The post In a 3×3 matrix, entries aij are selected randomly from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 with replacement where each element aij is a three digit number. Find the probability that each element in each row is divisible by 15. appeared first on Math Dart.

]]>**Solution**:

Given: Each aij is a three-digit number

So, the hundredth place can’t be zero (0).

The total number of each dement of arrangement of 3 digits no. is 9 X 10 X 10 =900

Also Given: Each element in each now is divisible by 15. That means each element is divisible by both 3 and 5.

We know that:

- The Numbers divisible 5 contain either 0 or 5 in the unit digit place.
- For divisibility of 3, the sum of digits of the numbers must be divisible by 3.

Now every element in each row is divisible by 15 must end with 0 and 5 also it must satisfy the criteria of the sum of each digit must be divisible by 3.

**Case 1**: When the unit digit is 0

S1={ { (1,2,0), (1,5,0), (1,8,0), (2,1,0), (2,4,0), (2,7,0), (3,0,0), (3,3,0), (3,6,0)…(9,0,0), (9,3.0), (9,6,0)}

No. of possibility is, 3×9 = 27

**Case 2**: When unit digit is 5.

S2 = {(1,0,5), (1,3,5), (1,8, 5), (2,2,5)… }

Number of possibilities: 27

Total no of possibility: n(S1) + n(S1) = 27 + 27 = 54

**Probability = (3 digit no. divisible by 15) / (Total 3 digit no,)**

So, Probability = 54/900 = 3/50.

Therefore, the probability that each element in each row is divisible by 15 is 3/50.

**Answer**: 3/50 = 0.06.

I hope you get the solution of “In a 3×3 matrix, entries aij are selected randomly from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 with replacement where each element aij is a three digit number. Find the probability that each element in each row is divisible by 15.” This question had asked in SEBA Class 12 Board Exam and you can expect such type of rare question in upcoming exams.

The post In a 3×3 matrix, entries aij are selected randomly from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 with replacement where each element aij is a three digit number. Find the probability that each element in each row is divisible by 15. appeared first on Math Dart.

]]>The post What is the Factorial of 100? appeared first on Math Dart.

]]>For one thing, what precisely is a factorial? The factorial is the aftereffect of increasing all entire numbers in a picked number (for this situation 100) right down to 1.

You will generally see factorials expressed with an exclamation mark (!) after the number, like so: 100!

The factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n: n!= n * (n-1) * (n-2) * … * 2 * 1

So let’s take 100! and calculate the factorial by multiplying each whole numbers:

**100 x 99 x 98 x 97 x 96 x 95 x 94 …** = 9.3326215443944E+157

Componendo and Dividendo: Proof & Example

**Factorial of Hundred (100!) is exactly:** 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

For this situation, the quantity of entire numbers in 100 is more than five. You can perceive how this can rapidly go crazy with bigger numbers. Alexa can give up to bigger numbers, but for a normal person this will never be possible.

Factorials are utilized in math a considerable amount while working out the number of potential mixes or stages of something. On the off chance that you ponder rearranging a deck of 52 cards, you can utilize factorials to compute the number of potential orders there are.

Ideally, this article has helped you in your journey to work out the factorial of 100. Go ahead and share with your companions, family, instructors, and anyone who may be keen on factorials of numbers (which is unquestionably everybody!).

The post What is the Factorial of 100? appeared first on Math Dart.

]]>The post today is monday after 61 days it will be appeared first on Math Dart.

]]>**Answer**: If today is Monday after 61 days it will be Saturday.

I don’t when you search this question on any search engine, it give result as Friday to me. Probably the Artificial Intelligent needs some more logic. To solve this question we will simply need to know the multiplication of seven (7).

We’ve given Today is Monday.

So, in the multiplication of any seven days will get another Monday.

Like we will get Monday after 7, 14, 21, 28, 35 … days.

So, hereafter 63 days, we will get again a Monday

Now we only need after 61 days so we will subtract by two i.e 63-2=61.

Since the 63rd day is Monday. So the 62nd day will be Sunday and similarly, 61 days will be Saturday.

**Answer**: Saturday

One more method to solve “today is monday after 61 days it will be ?”

**Step 1:** 61 divided by 7 gives us 8.71

**Step 2:** Now 8 * 7 = 56.

**Step 3:** 61 – 56 = 5.

**Step 4:** Monday + 5 = Saturday.

**If today is Friday, what would be the day after 100 days?**

Sunday. To begin with, you need to decide how long will have passed. Separation. 100 separated by 7 equivalents 14.3 weeks. Second, you need to adjust to the closest number of weeks to decide the real number of days that will have passed. Increase. 14 x 7 = 98. At last, you need to add any excess days from the first number given. Expansion. 98 + 2 = 100. Since in 14 weeks it will in any case be Friday, then, at that point, you should add 2 days, which would make it Sunday.

We will perform same steps as “today is monday after 61 days it will be” Here are the means, once more:

**Step 1:**100 partitioned by 7 = 14.3

**Step 2:**14 x 7 = 98

**Step 3:**100 – 98 = 2

**Step 4:**Friday + 2 = Sunday.

I hope this detailed article on the question “today is monday after 61 days it will be” had helped you to understand and to solve this type of question.

The post today is monday after 61 days it will be appeared first on Math Dart.

]]>The post Componendo and Dividendo: Proof & Example appeared first on Math Dart.

]]>This is especially useful when working with equations involving fractions or rational functions in math competitions, especially when you’re looking at fractions. i.e \frac{a+b}{a-b}=\frac{c+d}{c-d}

Proof of **Componendo and Dividendo** \frac{a+b}{a-b}=\frac{c+d}{c-d}:

Add one both sides of the left and right-hand side fractions.

\begin{array}{l}
\Rightarrow \frac{a}{b}+1=\frac{c}{d}+1 \\
\Rightarrow \frac{a+b}{b}=\frac{c+d}{d}
\end{array}

Let this equation be (A)

Step: 2

Subtract one both sices of the left and right hand side fractions:

\begin{array}{l}
\Rightarrow \frac{a}{b}-1=\frac{c}{d}-1 \\
\Rightarrow \frac{a-b}{b}=\frac{c-d}{d}
\end{array}

Let this equation be (B)

Now divide equation divide A by B:

\begin{array}{l}
\Rightarrow \frac{\left[\frac{a+b}{b}\right]}{\left[\frac{a-b}{b}\right]}=\frac{\left[\frac{c+d}{d}\right]}{\left[\frac{c-d}{d}\right]}\\
\Rightarrow\left[\frac{a+b}{b}\right] \times\left[\frac{b}{a-b}\right]=\left[\frac{c+d}{d}\right] \times\left[\frac{d}{c-d}\right]\\
\Rightarrow\left[\frac{a+b}{a-b}\right] \times\left[\frac{b}{b}\right]=\left[\frac{c+d}{c-d}\right] \times\left[\frac{d}{d}\right]\\
\Rightarrow\left[\frac{a+b}{a-b}\right] \times\left[\frac{\not b}{\not b}\right]=\left[\frac{c+d}{c-d}\right] \times\left[\frac{\not d}{\not d}\right]\\
\Rightarrow\left[\frac{a+b}{a-b}\right] \times 1=\left[\frac{c+d}{c-d}\right] \times 1\\
\therefore \quad \frac{a+b}{a-b}=\frac{c+d}{c-d}
\end{array}

If a / b=c / d

- Invertendo: b / a=d / c
- Alternendo : a / c=b / d
- Componendo: (a+b), b=(c+d) / d
- Dividendo : (a-b) / b=(c-d) / d
- Componendo – Dividendo: (a+b) /(a-b)=(c+d)/(c-d)

**Componendo and Dividendo Example:**

If \frac{a}{\delta}=\frac{c}{d}, prove that \frac{3 a-8 b}{2 a+8 b}=\frac{2 c-8 d}{3 c-8 d}

**Method 1**. We have

\begin{array}{l}
\frac{a}{b}=\frac{c}{d} \\
\frac{3 a}{8 b}=\frac{3 c}{8 d} \\
\frac{3 a+9 b}{2 a-8 b}=\frac{3 c+8 d}{3 c-8 d} \\
\frac{3 a-9 b}{2 a+8 b}=\frac{3 c-8 d}{3 c+8 d}
\end{array}

**Must Know**: How many times are the hands of a clock at right angle in a day?

**Method 2**. Alternatively, we can make use of the property that \frac{a+k h}{a-k b}=\frac{c+k d}{c-k d}

If \frac{a}{b}=\frac{c}{d}

Then,

\frac{a+\left(\frac{8}{3}\right) b}{a-\left(\frac{8}{3}\right) b}=\frac{c+\left(\frac{8}{3}\right) d}{c-\left(\frac{8}{3}\right) d}

Multiplying both the numerator and the denominator by 2 on both sides:

\begin{array}{l}
\frac{3 a+8 b}{3 a-8 b}=\frac{3 c+8 d}{2 c-8 d} \\
\frac{3 a-8 b}{3 a+5 b}=\frac{3 ec-8 d}{2 c+8 d}
\end{array}

(applying invertendo)

In this article, we have seen Componendo and Dividendo proof and example. Hope you like it, and please share with your maths nerds and helped them out to clear their doubts.

The post Componendo and Dividendo: Proof & Example appeared first on Math Dart.

]]>The post How many times are the hands of a clock at right angle in a day? appeared first on Math Dart.

]]>The simple mathematical approach can be taken as follows to solve “How many times are the hands of a clock at a right angle in a day?”: In a 12 hour period, the minute hand makes 12 turns while the hour hand makes one turn.

If you switch to a rotating coordinate system where the hour handstands, the minute hand only makes 11 turns and 22 times at right angles to the hour hand.

In 24 hours a day, you get 2 × 22 = 44, times the hands make right angle.

Number of Times When the hands of a clock are at the right angle in a day (hours hand and minute hand) in ante meridiem and post meridiem:

Ante Meridiem |
Post Meridiem |

12:16 AM | 12:16 PM |

12:49 AM | 12:49 PM |

1:21 AM | 1:21 PM |

1:54 AM | 1:54 PM |

2:27 AM | 2:27 PM |

3:00 AM | 3:00 PM |

3:32 AM | 3:32 PM |

4:05 AM | 4:05 PM |

4:38 AM | 4:38 PM |

5:11 AM | 5:11 PM |

5:43 AM | 5:43 PM |

6:16 AM | 6:16 PM |

6:49 AM | 6:49 PM |

7:22 AM | 7:22 PM |

7:54 AM | 7:54 PM |

8:27 AM | 8:27 PM |

9:00 AM | 9:00 PM |

9:33 AM | 9:33 PM |

10:05 AM | 10:05 PM |

10:38 AM | 10:38 PM |

11:11 AM | 11:11 PM |

11:44 AM | 11:44 PM |

Here you can see in detail where the hands of the clock are at the right angle. There are two times when the hour’s hand appears in the post and ante meridiem appears only one time literally and i.e 2 AM, 2 PM period and 8 AM, 8 PM period. This happens when the 4th quarter completes.

It happens twice every 12 hours, once every 15 after and once a quarter, sometimes two for twenty-four hours = 48. But this decision has one drawback, namely that the clock is progressing, the minute’s needle positioned 90 degrees later in the hour.

Let us check the wrong approach most of us did for the first time.

Here we took both 8:55 AM and 9:00 AM both of which will never be possible until someone discovers a time machine. And we also took 2:55 and 3:00 together, where we did our fourth mistake, hence we got two more than the exact answer. If we subtract our errors, then we got the correct answer that is 44 (=48-4). Hope this will make you understand easily.

For more logical maths nerds let’s take an example, a 3:XX alignment at 90 degrees occurs at 3:33 (not 3:30) because the hour hand on the dial has passed 90 degrees, so the minute hand passes after “6”. In fact, at exactly 9:00, the minute hand will expire from the hour hand, for example, a 7:xx sync occurs at 07:54, but no 8:xx sync – it actually happens at 9:00. So we only get 11 90 degrees orientations in 12 hours. And it can double quarter to quarter because of that.

So 2 x 11 for 12 hours = 22 for twelve hours and 44 for 24 hours.

**Answer**: 44 times the hands of a clock at right angle in a day.

We know that: 1440 minutes in 24 hours.

The angular speed of the hour hand is 0.5° per minute.

The angular speed of the minute hand is 6° per minute.

In the clockwise direction, the angular speed of the minute hand is 5.5° per minute.

The angle is zero when the time is 12:00.

=> 90 ° / 5.5 gives 16.3636363636 minutes

=> 1440 / 16.36 is 88 (rounded up)

This is for 4 quadrants.

Since we only need to look at 2 quadrants.

So the answer is 88/2 = 44

So, here we discuss two different ways to solve a problem “How many times are the hands of a clock at right angle in a day?” in two approaches: mathematical and applied physics.

The post How many times are the hands of a clock at right angle in a day? appeared first on Math Dart.

]]>The post GATE Real Analysis 2021 Solutions appeared first on Math Dart.

]]>**GATE Real Analysis 2021 Problem 1:**

Let f_{n}:[0,10]\rightarrow \mathbb{R} be given by f_{n}(x)=n x^{3} e^{-n x} for n=1,2,3, \ldots .

Consider the following statements:

P: \left(f_{n}\right) is equicontinuous on [0,10].

Q: \sum_{n=1}^{\infty} f_{n} does NOT converge uniformly on [0,10].

Then

(A) both \mathrm{P} and \mathrm{Q} are TRUE

(B) \mathrm{P} is TRUE and \mathrm{Q} is FALSE

(C) P is FALSE and Q is TRUE

(D) both \mathrm{P} and \mathrm{Q} are FALSE

**GATE Real Analysis 2021 Problem 2:**

Let f: \mathbb{R}^{2} \rightarrow \mathbb{R} be given by

\begin{array}{cl}
f(x, y)= & \left\{\begin{array}{cl}
\sqrt{x^{2}+y^{2}} \sin \left(y^{2} / x\right) & \text { ir } x \neq 0 \\
0 & \text { if } x=0
\end{array}\right.
\end{array}

Consider the following statements:

P: f is continuous at (0,0) but f is NOT differentiable at (0,0).

Q: The directional derivative D_{u} f(0,0) of f at (0,0) exists in the direction of every unit vector u \in \mathbb{R}^{2}.

Then

(A) both \mathrm{P} and \mathrm{Q} are TRUE

(B) \mathrm{P} is TRUE and \mathrm{Q} is FALSE

(C) P is FALSE and Q is TRUE

(D) both \mathrm{P} and \mathrm{Q} are FALSE

**GATE Real Analysis 2021 Problem 3:**

Let f:\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) \rightarrow \mathbb{R} be given by f(x)=\frac{\pi}{2}+x-\tan ^{-1} x. Consider the following statements:

P: |f(x)-f(y)|<|x-y| for all x, y \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)

Q: f has a fixed point.

(A) both \mathrm{P} and \mathrm{Q} are TRUE

(B) \mathrm{P} is TRUE and \mathrm{Q} is FALSE

(C) P is FALSE and Q is TRUE

(D) both \mathrm{P} and \mathrm{Q} are FALSE

**GATE Real Analysis 2021 Problem 4:**

Let f: \mathbb{R}^{2} \rightarrow \mathbb{E} be given by f(x, y)=4 x y-2 x^{2}-y^{4} . Then f has

(A) a point of local maximum and a saddle point

(B) a point of local minimum and a saddle point

(C) a point of local maximum and a point of local minimum

(D) two saddle points

**Check** : Numerical Analysis GATE Solutions 2021 & 2020, Syllabus, Weightage (MA)

**GATE RA 2021 Problem 5:**

Consider the following statements:

P: d_{1}(x, y)=\left|\log \left(\frac{x}{y}\right)\right| is a metric on (0,1).

Q: d_{2}(x, y)=\left\{\begin{array}{cl}|x|+|y| . & \text { if } x \neq y \\ 0, & \text { if } x=y\end{array} \quad\right. is a mefric on (0,1).

Then

(A) both \mathrm{P} and \mathrm{Q} are TRUE

(B) \mathrm{P} is TRUE and \mathrm{Q} is FALSE

(C) P is FALSE and Q is TRUE

(D) both \mathrm{P} and \mathrm{Q} are FALSE

I Hope, these GATE solutions have helped you, please comment on which previous year gate question paper solutions. mathdart.com brings you many helpful notes for GATE preparation.

The post GATE Real Analysis 2021 Solutions appeared first on Math Dart.

]]>The post Sylow’s theorems and their Applications – CSIR-NET & GATE Notes appeared first on Math Dart.

]]>In this article, we’ll try to understand the three of Sylow’s theorem and their application as simple we can. Here are the GATE notes and if you are looking for CSIR NET Maths Notes for Group Theory, you must check this article below. You’ll also get the skill to **how to find number of sylow p-subgroups**.

**p-group:** A group is said to be p-group if O(G)=p where p is prime.

Example Given: Let O(G)=49=7^{2} then G is 7 – Group

**Cauchy Theorem:** If G has a finite abelian group and P \mid O(G) then G has a subgroup of order p and which is isomorphic to \mathbb{Z}_{p}.

Remark: If G be a finite group and p \mid \mathrm{O}(G) then G has elements of order P.

**Sylow 1st Theorem:** If G be a finite group and p^{n}|0(G) then G has subgroup of order p^{n}. This theorem helps you to find **how to find Sylow p-subgroups** orders.

**Sylow 2nd Theorem**: Any two p-sylow subgroup of G are conjugate. i.e Let H_{1} and H_{2} are two p – sylow subgroup of G then \exists x \in G such that H_{1}=x{H}_{2} x^{-1}.

The 3rd Sylow’s Theorem is the key to how to find number of sylow p-subgroups.

**Sylow – 3rd Theorem:** The number of p-\operatorname{ssG}(n) in G is equal to 1+kp

such that 1+kp \mid o(G) and k=0,1,2,....

- If H is unique p-\operatorname{SSG} in G if and only if H \mathrel{\unlhd} G.
- Let G=GL\left(n, F_{q}\right). then q-SSG are also q^{\frac{n(n-1)}{2}}.
- Let G=SL\left(n, Z_{p}\right). then p-SSG are also p^{\frac{n(n-1)}{2}}.

Here I’m sharing my notes for Sylow’s Theorem from direct copy.

Sylow’s First Theorem Application: This theorem is generally applied when we have given the order of group G, then we can find the order of the subgroup of G.

Sylow’s 1st Theorem Problem: Let o(G)=24=8 \times 3.

Find the order of subgroups in group G. (**How to find sylow p-subgroups**)

Solution: Here o(G)=24=8 \times 3 = {2}^{3} \times 3

2 \mid \mathrm{o}(\mathrm{G}) then \mathrm{G} has a subgroup of order 2.

3 \mid \mathrm{o}(\mathrm{G}) then \mathrm{G} has a subgroup of order 3.

2^{2} \mid \mathrm{o}(\mathrm{G}) then \mathrm{G} has a subgroup of order 4.

2^{3} \mid \mathrm{o}(\mathrm{G}) then \mathrm{G} has a subgroup of order 8.

The second theorem is applied to find the conjugate subgroups of a given group G. And Sylow’s third theorem is applied to find the number of p-SSG in a group G (**how to find number of sylow p-subgroups **in a group G).

Hope you like this post on the GATE and NET important topic Sylow’s Theorem and the application. The problems are basically fact-based and comparatively easy than the proofs we learned in colleges.

The post Sylow’s theorems and their Applications – CSIR-NET & GATE Notes appeared first on Math Dart.

]]>The post Numerical Analysis GATE Solutions 2021 & 2020, Syllabus, Weightage (MA) appeared first on Math Dart.

]]>**Numerical Analysis GATE 2021 Problem 1**: Let f(x)=x^{4}+2 x^{3}-11 x^{2}-12 x+36 \text { for } x \in \mathbb{R}. The order of convergence of the Newton-Raphson method:

x_{n+1}=x_{\mathrm{m}}-\frac{f\left(x_{\mathrm{n}}\right)}{f'\left(x_{n}\right)}, \quad n \geq 0[\ with x_{0}=2.1, for finding the root \alpha=2 of the equation f(x)=0 is:

Solution:

**Numerical Analysis GATE 2021 Problem 2**: If the polynomial p(x)=\alpha+\beta(x+2)+\gamma(x+2)(x+1)+\delta(x+2)(x+1) x interpolates the data

x | -2 | -1 | 0 | 1 | 2 |

f(x) | 2 | -1 | 8 | 5 | -34 |

then \alpha+\beta+y+\delta=

Solution:

**Numerical Analysis GATE 2021 Problem 3**: The quadrature formula \int_{0}^{2} x f(x) d x \approx \alpha f(0)+\beta f(1)+\gamma f(2) is exact for all polynomials of degree \leq 2. Then 2 \beta-\gamma=

Check: Ordinary Differential Equation: ODE GATE Solutions 2021-2018

**Numerical Analysis GATE 2020 Problem 1**: Let a, b, c \in \mathbb{R} be such that the quadrature rule \int_{-1}^{1} f(x) d x \approx a f(-1)+b f(0)+c f^{\prime}(1) is exact for all polynomials of degree less than or equal to 2. Then b is equal to ___ (rounded off to two decimal places).

Solution:

**Numerical Analysis GATE 2020 Problem 2**: Let f(x)=x^{4} and let p(x) be the interpolating polynomial of f at nodes 1,2 and 3. Then p(0) is equal to:

Solution:

**Numerical Analysis GATE 2020 Problem 3**: Consider the iterative scheme x_{n}=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}, \quad n>1 with initial point x_{0}>0. Then the sequence \left\{x_{n}\right.}

(A) converges only if x_{n}>1.

(B) converges only if x_{n}<3.

(C) converges for any x_{0}<3.

(D) does not converge for any x_{0}.

Solution:

Hope you are helped on Numerical Analysis GATE Solutions from the years 2021 and 2020. If we missed any questions of any of the GATE Paper of Numerical Analysis please mail us at: mathdartwebsite@gmail.com.

The post Numerical Analysis GATE Solutions 2021 & 2020, Syllabus, Weightage (MA) appeared first on Math Dart.

]]>The post Ordinary Differential Equation: ODE GATE Solutions 2021-2018 appeared first on Math Dart.

]]>Here’s the syllabus of Ordinary Differential Equation in GATE Mathematics:

- First-order ordinary differential equations,
- Existence and uniqueness theorems for initial value problems
- Systems of linear first-order ordinary differential equations
- Linear ordinary differential equations of higher order with constant coefficients
- Linear second-order ordinary differential equations with variable coefficients
- Method of Laplace transforms for solving ordinary differential equations
- Series solutions (power series, Frobenius method)
- Legendre and Bessel functions and their orthogonal properties.

There are also topics in the GATE exam outside the syllabus. Just like there was a question in 2018 GATE of the topic regular singularity of ODE.

Q.1 The eigenvalues of the boundary value problem:

\frac{d^{2} y}{d x^{2}}+\lambda y=0, \quad x \in(0, \pi), \quad \lambda>0; y(0)=0, \quad y(\pi)-\frac{d y}{d x}(\pi)=0. Then

(A) \lambda=(n \pi)^{2}, \quad n=1,2,3, \ldots

(B) \lambda=n^{2}, \quad n=1,2,3, \ldots

(C) \lambda=k_{n}^{2}, where k_{n}, n=1,2,3, \ldots are the roots of k-\tan (k \pi)=0

(D) \lambda=k_{n}^{2}, where k_{n}, n=1,2,3, \ldots are the roots of k+\tan (k \pi)=0

Solution:

Q.2 Let y(x) be the solution of the following initial value problem:

x^{2} \frac{d^{2} y}{d x^{2}}-4 x \frac{d y}{d x}+6 y=0; x>0, y(2)=0, \frac{d y}{d x}(2)=4.

Then y(4) =.

Solution:

Q5. The initial value problem y^{\prime}=y^{\frac{3}{5}}, y(0)=b has

(A) a unique solution if b = 0

(B) no solution if b = 1

(C) infinitely many solutions if b = 2

(D) a unique solution if b = 1.

Solution:

Q.4 Let u be a solution of the differential equation y’ + xy = 0 and let \phi=u \psi be a solution of the differential cquation y^{\prime \prime}+2 x y^{\prime}+\left(x^{2}+2\right) y=0 satisfying \phi(0) = 1 and \phi{\prime}(0) = 0. Then \phi(x) is

Solution:

Q.5 Consider the boundary value problem (BVP): \frac{d^{2} y}{d x^{2}}+\alpha y(x)=0, \alpha \in \mathbb{R} (the set of all real numbers), with the boundary conditions) y(0)=0, y(\pi)=k (x is a non-zero real number).

Then which one of the following statements is TRUE?

(A) For \alpha=1, the BVP has infinitely many solutions

(B) For \alpha=1, the BVP has a unique solution

(C) For \alpha=-1, k<0, the BVP has a solution y(x) such that y(x)>0; for all x \in (0,\pi)

(D) For \alpha=-1, k>0, the BVP has a solution y(x) such that y(x)>0; for all x \in (0,\pi)

Solution:

Q.6 The general solution of the differential equation: xy'=y+\sqrt{x^{2}+y^{2}} \text { for } x > 0 \text {. }

is given by (with an arbitrary positive constant k)

(A) k y^{2}=x+\sqrt{x^{2}+y^{2}}

(B) k x^{2}=x+\sqrt{x^{2}+y^{2}}

(C) k x^{2}=y+\sqrt{x^{2}+y^{2}}

(D) k y^{2}=y+\sqrt{x^{2}+y^{2}}

Solution:

Q.7 If y_{1}(x)=e^{-x^{2}} is a solution of the differential equation x y^{\prime \prime}+\alpha y^{\prime}+\beta x^{3} y=0 for some real numbers \alpha and \beta , then \alpha \beta =

Solution:

Hope you are helped on ODE GATE Solutions from the years 2021, 2020, 2019, and 2018. If we missed any questions of any of the GATE Paper of ODE please mail us at: mathdartwebsite@gmail.com.

The post Ordinary Differential Equation: ODE GATE Solutions 2021-2018 appeared first on Math Dart.

]]>The post Harish-Chandra: Most Underrated Mathematician Ever appeared first on Math Dart.

]]>Some of Harish Chandra’s mathematical innovations and researches:

- Harish-Chandra’s c-function
- Harish-Chandra homomorphism
- Harish-Chandra isomorphism
- Harish-Chandra’s character formula
- Harish-Chandra module
- Harish-Chandra’s regularity theorem
- Harish-Chandra’s Schwartz space

A homomorphism from a subalgebra of the universal enveloping algebra of a semisimple Lie algebra to the **universal enveloping algebra** (algebra that contains all representations of a Lie algebra) of a subalgebra. A particularly special case is the Harish-Chandra isomorphism identifying the center of the universal enveloping algebra with the never-changing polynomials on a Cartan subalgebra.

A representation of a real Lie group, connected to a general representation, with regularity and limit conditions. When the associated representation is a (g,K)-module, then its Harish-Chandra module is a representation with desirable factorization properties.

**Check**: Ramanujan Magic Square and What’s Unique in its Algorithm

Chandra’s regularity theorem states that: “Every invariant edge distribution on a semisimple Lie group, and in particular every character of an irreducible unitary representation on a Hilbert space, is given by a locally integrable function.”

After receiving his MSc in Physics in 1943, he moved to the IISc – Indian Institute of Science, Bangalore for further studies in theoretical physics and worked with Homi J. Bhabha (one of the ISRO’s founders). Professor Harish-Chandra began his work in the theory of elementary particles, but he turned in 1949 from physics to pure mathematics.

He was the recipient of the Cole Prize of the American Mathematical Society, in 1954. The INS Academy honored him with the Srinivasa Ramanujan Medal in 1974. In 1981, Harish received an honorary degree from Yale University.

Hence, Professor Harish should be considered the most underrated mathematician ever.

The post Harish-Chandra: Most Underrated Mathematician Ever appeared first on Math Dart.

]]>