Deep Learning · MT2 · section

Activation functions and how to calculate them

Neural-network foundations

What an activation function does, and why it is needed

A neuron first produces the weighted sum z. The activation function f is a rule that transforms that sum into the output a = f(z). Applying an activation does not change the stored weights; it calculates a new output from the current sum.

If every layer only multiplies and adds, several layers still combine into a single affine transformation. A nonlinear activation lets the network represent more complicated relationships. The output activation must also suit the task: sigmoid for a binary-class probability, softmax for mutually exclusive class probabilities, and often a linear output for unrestricted regression.

The activation value says what the neuron outputs. The activation derivative says how quickly that output changes when z changes, which is needed during backpropagation. For sigmoid at z = 0, the value is 0.5 but the derivative is 0.25. They serve different purposes.

Yaad rakhne ka cue: z = activation se pehle ka sum; a = uske baad ka answer; f′ = answer kitni tezi se badlega.

Definition. An activation function maps a neuron's preactivation to its output. Nonlinear activations allow a multilayer network to represent nonlinear functions; the output activation also encodes the prediction type. Without a nonlinear activation, a composition of affine layers remains affine.

Explanation and working

\sigma(z)=\frac{1}{1+e^{-z}},\quad \sigma'(z)=\sigma(z)[1-\sigma(z)]LaTeX: \sigma(z)=\frac{1}{1+e^{-z}},\quad \sigma'(z)=\sigma(z)[1-\sigma(z)]

Sigmoid converts any finite real input into a value strictly between 0 and 1. At zero it is 0.5. It is useful for a binary-class probability or independent multilabel outputs. At very positive or negative inputs it saturates, so its derivative becomes small. Its derivative is at most 0.25.

Trace the output of each curve as the same raw input moves from negative to positive. Tap the diagram to enlarge.
\tanh(z)=\frac{e^{2z}-1}{e^{2z}+1},\quad \tanh'(z)=1-\tanh^2(z)LaTeX: \tanh(z)=\frac{e^{2z}-1}{e^{2z}+1},\quad \tanh'(z)=1-\tanh^2(z)

Tanh lies between −1 and +1 and is zero-centered. It is common in recurrent hidden states. It also saturates. “Zero-centered” does not mean “no vanishing gradients”.

\operatorname{ReLU}(z)=\max(0,z),\quad f'(z)=\begin{cases}1&z>0\\0&z<0\end{cases}LaTeX: \operatorname{ReLU}(z)=\max(0,z),\quad f'(z)=\begin{cases}1&z>0\\0&z<0\end{cases}

ReLU returns the input if positive and zero if negative. It is cheap to compute and has derivative 1 on its positive side. Units can become inactive on all relevant inputs, known as dying ReLU. At z = 0 the mathematical derivative is undefined; a convention such as 0 is used in implementations and this course’s exercises when needed.

\operatorname{LeakyReLU}(z)=\begin{cases}z&z\ge0\\az&z<0\end{cases},\qquad a=0.01\ \text{unless specified}LaTeX: \operatorname{LeakyReLU}(z)=\begin{cases}z&z\ge0\\az&z<0\end{cases},\qquad a=0.01\ \text{unless specified}

The small negative slope a lets some gradient through for negative inputs. At z = −2 and a = 0.01, output = −0.02. Do not take the absolute value. PReLU learns the negative slope; Leaky ReLU normally fixes it.

p_i=\frac{e^{z_i}}{\sum_j e^{z_j}}=\frac{e^{z_i-c}}{\sum_j e^{z_j-c}},\qquad c=\max_jz_jLaTeX: p_i=\frac{e^{z_i}}{\sum_j e^{z_j}}=\frac{e^{z_i-c}}{\sum_j e^{z_j-c}},\qquad c=\max_jz_j

Softmax acts on a group of class scores, called logits. All probabilities are positive and sum to 1. Subtracting the same maximum c from all logits leaves the result unchanged and avoids very large exponentials. It is suited to mutually exclusive multiclass output.

Full softmax example. For [−1, 0, 1], exponentials are [0.367879, 1, 2.718282]. Sum = 4.086161. Divide each by that same sum: [0.090031, 0.244728, 0.665241]. Check their sum is approximately 1. Three separate sigmoids would give [0.268941, 0.5, 0.731059], which are different and do not sum to 1.

Derivation of the sigmoid derivative

Write sigmoid as a negative power. The outer derivative contributes −1, and differentiating the inner exponential contributes a second negative sign:

\sigma(z)=(1+e^{-z})^{-1}LaTeX: \sigma(z)=(1+e^{-z})^{-1}
\sigma'(z)=-(1+e^{-z})^{-2}(-e^{-z})=\frac{e^{-z}}{(1+e^{-z})^2}LaTeX: \sigma'(z)=-(1+e^{-z})^{-2}(-e^{-z})=\frac{e^{-z}}{(1+e^{-z})^2}

Separate the expression into two factors:

\sigma'(z)=\frac{1}{1+e^{-z}}\frac{e^{-z}}{1+e^{-z}}=\sigma(z)[1-\sigma(z)]LaTeX: \sigma'(z)=\frac{1}{1+e^{-z}}\frac{e^{-z}}{1+e^{-z}}=\sigma(z)[1-\sigma(z)]

The second factor is 1−σ(z) because subtracting 1/(1+e⁻ᶻ) from 1 leaves e⁻ᶻ/(1+e⁻ᶻ). Once a forward pass has supplied σ(z), its derivative requires no further exponential evaluation.

Derivation of the tanh derivative

Use the quotient rule on the exponential definition. Here u = e²ᶻ−1 and v = e²ᶻ+1, with u′ = v′ = 2e²ᶻ.

\frac{d}{dz}\tanh z=\frac{2e^{2z}(e^{2z}+1)-2e^{2z}(e^{2z}-1)}{(e^{2z}+1)^2}=\frac{4e^{2z}}{(e^{2z}+1)^2}LaTeX: \frac{d}{dz}\tanh z=\frac{2e^{2z}(e^{2z}+1)-2e^{2z}(e^{2z}-1)}{(e^{2z}+1)^2}=\frac{4e^{2z}}{(e^{2z}+1)^2}

The identity 1−tanh²(z) gives the same expression, since the numerator becomes (e²ᶻ+1)²−(e²ᶻ−1)² = 4e²ᶻ. Therefore the backward-pass factor is 1−a² when a = tanh(z).

Why nonlinear activations are required

For two affine layers without a nonlinear activation:

h=W_1x+b_1,\qquad \hat y=W_2h+b_2=W_2W_1x+(W_2b_1+b_2)LaTeX: h=W_1x+b_1,\qquad \hat y=W_2h+b_2=W_2W_1x+(W_2b_1+b_2)

The composition is still one affine transformation. A nonlinear activation between layers changes the family of functions the network can represent. This is the mathematical reason for using nonlinear hidden activations; merely adding affine layers does not provide this expressive advantage.

sigmoid example 1

Question. For preactivation z = -2, calculate the sigmoid output and state its derivative.

Given

z = -2; activation = sigmoid.

To find

The output a and the derivative with respect to z.

Formula — which one and why

Use the activation named in the question. The derivative is a second calculation required for a backward pass.

z is the raw sum; a is the activated output; f′(z) is the local slope.

a=\frac1{1+e^{-z}}LaTeX: a=\frac1{1+e^{-z}}
f\prime(z)=a(1-a)LaTeX: f\prime(z)=a(1-a)

Yaad rakhne ka cue: One over one plus e ki minus-z power; slope = output × (1−output).

Solution — step by step

Step 1: negate the whole input.

-z=-(-2)=2LaTeX: -z=-(-2)=2

Step 2: calculate the exponential.

e^{2}\approx7.389056099LaTeX: e^{2}\approx7.389056099

Step 3: add one in the denominator.

1+7.389056099\approx8.389056099LaTeX: 1+7.389056099\approx8.389056099

Step 4: divide one by the complete denominator.

a=\frac1{8.389056099}\approx0.119202922LaTeX: a=\frac1{8.389056099}\approx0.119202922

Step 5: use the activated output in the slope.

f\prime(-2)=(0.119202922)(1-0.119202922)\approx0.104993585LaTeX: f\prime(-2)=(0.119202922)(1-0.119202922)\approx0.104993585

Final answer: a ≈ 0.119202922. Derivative ≈ 0.104993585.

sigmoid example 2

Question. For preactivation z = 0, calculate the sigmoid output and state its derivative.

Given

z = 0; activation = sigmoid.

To find

The output a and the derivative with respect to z.

Formula — which one and why

Use the activation named in the question. The derivative is a second calculation required for a backward pass.

z is the raw sum; a is the activated output; f′(z) is the local slope.

a=\frac1{1+e^{-z}}LaTeX: a=\frac1{1+e^{-z}}
f\prime(z)=a(1-a)LaTeX: f\prime(z)=a(1-a)

Yaad rakhne ka cue: One over one plus e ki minus-z power; slope = output × (1−output).

Solution — step by step

Step 1: negate the whole input.

-z=-(0)=0LaTeX: -z=-(0)=0

Step 2: calculate the exponential.

e^{0}\approx1.000000000LaTeX: e^{0}\approx1.000000000

Step 3: add one in the denominator.

1+1.000000000\approx2.000000000LaTeX: 1+1.000000000\approx2.000000000

Step 4: divide one by the complete denominator.

a=\frac1{2.000000000}\approx0.500000000LaTeX: a=\frac1{2.000000000}\approx0.500000000

Step 5: use the activated output in the slope.

f\prime(0)=(0.500000000)(1-0.500000000)\approx0.250000000LaTeX: f\prime(0)=(0.500000000)(1-0.500000000)\approx0.250000000

Final answer: a ≈ 0.500000000. Derivative ≈ 0.250000000.

sigmoid example 3

Question. For preactivation z = 2, calculate the sigmoid output and state its derivative.

Given

z = 2; activation = sigmoid.

To find

The output a and the derivative with respect to z.

Formula — which one and why

Use the activation named in the question. The derivative is a second calculation required for a backward pass.

z is the raw sum; a is the activated output; f′(z) is the local slope.

a=\frac1{1+e^{-z}}LaTeX: a=\frac1{1+e^{-z}}
f\prime(z)=a(1-a)LaTeX: f\prime(z)=a(1-a)

Yaad rakhne ka cue: One over one plus e ki minus-z power; slope = output × (1−output).

Solution — step by step

Step 1: negate the whole input.

-z=-(2)=-2LaTeX: -z=-(2)=-2

Step 2: calculate the exponential.

e^{-2}\approx0.135335283LaTeX: e^{-2}\approx0.135335283

Step 3: add one in the denominator.

1+0.135335283\approx1.135335283LaTeX: 1+0.135335283\approx1.135335283

Step 4: divide one by the complete denominator.

a=\frac1{1.135335283}\approx0.880797078LaTeX: a=\frac1{1.135335283}\approx0.880797078

Step 5: use the activated output in the slope.

f\prime(2)=(0.880797078)(1-0.880797078)\approx0.104993585LaTeX: f\prime(2)=(0.880797078)(1-0.880797078)\approx0.104993585

Final answer: a ≈ 0.880797078. Derivative ≈ 0.104993585.

tanh example 1

Question. For preactivation z = -2, calculate the tanh output and state its derivative.

Given

z = -2; activation = tanh.

To find

The output a and the derivative with respect to z.

Formula — which one and why

Use the activation named in the question. The derivative is a second calculation required for a backward pass.

z is the raw sum; a is the activated output; f′(z) is the local slope.

a=\frac{e^{2z}-1}{e^{2z}+1}LaTeX: a=\frac{e^{2z}-1}{e^{2z}+1}
f\prime(z)=1-a^2LaTeX: f\prime(z)=1-a^2

Yaad rakhne ka cue: Tanh ka answer −1 se 1; slope one minus output squared.

Solution — step by step

Step 1: double z before exponentiation.

2z=2(-2)=-4LaTeX: 2z=2(-2)=-4

Step 2: calculate the exponential.

e^{-4}\approx0.018315639LaTeX: e^{-4}\approx0.018315639

Step 3: calculate numerator and denominator separately.

0.018315639-1\approx-0.981684361LaTeX: 0.018315639-1\approx-0.981684361
0.018315639+1\approx1.018315639LaTeX: 0.018315639+1\approx1.018315639

Step 4: divide.

a=\frac{-0.981684361}{1.018315639}\approx-0.964027580LaTeX: a=\frac{-0.981684361}{1.018315639}\approx-0.964027580

Step 5: square the activated output for its slope.

f\prime(-2)=1-(-0.964027580)^2\approx0.070650825LaTeX: f\prime(-2)=1-(-0.964027580)^2\approx0.070650825

Final answer: a ≈ -0.964027580. Derivative ≈ 0.070650825.

tanh example 2

Question. For preactivation z = 0, calculate the tanh output and state its derivative.

Given

z = 0; activation = tanh.

To find

The output a and the derivative with respect to z.

Formula — which one and why

Use the activation named in the question. The derivative is a second calculation required for a backward pass.

z is the raw sum; a is the activated output; f′(z) is the local slope.

a=\frac{e^{2z}-1}{e^{2z}+1}LaTeX: a=\frac{e^{2z}-1}{e^{2z}+1}
f\prime(z)=1-a^2LaTeX: f\prime(z)=1-a^2

Yaad rakhne ka cue: Tanh ka answer −1 se 1; slope one minus output squared.

Solution — step by step

Step 1: double z before exponentiation.

2z=2(0)=0LaTeX: 2z=2(0)=0

Step 2: calculate the exponential.

e^{0}\approx1.000000000LaTeX: e^{0}\approx1.000000000

Step 3: calculate numerator and denominator separately.

1.000000000-1\approx0.000000000LaTeX: 1.000000000-1\approx0.000000000
1.000000000+1\approx2.000000000LaTeX: 1.000000000+1\approx2.000000000

Step 4: divide.

a=\frac{0.000000000}{2.000000000}\approx0.000000000LaTeX: a=\frac{0.000000000}{2.000000000}\approx0.000000000

Step 5: square the activated output for its slope.

f\prime(0)=1-(0.000000000)^2\approx1.000000000LaTeX: f\prime(0)=1-(0.000000000)^2\approx1.000000000

Final answer: a ≈ 0.000000000. Derivative ≈ 1.000000000.

tanh example 3

Question. For preactivation z = 2, calculate the tanh output and state its derivative.

Given

z = 2; activation = tanh.

To find

The output a and the derivative with respect to z.

Formula — which one and why

Use the activation named in the question. The derivative is a second calculation required for a backward pass.

z is the raw sum; a is the activated output; f′(z) is the local slope.

a=\frac{e^{2z}-1}{e^{2z}+1}LaTeX: a=\frac{e^{2z}-1}{e^{2z}+1}
f\prime(z)=1-a^2LaTeX: f\prime(z)=1-a^2

Yaad rakhne ka cue: Tanh ka answer −1 se 1; slope one minus output squared.

Solution — step by step

Step 1: double z before exponentiation.

2z=2(2)=4LaTeX: 2z=2(2)=4

Step 2: calculate the exponential.

e^{4}\approx54.598150033LaTeX: e^{4}\approx54.598150033

Step 3: calculate numerator and denominator separately.

54.598150033-1\approx53.598150033LaTeX: 54.598150033-1\approx53.598150033
54.598150033+1\approx55.598150033LaTeX: 54.598150033+1\approx55.598150033

Step 4: divide.

a=\frac{53.598150033}{55.598150033}\approx0.964027580LaTeX: a=\frac{53.598150033}{55.598150033}\approx0.964027580

Step 5: square the activated output for its slope.

f\prime(2)=1-(0.964027580)^2\approx0.070650825LaTeX: f\prime(2)=1-(0.964027580)^2\approx0.070650825

Final answer: a ≈ 0.964027580. Derivative ≈ 0.070650825.

ReLU example 1

Question. For preactivation z = -2, calculate the ReLU output and state its derivative.

Given

z = -2; activation = ReLU.

To find

The output a and the derivative with respect to z.

Formula — which one and why

Use the activation named in the question. The derivative is a second calculation required for a backward pass.

z is the raw sum; a is the activated output; f′(z) is the local slope.

a=\max(0,z)LaTeX: a=\max(0,z)

Yaad rakhne ka cue: Positive side slope 1; negative side ReLU 0, Leaky α. Zero par convention likho.

Solution — step by step

Step 1: select the branch from the sign of z.

-2 is negative, so use the negative branch.

Step 2: calculate the output.

a=0LaTeX: a=0

Step 3: state the slope on that branch.

f\prime(-2)=0LaTeX: f\prime(-2)=0

Final answer: a ≈ 0.000000000. Derivative ≈ 0.000000000.

ReLU example 2

Question. For preactivation z = 0, calculate the ReLU output and state its derivative.

Given

z = 0; activation = ReLU.

To find

The output a and the derivative with respect to z.

Formula — which one and why

Use the activation named in the question. The derivative is a second calculation required for a backward pass.

z is the raw sum; a is the activated output; f′(z) is the local slope.

a=\max(0,z)LaTeX: a=\max(0,z)

Yaad rakhne ka cue: Positive side slope 1; negative side ReLU 0, Leaky α. Zero par convention likho.

Solution — step by step

Step 1: select the branch from the sign of z.

Both branches meet at zero. The value is zero, but the left and right slopes differ.

Step 2: calculate the output.

a=0LaTeX: a=0

Step 3: state the slope on that branch.

The ordinary derivative at zero does not exist. A software subgradient convention may choose a value such as 0; state it if required.

Final answer: a ≈ 0.000000000. Ordinary derivative is undefined at zero.

ReLU example 3

Question. For preactivation z = 2, calculate the ReLU output and state its derivative.

Given

z = 2; activation = ReLU.

To find

The output a and the derivative with respect to z.

Formula — which one and why

Use the activation named in the question. The derivative is a second calculation required for a backward pass.

z is the raw sum; a is the activated output; f′(z) is the local slope.

a=\max(0,z)LaTeX: a=\max(0,z)

Yaad rakhne ka cue: Positive side slope 1; negative side ReLU 0, Leaky α. Zero par convention likho.

Solution — step by step

Step 1: select the branch from the sign of z.

2 is positive, so retain it.

Step 2: calculate the output.

a=2LaTeX: a=2

Step 3: state the slope on that branch.

f\prime(2)=1LaTeX: f\prime(2)=1

Final answer: a ≈ 2.000000000. Derivative ≈ 1.000000000.

Leaky ReLU example 1

Question. For preactivation z = -2, calculate the Leaky ReLU output and state its derivative. Use negative-side slope α = 0.1.

Given

z = -2; activation = Leaky ReLU.

α = 0.1.

To find

The output a and the derivative with respect to z.

Formula — which one and why

Use the activation named in the question. The derivative is a second calculation required for a backward pass.

z is the raw sum; a is the activated output; f′(z) is the local slope.

a=\begin{cases}z&z\ge0\\0.1z&z<0\end{cases}LaTeX: a=\begin{cases}z&z\ge0\\0.1z&z<0\end{cases}

Yaad rakhne ka cue: Positive side slope 1; negative side ReLU 0, Leaky α. Zero par convention likho.

Solution — step by step

Step 1: select the branch from the sign of z.

-2 is negative, so use the negative branch.

Step 2: calculate the output.

a=-0.2LaTeX: a=-0.2

Step 3: state the slope on that branch.

f\prime(-2)=0.1LaTeX: f\prime(-2)=0.1

Final answer: a ≈ -0.200000000. Derivative ≈ 0.100000000.

Leaky ReLU example 2

Question. For preactivation z = 0, calculate the Leaky ReLU output and state its derivative. Use negative-side slope α = 0.1.

Given

z = 0; activation = Leaky ReLU.

α = 0.1.

To find

The output a and the derivative with respect to z.

Formula — which one and why

Use the activation named in the question. The derivative is a second calculation required for a backward pass.

z is the raw sum; a is the activated output; f′(z) is the local slope.

a=\begin{cases}z&z\ge0\\0.1z&z<0\end{cases}LaTeX: a=\begin{cases}z&z\ge0\\0.1z&z<0\end{cases}

Yaad rakhne ka cue: Positive side slope 1; negative side ReLU 0, Leaky α. Zero par convention likho.

Solution — step by step

Step 1: select the branch from the sign of z.

Both branches meet at zero. The value is zero, but the left and right slopes differ.

Step 2: calculate the output.

a=0LaTeX: a=0

Step 3: state the slope on that branch.

The ordinary derivative at zero does not exist. A software subgradient convention may choose a value such as 0; state it if required.

Final answer: a ≈ 0.000000000. Ordinary derivative is undefined at zero.

Leaky ReLU example 3

Question. For preactivation z = 2, calculate the Leaky ReLU output and state its derivative. Use negative-side slope α = 0.1.

Given

z = 2; activation = Leaky ReLU.

α = 0.1.

To find

The output a and the derivative with respect to z.

Formula — which one and why

Use the activation named in the question. The derivative is a second calculation required for a backward pass.

z is the raw sum; a is the activated output; f′(z) is the local slope.

a=\begin{cases}z&z\ge0\\0.1z&z<0\end{cases}LaTeX: a=\begin{cases}z&z\ge0\\0.1z&z<0\end{cases}

Yaad rakhne ka cue: Positive side slope 1; negative side ReLU 0, Leaky α. Zero par convention likho.

Solution — step by step

Step 1: select the branch from the sign of z.

2 is positive, so retain it.

Step 2: calculate the output.

a=2LaTeX: a=2

Step 3: state the slope on that branch.

f\prime(2)=1LaTeX: f\prime(2)=1

Final answer: a ≈ 2.000000000. Derivative ≈ 1.000000000.

Softmax for [0, 0, 0]

Question. Find stable softmax probabilities for logits [0, 0, 0].

Given

Logits = [0, 0, 0]; there are three mutually exclusive classes.

To find

All three class probabilities.

Formula — which one and why

Subtracting one common maximum preserves probabilities and keeps exponentials well scaled. Normalize using one shared sum.

A logit is a raw class score. c is the largest score; pᵢ is probability for class i.

c=\max_j z_jLaTeX: c=\max_j z_j
p_i=\frac{e^{z_i-c}}{\sum_j e^{z_j-c}}LaTeX: p_i=\frac{e^{z_i-c}}{\sum_j e^{z_j-c}}

Yaad rakhne ka cue: Max minus, e-power, same denominator se sab divide.

Solution — step by step

Step 1: subtract the largest score from every entry.

z-c=\begin{bmatrix}0&0&0\end{bmatrix}LaTeX: z-c=\begin{bmatrix}0&0&0\end{bmatrix}

Step 2: exponentiate shifted entry 1.

e^{0}\approx1.000000000LaTeX: e^{0}\approx1.000000000

Step 3: exponentiate shifted entry 2.

e^{0}\approx1.000000000LaTeX: e^{0}\approx1.000000000

Step 4: exponentiate shifted entry 3.

e^{0}\approx1.000000000LaTeX: e^{0}\approx1.000000000

Step 5: add all exponentials into one denominator.

D=1.000000000+1.000000000+1.000000000\approx3.000000000LaTeX: D=1.000000000+1.000000000+1.000000000\approx3.000000000

Step 6: divide entry 1 by the same denominator.

p_1=\frac{1.000000000}{3.000000000}\approx0.333333333LaTeX: p_1=\frac{1.000000000}{3.000000000}\approx0.333333333

Step 7: divide entry 2 by the same denominator.

p_2=\frac{1.000000000}{3.000000000}\approx0.333333333LaTeX: p_2=\frac{1.000000000}{3.000000000}\approx0.333333333

Step 8: divide entry 3 by the same denominator.

p_3=\frac{1.000000000}{3.000000000}\approx0.333333333LaTeX: p_3=\frac{1.000000000}{3.000000000}\approx0.333333333

Final answer: Probabilities ≈ [0.333333333, 0.333333333, 0.333333333]; they sum to 1 within rounding.

Softmax for [1, 2, 3]

Question. Find stable softmax probabilities for logits [1, 2, 3].

Given

Logits = [1, 2, 3]; there are three mutually exclusive classes.

To find

All three class probabilities.

Formula — which one and why

Subtracting one common maximum preserves probabilities and keeps exponentials well scaled. Normalize using one shared sum.

A logit is a raw class score. c is the largest score; pᵢ is probability for class i.

c=\max_j z_jLaTeX: c=\max_j z_j
p_i=\frac{e^{z_i-c}}{\sum_j e^{z_j-c}}LaTeX: p_i=\frac{e^{z_i-c}}{\sum_j e^{z_j-c}}

Yaad rakhne ka cue: Max minus, e-power, same denominator se sab divide.

Solution — step by step

Step 1: subtract the largest score from every entry.

z-c=\begin{bmatrix}-2&-1&0\end{bmatrix}LaTeX: z-c=\begin{bmatrix}-2&-1&0\end{bmatrix}

Step 2: exponentiate shifted entry 1.

e^{-2}\approx0.135335283LaTeX: e^{-2}\approx0.135335283

Step 3: exponentiate shifted entry 2.

e^{-1}\approx0.367879441LaTeX: e^{-1}\approx0.367879441

Step 4: exponentiate shifted entry 3.

e^{0}\approx1.000000000LaTeX: e^{0}\approx1.000000000

Step 5: add all exponentials into one denominator.

D=0.135335283+0.367879441+1.000000000\approx1.503214724LaTeX: D=0.135335283+0.367879441+1.000000000\approx1.503214724

Step 6: divide entry 1 by the same denominator.

p_1=\frac{0.135335283}{1.503214724}\approx0.090030573LaTeX: p_1=\frac{0.135335283}{1.503214724}\approx0.090030573

Step 7: divide entry 2 by the same denominator.

p_2=\frac{0.367879441}{1.503214724}\approx0.244728471LaTeX: p_2=\frac{0.367879441}{1.503214724}\approx0.244728471

Step 8: divide entry 3 by the same denominator.

p_3=\frac{1.000000000}{1.503214724}\approx0.665240956LaTeX: p_3=\frac{1.000000000}{1.503214724}\approx0.665240956

Final answer: Probabilities ≈ [0.090030573, 0.244728471, 0.665240956]; they sum to 1 within rounding.