF. If the overall heat transfer coefficient based on the refrigerant side is 30.0 Btu/hr. ft2 °F, determine the heat transfer area on the refrigerant side. (c) The air volumetric flow rate. R-134a
R-134a 95°F Ain 77℉ .90rp.- ooE

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##### 1. A solution prepared by dissolving 15.0 of thiophene (C4H,S) in 400.0 mL of toluene (CzHs). The density of thiophene is 1.065 g/mL and the density of toluene is 0.867 g/mL_ points) Complete the table below _ Assume that volumes are additive.maSSvolume (mL)amount (mol )C,H,s15.0C-Hs400,0Solution
1. A solution prepared by dissolving 15.0 of thiophene (C4H,S) in 400.0 mL of toluene (CzHs). The density of thiophene is 1.065 g/mL and the density of toluene is 0.867 g/mL_ points) Complete the table below _ Assume that volumes are additive. maSS volume (mL) amount (mol ) C,H,s 15.0 C-Hs 400,0 Sol...
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##### . Increased levels of circulating ATP would likely inhibit which of the following pathways? a. Protein...
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##### 3_ (20 points) Compute the SVD of the matrix2 1 _1 ~2A =
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##### Which of the following is FALSE regarding the optical activity? Enantiomers rotate plane polarize light to...
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product for the dehydration reaction shown below? What CH,CHOH esos CH,CH_OH CH,OCH, CHz-CHz None CH,CH,OCH;Cha c - 2What is the multiplicity (splitting pattern) of the methine hydrogen (CH) indicated in the proton NMR of the following compound? CH,CHCHGCHs A. singlet B. doublet C. triplet D. quart...
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##### Two forcesand F act on 9.20-kg object: F = 35.0 N and Fz 16.0 N:TeeMnJheneneehanrh90.0"60.09(a) Find the acceleration of the object for the configuration of forces shown in Figure (a) magnitude 183 m/s2 direction 24.57 (counterclockwise trom(b) Find the acceleration of the object for the configuration of forces shown in Figure (b). 9.52 magnitude You find the magnitude of force in the same manner you find the magnitude of any other vector: first resolve each vector into components. m/s2 299
Two forces and F act on 9.20-kg object: F = 35.0 N and Fz 16.0 N: TeeMn Jheneneehanrh 90.0" 60.09 (a) Find the acceleration of the object for the configuration of forces shown in Figure (a) magnitude 183 m/s2 direction 24.57 (counterclockwise trom (b) Find the acceleration of the object for the...
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##### Use translating to graph each function and state the domain and range. See Example 2. $f(x)=\sqrt{x+6}$
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##### ThinkingSono follouing calculations With Problors, shorlng &lf Ihe oquations usod and Ihe rulos correct unlls_ Somcaut Express Ihe arswers occording Dquss8 ^ 47 4 sample of a motal Is nealed ofwuler 418 C) at 21,5"C. 95,.2"C and then placed The final lemperature calorimoter containing 120.0 spocific heat capacity of thc metal useo? the water What 24.5"C. What the ntetal?chunk of mcrcury 80.89C was added 200.0 0l waler '5"C . The pecilc heat ol mercury is 0.14 Jg &
Thinking Sono follouing calculations With Problors, shorlng &lf Ihe oquations usod and Ihe rulos correct unlls_ Somcaut Express Ihe arswers occording Dqus s8 ^ 47 4 sample of a motal Is nealed ofwuler 418 C) at 21,5"C. 95,.2"C and then placed The final lemperature calorimoter containin...
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##### Question 1. Determine the value of that will draw the maximum power from the rest of...
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Brief Exercise 15-11 Cheyenne Inc. owns shares of Ayayai Corporation stock. At December 31, 2017, the securities were carried in Cheyenne’s accounting records at their cost of $979,000, which equals their fair value. On September 21, 2018, when the fair value of the securities was$1,092,000, ...


text:
# rcond

Reciprocal condition number

## Syntax

``C = rcond(A)``

## Description

example

````C = rcond(A)` returns an estimate for the reciprocal condition of `A` in 1-norm. If `A` is well conditioned, `rcond(A)` is near 1.0. If `A` is badly conditioned, `rcond(A)` is near 0. ```

## Examples

collapse all

Examine the sensitivity of a badly conditioned matrix.

A notable matrix that is symmetric and positive definite, but badly conditioned, is the Hilbert matrix. The elements of the Hilbert matrix are $H\left(i,j\right)=1/\left(i+j-1\right)$.

Create a 10-by-10 Hilbert matrix.

`A = hilb(10);`

Find the reciprocal condition number of the matrix.

`C = rcond(A)`
```C = 2.8286e-14 ```

The reciprocal condition number is small, so `A` is badly conditioned.

The condition of `A` has an effect on the solutions of similar linear systems of equations. To see this, compare the solution of $Ax=b$ to that of the perturbed system, $Ax=b+0.01$.

Create a column vector of ones and solve $Ax=b$.

```b = ones(10,1); x = A\b;```

Now change $b$ by `0.01` and solve the perturbed system.

```b1 = b + 0.01; x1 = A\b1;```

Compare the solutions, `x` and `x1`.

`norm(x-x1)`
```ans = 1.1250e+05 ```

Since `A` is badly conditioned, a small change in `b` produces a very large change (on the order of 1e5) in the solution to `x = A\b`. The system is sensitive to perturbations.

Examine why the reciprocal condition number is a more accurate measure of singularity than the determinant.

Create a 5-by-5 multiple of the identity matrix.

`A = eye(5)*0.01;`

This matrix is full rank and has five equal singular values, which you can confirm by calculating `svd(A)`.

Calculate the determinant of `A`.

`det(A)`
```ans = 1.0000e-10 ```

Although the determinant of the matrix is close to zero, `A` is actually very well conditioned and not close to being singular.

Calculate the reciprocal condition number of `A`.

`rcond(A)`
```ans = 1 ```

The matrix has a reciprocal condition number of `1` and is, therefore, very well conditioned. Use `rcond(A)` or `cond(A)` rather than `det(A)` to confirm singularity of a matrix.

## Input Arguments

collapse all

Input matrix, specified as a square numeric matrix.

Data Types: `single` | `double`

## Output Arguments

collapse all

Reciprocal condition number, returned as a scalar. The data type of `C` is the same as `A`.

The reciprocal condition number is a scale-invariant measure of how close a given matrix is to the set of singular matrices.

• If `C` is near 0, the matrix is nearly singular and badly conditioned.

• If `C` is near 1.0, the matrix is well conditioned.

## Tips

• `rcond` is a more efficient but less reliable method of estimating the condition of a matrix compared to the condition number, `cond`.

## Version History

Introduced before R2006a


text:
Problem with differential equation (DESOLVE)
03-03-2015, 05:29 PM
Post: #1
 ZellAllon Junior Member Posts: 4 Joined: Feb 2015
Problem with differential equation (DESOLVE)
Hello,

I don't know why when I try to solve a Homogeneous differential equation my hp prime I get something like:

desolve((y') = ((y/x)+(x/y)),y) = [pnt[G_0*e^((1/2)*_(t38)^2),G_0*_(t38)*e^((1/2)*_(t38)^2)]]

the solution must be Y^2=X^2*LN(X^2)+C*X^2

03-03-2015, 06:59 PM (This post was last modified: 03-03-2015 07:12 PM by salvomic.)
Post: #2
 salvomic Senior Member Posts: 1,394 Joined: Jan 2015
RE: Problem with differential equation (DESOLVE)
(03-03-2015 05:29 PM)ZellAllon Wrote:  Hello,

I don't know why when I try to solve a Homogeneous differential equation my hp prime I get something like:

desolve((y') = ((y/x)+(x/y)),y) = [pnt[G_0*e^((1/2)*_(t38)^2),G_0*_(t38)*e^((1/2)*_(t38)^2)]]

the solution must be Y^2=X^2*LN(X^2)+C*X^2

see here: Parisse replied to me few time ago...

the "strange" expression should be like
$y=c*e^{\frac{t^{2}}{2}} \ AND \ y=c*t*e^{\frac{t^{2}}{2}}$

G_0 ok for "c", but, yes, "_t38" is a bit bizzarre, and we are lucky that it is not "p38"

Note also that $$e^{\frac{t^{2}}{2}}$$ is simply $$\sqrt{e^{t^{2}}}$$ ...

∫aL√0mic (IT9CLU) :: HP Prime 50g 41CX 71b 42s 39s 35s 12C 15C - DM42, DM41X - WP34s Prime Soft. Lib
03-04-2015, 10:07 AM
Post: #3
 parisse Senior Member Posts: 1,208 Joined: Dec 2013
RE: Problem with differential equation (DESOLVE)
You get parametric solutions currently. With Xcas current CAS version, you would get
[√2*x*√(ln(x/G_0)),-√2*x*√(ln(x/G_0))]
03-04-2015, 04:49 PM
Post: #4
 ZellAllon Junior Member Posts: 4 Joined: Feb 2015
RE: Problem with differential equation (DESOLVE)
(03-03-2015 06:59 PM)salvomic Wrote:
(03-03-2015 05:29 PM)ZellAllon Wrote:  Hello,

I don't know why when I try to solve a Homogeneous differential equation my hp prime I get something like:

desolve((y') = ((y/x)+(x/y)),y) = [pnt[G_0*e^((1/2)*_(t38)^2),G_0*_(t38)*e^((1/2)*_(t38)^2)]]

the solution must be Y^2=X^2*LN(X^2)+C*X^2

see here: Parisse replied to me few time ago...

the "strange" expression should be like
$y=c*e^{\frac{t^{2}}{2}} \ AND \ y=c*t*e^{\frac{t^{2}}{2}}$

G_0 ok for "c", but, yes, "_t38" is a bit bizzarre, and we are lucky that it is not "p38"

Note also that $$e^{\frac{t^{2}}{2}}$$ is simply $$\sqrt{e^{t^{2}}}$$ ...
ok, understood, thanks.But how can I take these parametric solutions and get y as a function of x?
03-04-2015, 05:44 PM
Post: #5
 ZellAllon Junior Member Posts: 4 Joined: Feb 2015
RE: Problem with differential equation (DESOLVE)
(03-04-2015 10:07 AM)parisse Wrote:  You get parametric solutions currently. With Xcas current CAS version, you would get
[√2*x*√(ln(x/G_0)),-√2*x*√(ln(x/G_0))]

Hello,

How do you get that solution? explain to me please!
03-04-2015, 06:09 PM
Post: #6
 Tim Wessman Senior Member Posts: 2,280 Joined: Dec 2013
RE: Problem with differential equation (DESOLVE)
(03-04-2015 05:44 PM)ZellAllon Wrote:  How do you get that solution? explain to me please!

That is the author of the CAS inside Prime. He is using the pc version (which is newer then the current version in Prime) and that result is returned. If/Until that code is put into the Prime firmware the calculator will continue to return the result you posted.

TW

Although I work for HP, the views and opinions I post here are my own.
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text:
2019
Том 71
№ 10

### All Issues

Article (Russian)

### Approximation of continuous vector functions

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1435–1448

We study the possibility of uniform approximation of continuous mappings of metric compact sets into metric spaces. Notions of “weak dimension” and “weak Kolmogorov width” are introduced to compare approximating properties of infinite-dimensional subspaces. For classes of mappings specified by the majorant of the modulus of continuity, we present bilateral estimates of “weak” widths that may coincide under certain conditions.

Article (Russian)

### Diffusion approximation of normalized integrals of weakly dependent processes and its applications

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1449–1466

We obtain exact estimates for the rate of convergence of normalized integrals of weakly dependent stationary processes to the standard Wiener process in the uniform metric in probability. These estimates are then applied to the investigation of the behavior of stochastic systems with curvilinear boundaries subjected to the action of weakly dependent random perturbations.

Article (Ukrainian)

### Pointwise estimation of comonotone approximation

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1467–1472

We prove that, for a continuous function f(x) defined on the interval [−1,1] and having finitely many intervals where it is either nonincreasing or nondecreasing, one can always find a sequence of polynomials P n (x) with the same local properties of monotonicity as the function f(x) and such that ¦f(x)P n (x) ¦≤Cω2(f;n−2+n −1√1−x 2), whereC is a constant that depends on the length of the smallest interval.

Article (Russian)

### Factorization of matrices of permutation type

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1473–1478

By using the methods of the theory of algebraic functions, we present an explicit construction of the canonical factorization of matrices of permutation type given on an open contour.

Article (Ukrainian)

### On some classes of regular linear extensions of dynamical systems on a torus

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1479–1485

We select some classes of linear extensions of dynamical systems on a torus for which weak regularity implies regularity.

Article (Russian)

### On sets of regular growth for functions analytic in an open half plane

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1486–1501

We introduce the notion of the set of regular growth for functions analytic in an open half plane. In particular, for a function of completely regular growth in an open half plane, the entire half plane is its set of regular growth. Developed theory is applied to the solution of a problem of Hermitian interpolation in a class of functions of completely regular growth in a half plane with given indicator.

Article (Russian)

### Asymptotics of a solution of a discontinuous singularly perturbed Cauchy problem

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1502–1508

We construct the asymptotic expansion of a solution of the Cauchy problem for a singularly perturbed system of differential equations whose right-hand side is discontinuous on a certain surface. We consider the case where the surface of discontinuity is crossed and estimate the remainder of the constructed asymptotic expansion.

Article (English)

### Bogolyubov averaging and normalization procedures in nonlinear mechanics. II

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1509–1526

By using a new method suggested in the first part of the present work, we study systems which become linear in the zero approximation and have perturbations in the form of polynomials. This class of systems has numerous applications. The following fact is even more important: Our technique demonstrates how to generalize the classical method of Poincaré-Birkhoff normal forms and obtain new results by using group-theoretic methods. After a short exposition of the general theory of the method of asymptotic decomposition, we illustrate the new normalization technique as applied to models based on the Lotka-Volterra equations.

Article (Ukrainian)

### Information complexity of weakly singular integral equations

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1527–1533

We establish the exact power order of information complexity for integral equations whose kernels have power singularities and free terms belong to the corresponding Hölder space.

Article (Russian)

### Optimization of algorithms for the approximate solution of the Volterra equations with infinitely differentiable kernels

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1534–1545

For the Volterra equations with analytic kernels, we establish the exact power order of complexity of their approximate solutions and show that the optimal power order is realized by the method of simple iterations based on the use of information in the form of the values of kernels and free terms at certain points. In addition, for the Volterra equations with infinitely differentiable kernels, we determine the minimal order of the error of direct methods and construct a method which realizes this order.

Article (Russian)

### On the existence of interpolating $SK$-splines

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1546–1553

We establish new sufficient conditions for the existence and uniqueness of generalized interpolatingSK-splines with uniformly distributed interpolation nodes. Our results include all known important assertions obtained in this field as special cases.

Article (Ukrainian)

### On the error of the interpolation by bilinear splines

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1554–1560

We study the problem of approximation of functions from the classes Wr,s H ω and Wr,s H ω,2by bilinear splines. For some values of r ands, we obtain exact estimates of the error.

Brief Communications (Ukrainian)

### On the Nth diameters of continua

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1561–1563

We study an analog of the problem on then th diameters of continua in the complex plane.

Brief Communications (Ukrainian)

### On the uniqueness of solutions of the Dirichlet and Neumann problems for an elliptic second-order differential equation on a semiaxis

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1564–1567

For an elliptic second-order differential equation in a Banach space, we give a description of all solutions of the homogeneous Dirichlet and Neumann problems and establish conditions under which these problems are uniquely solvable.

Brief Communications (Russian)

### Stochastic analysis and one minimization problem

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1568–1571

We consider the problem of finding the form of a functional of an infinite-dimensional argument for which a certain given expression takes the minimum value for a fixed value of the parameter. The equation obtained for an unknown functional resembles equations with extended stochastic integral.

Brief Communications (Russian)

### Sufficient conditions for the complementability of maximal cyclic subgroups in a finite 2-group

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1572–1575

We present sufficient conditions for each maximal cyclic subgroup of a finite 2-group to be complementable.

Brief Communications (Russian)

### On conditions for convergence of Taylor-Dirichlet series in a convex domain

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1576-1581

We establish necessary and sufficient conditions for the absolute convergence of the series $$\mathop \sum \limits_{v = 1}^\infty \sum\limits_{k = 0}^{m_v - I} {a_{v,k} z^k \exp (\lambda _v z)}$$ in an open region. We also give conditions under which an arbitrary function analytic in a closed region (analytic in an open region and continuous in a closed region) can be represented by a series of this type.

Brief Communications (Ukrainian)

### On a mixed problem for systems of definite quasilinear partial differential equations with deviating argument

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1581–1585

For a mixed problem for a system of definite quasilinear pseudoparabolic equations with deviating argument, we prove a theorem on differential inequalities and existence of a unique regular solution and a comparison theorem and give sufficient conditions of existence of solutions with constant sign.

Brief Communications (Russian)

### Asymptotic inequalities for the distribution of the time of stay of a semi-Markov process in an expanding set of states

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1586–1590

We establish asymptotic estimates for the behavior of the distribution of the time of the first hit of an infinitely remote level by a semi-Markov process on a semiaxis of integer numbers.

Brief Communications (Ukrainian)

### On the solvability of the Riccati matrix algebraic equation

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1591–1593

We obtain conditions of solvability of the Riccati matrix algebraic equation.

Brief Communications (Russian)

### Two theorems on imbeddings of 0-dimensional groups

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1594-1596

In the class of 0-dimensional groups with infinite weight, the universal group is constructed. We prove that a 0-dimensional group can be imbedded into a multiplicative subgroup of a topological ring.

Brief Communications (Russian)

### Solutions of systems of nonlinear difference-differential equations of neutral type asymptotically bounded in the entire axis

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1597-1601

For a certain class of systems of difference-differential equations of neutral type, we study the properties of solutions asymptotically bounded on the entire axis.

Brief Communications (Russian)

### New condition of harmonicity of functions of infinitely many variables (translation nonpositive case)

Ukr. Mat. Zh. - 1994νmber=3. - 46, № 11. - pp. 1602-1605

A criterion of harmonicity of functions in a Hilbert space is given in the case of nonnegative second derivatives without using an assumption that they are mutually independent. This assumption is replaced by a weaker condition.


text:
Question 6

# In the following question, select the related letters from the given alternatives. MILK : PMOO :: TIPS : ?

Solution

The logic here is

M + 3 = P

I + 4 = M

L + 3 = O

K + 4 = O

Similarly,

T + 3 = W

I + 4 = M

P + 3 = S

S + 4 = W

$$\therefore\$$TIPS is related to WMSW in the same way MILK is related to PMOO.

Hence, the correct answer is Option D

• Free SSC Study Material - 18000 Questions
• 230+ SSC previous papers with solutions PDF

OR


text:
• question_answer Directions From the four answers given, shade the appropriate answer in the space provided for it on the OW Answer Sheet. Change 1/8 into percentage A) 12.5%              B) 15% C) 8%                               D) 25%

Ans.     Fro changing it into percentage $\Rightarrow \,\,\,\,\,\frac{1}{8}\times 100=12.5%$


text:
# calibration calibration

Author Comment Last activity Moderation
alejobonifacio "Hi @warren! Yes, 0.1-1 g/L is a useful range at least in rivers in this part of the world, in fact for this river test it worked well. I still have..." | Read more » 3 months ago
liz "Thank you for this close and careful work <3 " | Read more » 3 months ago
warren "Hi @alejobonifacio i wonder if there is an 'ideal' range if we can use resistors or capacitor sizes to shift that to our looked-for "search range" ..." | Read more » 3 months ago
warren "Also noting this page with good notes on blue filters: https://publiclab.org/questions/patalbright/10-20-2020/how-to-find-the-sweet-spot-for-manual..." | Read more » 4 months ago
PeterBella "@fongvania @stevie Thank you for the EPA Colocation Instruction Guide slide deck. @TravisLondon The info linked above in the slide deck does a grea..." | Read more » 7 months ago
stevie "Great info! thank you! " | Read more » 7 months ago
fongvania "From this Guidebook for Developing a Community Air Monitoring Network: If possible, permanently install a community monitor at the colocation site..." | Read more » 7 months ago
eustatic "I think it all depends on the sampling rate and whether the data converge, ha. For electronic monitors sampling every few seconds, two weeks isn't..." | Read more » 7 months ago
Ag8n "We were in an medical environment, usually class 100,000. These instruments are different from the ones used by public lab, but based on the same ..." | Read more » 7 months ago
Luniversdelalicorne "totally agree " | Read more » 12 months ago
Ag8n "A possible light source for a spectrometer--the light on the cell phone. But what is the spectrum like? I didn't know. Here is an answer from qu..." | Read more » 12 months ago
Ag8n "Look up didymium glass filters,PNB586 or BG20. That will be a start. Usually, if you find a good vendor, they will have a copy of the UV/Vis spec..." | Read more » 12 months ago
Ag8n "One of the alternate calibration methods, posted on a different thread, was to use photographic filters that were made of didymium. It sounds like..." | Read more » 12 months ago
dcanprdn "R G B colored low-price lasers are everywhere ??!!! Ebay, Aliexpress etc ADP " | Read more » 12 months ago
dcanprdn "Why not use low price R G B office laser pointers for visible light calibration ? ADP " | Read more » 12 months ago
Ag8n "Sorry, got a phone call. The same line listed for this spectrometer should be usable for the spectrum lab instrumenent. The lines are only listed..." | Read more » over 1 year ago
Ag8n "Checked on amazon. Found three spectrometers, similar in function to the spectrum lab. They are priced at $7,$9, and $37. The$37 is the best. I..." | Read more » over 1 year ago
givemehits "ok " | Read more » over 1 year ago
warren "Can you link us to the calibrated spectrum you're referring to? You can calibrate both from the Capture interface and from the Analysis page (where..." | Read more » over 1 year ago
stef "select the spectrum to be calibrated in SW click on copy calibration, put in the number of the spectrum to copy. that should do it. Stef " | Read more » over 1 year ago
amerikangold "Correct ! " | Read more » over 1 year ago
Ag8n "Or of holmium chloride solution, for that matter. Probably the solution would work better, but there would be done handling issues. " | Read more » over 1 year ago
Ag8n "Thanks for the alternate source of didymium glass. Do you know of any less expensive sources of holmium oxide glass? Thank you. " | Read more » over 1 year ago


text:
# Efficient way to storage large lists in a file(s)

Let us assume that I need to storage in a file, or files, data with the following form

alist =
Table[Table[RandomReal[{0, 1}, 2], {i, 1, 200}], {j, 1, 10000}];
blist =
Table[Table[{RandomReal[{0, 1}, 6], RandomReal[],
RandomReal[{0, 1}, 6], RandomReal[{0, 1}, 6]}, {i, 1, 200}], {j,
1, 10000}];
clist = RandomReal[{0, 1}, 10000];

In a few seconds, Mathematica is able to create these lists.

However, when I try to export this data to a .txt file, using

Export["C:\\pathname\\abclists.txt", {alist, blist, clist}];

My computer starts to freeze, and I have to abort the evalutation. When I check the result, Mathematica has created a 600MB .txt file.

Is there a more efficient way to storage data in a file or files? I have no preference for the file type, as long as I can easily rebuild the original data, preserving the list levels.

• Have you tried DumpSave? – Roman Jan 9 at 16:27

Using ByteCount will tell you to expect more than 1 GB:

ByteCount @ { alist, blist, clist }

(* 1 136 880 448 *)

Compress

I would thus try Compress:

cexpr = Compress @ { alist, blist, clist };
(* Export["pathname\\data.m", cexpr ]; *)
ByteCount @ cexpr

(* 437 359 280 *)

I am getting a 433 MB file (which roughly matches ByteCount). You can Uncompress the expression after loading.

BinarySerialize

Another possibility as of Version 11.1 or later is BinarySerialize:

bexpr = BinarySerialize @ { alist, blist, clist };
ByteCount @ bexpr

(* 378 170 128 *)

So we are down to about 378 MB (the file is 312 MB on my computer). You can use BinaryDeserialize to get the original expression again (see below for explicit instructions for writing/reading binary data).

If we give the option PerformanceGoal -> "Size"

bexpr = BinarySerialize[ {alist,blist,clist}, PerformanceGoal -> "Size" ];
ByteCount @ bexpr

(* 327 494 245 *)

we are down to about 327 MB.

The documentation tells you how to write/read binary data:

stream = OpenWrite[ "pathname\\data.mx", BinaryFormat -> True ];
BinaryWrite[ stream, bexpr ];
Close @ stream;

data = BinaryDeserialize @ ByteArray @ BinaryReadList[ "pathname\\data.mx", "Byte" ];

• Thanks for your answer. To load the data, would we use ReadList or Import? If it's with import, could you show me how? The documentation talks about 'elements', but I don't know what they are, and how they relate to our compressed data. I tried using ReadList, but I had to abort the evaluation... – An old man in the sea. Jan 9 at 14:36
• Considering that the structure contains $4.2\times10^7$ fully random elements with 8 bytes each, I don't think you can do much better, even with an approach that stores the structure of the data more efficiently – Lukas Lang Jan 9 at 14:37
• @LukasLang Yes, that should indeed be mentioned: The random data here are a "worst case" - but the problem still is a general one. – gwr Jan 9 at 14:38
• Thanks gwr. I've tried BinarySerialize, but Mathematica doesn't recognize it as a built-in function... Also, I've noticed that if we separate the file into 3, for each list, then the combined time for import, export, compress, and uncompress are much lower in total... – An old man in the sea. Jan 9 at 20:29
• @Anoldmaninthesea. One has to play around with it. I added the information, that BinarySerialize is only available for Version 11.1 or later. – gwr Jan 9 at 21:34

Old (OS and Mma version dependent!) "MX" format is also pretty fast and good for storing.

abcList = {alist, blist, clist};
ByteCount[abcList]

Out[2]= 1136880448

Export[
FileNameJoin[{NotebookDirectory[EvaluationNotebook[]],
"test.mx"}], abcList, "MX"];

FileByteCount[
FileNameJoin[{NotebookDirectory[EvaluationNotebook[]], "test.mx"}]]

Out[4]= 434116760

abcListRead =
Import[FileNameJoin[{NotebookDirectory[EvaluationNotebook[]],
"test.mx"}]];


text:
# A recreational math problem, integers in a grid

I was thinking of the following recreational math problem:

We have a $4\times 4$ square filled with integers $a_{1,1},...,a_{4,4}$. It has $30$ sub-squares $A_{i,j,k}$, corners of the form $a_{i,j},a_{i,j+k},a_{i+k,j}, a_{i+k,j+k}$, such that sum of the elements of $A_{i,j,k}$ are denoted by $s_1,\cdots ,s_{30}$.

How can I show that it is possible to choose the integers $a_{i,j}$ such that

$\{1,2,\cdots,24\}\subset \{s_1,\cdots,s_{30}\}$?

And is it possible to prove that always $\{1,2,\cdots,25\}\not\subset \{s_1,\cdots,s_{30}\}$?

And what kind of methods there are to find those numbers $a_{i,j}$ or to prove that their absolute value can't be too large?

My thoughts are that one can form a system of linear equations to solve the problem but there are over $24!$ cases to check whether the system has a solution.

-
So your sub-squares include k=1 case $A_{i,j,1}=[a_{i,j}]$ – Nate Iverson Jun 4 '12 at 14:42
Yes, that is right. – student Jun 4 '12 at 14:47